text:
\section{Introduction}\label{Section1}
R. van Luijk~\cite{Luijk1} recently solved an old problem attributed to D. Mumford by
constructing a quartic K3 surface with geometric Picard number equal to one. Moreover,
he was able to solve a question posed by P. Swinnerton-Dyer and B. Poonen at an AIM workshop
in December 2002, who asked whether
there exists a K3 surface over a number field with Picard number one that contains infinitely many
rational points. Van Luijk was able to show that the set of quartic surfaces with Picard number
one and infinitely many rational points is dense in the set of all K3 surfaces in the
Zariski topology.
Here, we consider a related problem. More precisely, we are interested
in finding the simplest possible form of an equation defining a quartic surface $\mathcal{S}$ over
$\mathbb{Q}$ with Picard number one, and possessing infinitely many rational points. Here, we are talking about
the Picard number $\rho(S)=\mbox{rank of the Picard group of }\mathcal{S}$, rather than the geometric
Picard number, which is defined as the Picard number of $\overline{S}=S\times_{\mathbb{Q}}\overline{\mathbb{Q}}$.
We believe that this is provided by a surface with diagonal equation of the form
\begin{equation}
\label{generalquar}
\mathcal{S}:\;ax^4+by^4+cz^4+dw^4=0,
\end{equation}
with $a,b,c,d\in\mathbb{Z}\setminus\{0\}$.
It is not clear a priori whether such an example exists, and we were unable to find
a suitable reference in the existing literature.  The big advantage of considering surfaces
in diagonal quartic form follows from the fact that the Picard number of $\mathcal{S}$ may be easily
computed.
Pinch and Swinnerton-Dyer~\cite{PiSwi}, as well as Swinnerton-Dyer~\cite{SwiD2},
investigate in some generality the arithmetic of the diagonal quartic surface,
and in particular, the former give tables allowing one to read off the Picard rank of any given
diagonal quartic surface over $\mathbb{Q}$. We thus need to find many (or better, infinitely many)
examples of diagonal surfaces with infinitely many rational points.
Having these examples in hand, we can check whether some have Picard number
equal to one.
The literature concerning the arithmetic of diagonal quartic surfaces is large and starts with
investigations of the surface $x^4+y^4=z^4+w^4$ considered by Euler (see Mordell~\cite{Mor} and
Dem'janenko~\cite{Dem}). Swinnerton-Dyer~\cite{SwiD1} studied the surface in detail, describing
all curves of arithmetic genus 0 lying on the surface.  A classical paper of Richmond~\cite{Rich}
shows how one can construct a new point on (\ref{generalquar}) from a given non-trivial point when
$abcd=\square$.  Bright has written extensively on the subject, starting with his
thesis \cite{Bri}.  Richmond's idea was used by Logan, van Luijk, and
MacKinnon~\cite{LoLuM} to prove density (in the Euclidean topology) of the rational points
on (\ref{generalquar}) when $abcd=\square$.
In a recent paper van Luijk~\cite{Luijk2} extended ideas of Swinnerton-Dyer \cite{SwiD3} and
proved that if the surface $V$ defined over a number field $k$ admits two elliptic fibrations
then there exists an explicitly computable closed subset $Z$ of $V$, not equal to $V$, such that
for each field extension $K$ of $k$ of degree at most $d$ over the field of rational numbers,
the set $V(K)$ is Zariski dense as soon as it contains any point outside $Z$.
In particular, his theorem implies that if the quartic surface $\mathcal{S}$ admits two elliptic
fibrations, the set $\mathcal{S}(\mathbb{Q})$ is infinite, and a suitably chosen point lies outside the easily
computable set $Z$, then necessarily $\mathcal{S}(\mathbb{Q})$ is dense in the Zariski topology. More
specifically, the elliptic fibrations do not necessarily need to
possess a section: it is enough to have one properly chosen specialization which is of positive rank.
We have (maybe somewhat arbitrarily) restricted attention to surfaces (\ref{generalquar})
where $abcd \neq \square$, since many existing results depend upon $abcd=\square$, and in which
situation the Picard number is at least 2.

We know relatively few examples of surfaces (\ref{generalquar}) with
infinitely many points, satisfying the condition $abcd\neq \square$. These include
the quartic $x^4+y^4+z^4=w^4$ (of Picard rank 4) considered by Elkies~\cite{Elk1}, and the
quartics $x^4+y^4+z^4=2t^4$, $x^4+y^4+4z^4=t^4$, $x^4+2y^4+2z^4=t^4$, $x^4+8y^4+8z^4=t^4$,
(of ranks 4, 8, 3, 3, respectively) in Carmichael~\cite{Carm}.
In fact, the method used by Carmichael also yields parametric solutions of the Diophantine equation
(\ref{generalquar}) whenever $a+b=0$ and $4a^2cd$ is a perfect fourth power. As this solution is not
given explicitly in \cite{Carm}, it is stated below:
\[
x=k(8as^4-ct^4), \quad y=k(8as^4+ct^4), \quad z=8kas^3t, \quad w=4acst^3
\]
where $s,\,t$ are arbitrary parameters and $k$ is determined by the relation $4a^2cd=k^4.$
Further, when
$(a,\,b,\,c,\,d)=(1,1,1,-2k^2)$, and the Diophantine equation $X_1^2+3X_2^2=k X_3^2$ is
solvable, a solution of (\ref{generalquar}) in terms of parameters $p,\,q$ is  given by
\[x = \phi_1(p,\,q)-\phi_2(p,\,q),\, y = 2\phi_2(p,\,q),\, z = \phi_1(p,\,q)+\phi_2(p,\,q),\, w = \phi_3(p,\,q),
\]
where $X_i=\phi_i(p,\,q),\,i=1,\,2,\,3,$ is a parametric solution of the equation $X_1^2+3X_2^2=k X_3^2$.

We also note that some believe a small Picard number of $\mathcal{S}$ has a strong influence on
the set of rational points on $\mathcal{S}$. This belief was questioned in a recent paper of
Elsenhans~\cite{Els}, who considered the set of surfaces $ax^4+by^4=cz^4+dw^4$  with
$1\leq a, b, c, d\leq 15$ from a computational point of view. He noted that in the considered
range there is no great difference between the number of unsolved cases for differing
Picard rank. In light of this
remark it is natural to look for examples of diagonal quartic surfaces with infinitely many
rational points and small Picard number. \\ \\
We briefly describe the content of the paper.
In Section \ref{sec2} the first construction of diagonal quartic surfaces with $abcd\neq \square$
is given. The construction is to intersect the quartic surface with a quadric surface chosen in
such a way that the intersection splits into quadric curves.
In Section \ref{sec3} the second construction is given. This method is equivalent to finding
curves of low degree on the affine variety $aX^4+bY^4+cZ^2+dW^2=0$, in particular curves
where $Z,W$ are of degree at most 2. Demanding then that $Z,W$ be squares leads to a condition
governed by a curve of genus 1 that may be chosen to have positive rank.
The third construction is presented in Section \ref{sec4}. It is based on
diagonal quartic curves (of genus 3) of the form $Px^4+Qy^4+Rz^4=0$ with a given rational point.
In Section \ref{sec5} we use the same idea as in Section \ref{sec3} by seeking curves of low
degree on affine varieties
of the form $Px^6+Qy^6+Rz^3+Sw^2=0$ and $Px^6+Qy^6+Rz^3+Sw^3=0$. These constructions allow us
to find infinitely many sextic surfaces of the form $Px^6+Qy^6+Rz^6+Sw^2=0$
and $Px^6+Qy^6+Rz^6+Sw^3=0$ with infinitely many points. This is motivated by earlier
research~\cite{BreUl2} concerning the existence of rational points on $a(x^2-y^6)=b(z^6-w^6)$.
Finally, in the last section we give some remarks concerning possible extensions of these
approaches.

\section{First construction} \label{sec2}

Consider a diagonal quartic surface $c_1 x^4+c_2 y^4+c_3 z^4+w^4 =0$,
and demand that its intersection with a quadric surface of the type
\[ w^2 = a_1 x^2+a_2 x y+a_3 x z+a_4 y^2+a_5 y z+a_6 z^2 \]
be singular, splitting into two quadrics (each with multiplicity two).
This is equivalent to the demand that
\begin{equation}
\label{intersectionquartic}
c_1 x^4+c_2 y^4+c_3 z^4+(a_1 x^2+a_2 x y+a_3 x z+a_4 y^2+a_5 y z+a_6 z^2)^2
\end{equation}
split into two quadratic factors.  We let
\begin{equation}
\label{quadfactor}
z^2 + (b_1 x +b_2 y) z + b_3 x^2 + b_4 x y + b_5 y^2
\end{equation}
be one of the quadratic factors.

Write out the system of nine equations resulting from equating to zero
the coefficients of the remainder on dividing (\ref{intersectionquartic})
by (\ref{quadfactor}). We successively eliminate variables developing
a branching tree of cases, according as to vanishing or non-vanishing of
denominators arising from the elimination. A full analysis is too tiresome:
Groebner bases of some ideals arising in this process contain many hundreds
of polynomials of high degree. In order to find non-trivial examples, we
solved in turn for $c_1,c_2,a_1,a_3,b_5,b_3,b_1$, leading to two equations
in $a_2,a_4,a_5,a_6,b_2,b_4,c_3$ to be satisfied. The denominators vanishing
provided four cases to consider. Otherwise, the numerators have respective degrees
$4$, $3$, in $a_2$, and the resultant with respect to $a_2$ contains eight factors
(excluding the factor $c_3$). We were able successfully to investigate seven of
these factors (the seventh is quadratic in $a_4$, allowing parameterization; the eighth
has degrees $4,8,12,8,6$ in $a_4,a_5,a_6,b_2,c_3$).   \\

\noindent
When $a_3=a_5=b_1=b_2=0$, essentially just two examples arise. \\
First, we obtain the surface
\[ V_1: \; \frac{a^2(s^2-2r t)}{s^2}x^4+\frac{a^2 (s^2-2 r t)t^2}{r^2 s^2}y^4+\frac{2a^2 t}{r s^2}z^4-w^4=0, \]
which when intersected with the quadric
\[ w^2 = a x^2+\frac{2 a t}{s}x y +\frac{a t}{r} y^2 \]
results in the identity
\begin{align}
\label{quadint1}
& 2a^2rt(rx^2 + sxy + ty^2 + z^2)(rx^2 + sxy + ty^2 - z^2) -  \nonumber \\
&  (arsx^2+2artxy+atsy^2+rsw^2)(arsx^2+2artxy+atsy^2-rsw^2) =  \nonumber \\
& a^2(-s^2+2r t)r^2 x^4+a^2 (-s^2+2 r t)t^2y^4-2a^2rtz^4+r^2s^2w^4.
\end{align}
The surface $V_1$ has generic Picard rank equal to 2.
We find surfaces $V_1$ with infinitely many points by demanding for example that the intersection
of the two quadrics
\[r x^2 + s x y + t y^2 + z^2 = 0, \quad a r s x^2+2a r t x y+a s t y^2+r s w^2 = 0, \]
represents an elliptic curve of positive rational rank.
If we take $a=r=-1$, then the intersection becomes
\[ -x^2 + s x y + t y^2 + z^2=0, \quad s x^2 + 2 t x y - s t y^2 - s w^2 = 0, \]
equivalently, on setting $2t=u s^2$, $(X,Y,Z,W)=(x,s y/2,z,w)$,
\begin{equation}\label{model1}
 X^2 - 2 X Y - 2 u Y^2 = Z^2, \quad X^2 + 2 u X Y - 2 u Y^2 = W^2.
 \end{equation}
This is an elliptic curve on taking $(X,Y,Z,W)=(1,0,1,1)$ as origin. The point
$Q(X,Y,Z,W)=(1,0,1,-1)$ is of infinite order, and the multiples $mQ$, $m=2,3,...$ give
infinitely many points on the surface
\begin{equation}
\label{ex1}
S: \; (1+u) X^4 + 4u^2(1+u) Y^4 - u Z^4 - W^4 = 0
\end{equation}
of generic Picard rank 2.
For example, when $m=2$ we obtain
\[ (X,Y,Z,W)=(1+10u+u^2, \; -4(1-u), \; -3-10u+u^2, \; 1-10u-3u^2), \]
and when $m=3$
\begin{align*}
(X,Y,Z,W)= & (-(1+10u+u^2)(1+52u+38u^2+52u^3+u^4),  \\
           &  4(1-u)(-3-10u+u^2)(-1+10u+3u^2), \\
           &  -5-114u+233u^2+1140u^3+381u^4+94u^5-u^6, \\
           &  1-94u-381u^2-1140u^3-233u^4+114u^5+5u^6);
\end{align*}
etc. etc. \\
In light of the identity (\ref{quadint1}), the surface $S$ at (\ref{ex1}) contains the
pair of fibrations
\[ Q_1 = \lambda Q_3, \quad \lambda Q_2 = Q_4, \qquad \mbox{ and } \qquad Q_1 = \lambda Q_4, \qquad \lambda Q_2 = Q_3, \]
where
\begin{align*}
Q_1=u (-X^2+2X Y+2u Y^2+Z^2), \quad & Q_2=s(-X^2+2X Y+2u Y^2-Z^2), \\
Q_3=-X^2-2u X Y+2u Y^2+W^2, \quad & Q_4=s(X^2+2u X Y-2 u Y^2+W^2).
\end{align*}
The existence of infinitely many rational points, together with the
result of van Luijk (and computations to ensure the applicability of van Luijk's result), now
implies the following:

\begin{thm}\label{pic}
The set of diagonal quartic surfaces defined over $\mathbb{Q}$ with Picard number 2 and Zariski
dense set of rational points is infinite.
\end{thm}
\begin{proof} Due to the large number of parameters it is a difficult task to perform all the necessary
computations to apply Van Luijk's theorem to the surface $V_{1}$. We thus present the proof in the
simplest possible case. More precisely, consider the surface $S$ at (\ref{ex1}), which is a special case
of the surface $V_{1}$. The surface $S$ contains a genus one curve, say $C$, defined by the intersection
of the two quadrics at (\ref{model1}). The curve $C$ is smooth provided that $u\not\in\{-2,-1,-1/2,0\}$.
This model contains the rational point $(1,0,1,1)$, which can be used as a point at infinity.
Then there exists a birational map $\phi: \; C \rightarrow \mathcal{E}$, where $\mathcal{E}$ is the cubic curve over
$\mathbb{Q}(u)$ with Weierstrass equation
\begin{equation*}
\mathcal{E}:\;Y^2=X^3+27 u(6+11u+6u^2)X+54u^2(18 + 37 u + 18 u^2).
\end{equation*}
Note that the point $(1,0,1,-1)$ does not lie on any of the 48 lines lying on $S$. We need to check
that $(1,0,1,-1)$ corresponds to a non-torsion point on $\mathcal{E}$. Indeed, the curve $\mathcal{E}$ contains the
point $U=\phi((1,0,1,-1))=(3u(3u+4),27u(u+1)(u+2))$. By specialization at $u=2$, the point $U_{2}$, lying on
the curve $\mathcal{E}_{2}$, is of infinite order on $\mathcal{E}_{2}$ (for $2U_{2}$ is not integral, and the
Lutz-Nagell theorem applies).
This immediately implies that $U$ is of infinite order on $\mathcal{E}$. However, we are interested in
characterizing those $u=u_{0}\in\mathbb{Q}$ such that the specialized point $U_{u_{0}}$ is of infinite order on
$\mathcal{E}_{u_{0}}$. By Mazur's theorem we know that the order of a torsion point on an elliptic curve defined over
$\mathbb{Q}$ is at most $12$. A quick computation of the multiplicities $mU$ for $m=1,2,\ldots, 12$, reveals that
$mU_{u_{0}}=\mathcal{O}$ only for $m=4$ with $u_{0}=1$. Thus for $u_{0}\in\mathbb{Q}\setminus\{-2,-1,-1/2,0,1\}$ the point
$U_{u_{0}}$ is of infinite order on $\mathcal{E}_{u_{0}}$. Applying now Van Luijk's theorem, we get that for each
$u\in\mathbb{Q}\setminus\{-2,-1,-1/2,0,1\}$ the set of rational points on $S$ is Zariski dense.
\end{proof}

\begin{rem}
{\rm The reason why the above proof is simple is clear. We have an explicit point (in our case $(1,0,1,-1)$)
which lies on the surface, and thus on the corresponding elliptic curve (in our case $\mathcal{E}$). Thus, all
computations can be made explicit. In the general case, on the surface $V_{1}$ we do not have a generic
rational point and thus reasoning depends on the possible form of the required point(s) with (unknown)
coefficients $P=(X_{0},Y_{0},Z_{0},W_{0})$. In particular, it is very difficult to characterize the
torsion subgroup of those smooth fibers containing the point $P$.}
\end{rem}

\noindent
Second, we obtain the surface
\[ V_2: \; a^2 x^4+d^2 y^4 +2 a d s^2 z^4 - w^4 = 0, \]
which when intersected with the quadric
\[ w^2 = a x^2 + d y^2 \]
results in the identity
\begin{align*}
& (a x^2+d y^2 +w^2)(a x^2+d y^2-w^2) -2 a d (x y - s z^2) (x y + s z^2) = \\
& a^2 x^4+d^2 y^4 +2 a d s^2 z^4 - w^4.
\end{align*}
The surface has generic Picard rank 2.
We find surfaces $V_2$ with infinitely
many points by demanding for example that the intersection of the two quadrics
\[ a x^2+d y^2 = w^2 , \quad x y = s z^2 \]
represents an elliptic curve of positive rational rank.
Set $(x,y,z,w)=(s X,Y,Z,s u W)$, and take $(a,d)=(\alpha u^2, (1-\alpha)s^2 u^2)$,
with $(X,Y,Z,W)=(1,1,1,1)$ as elliptic curve origin.
The point $Q(X,Y,Z,W)=(1,1,-1,1)$ is of infinite order and the multiples $mQ$, $m=2,3,...$
give infinitely many points on the surface of generic rank 2
\begin{equation}
\label{ex2}
\alpha^2 X^4 + (1-\alpha)^2 Y^4 +2 \alpha(1-\alpha)Z^4 - W^4 = 0.
\end{equation}
For example with $m=2$:
\begin{align*}
(X,Y,Z,W) = & ((-3+4\alpha^2)^2, \; (1-8\alpha+4\alpha^2)^2, \; (-3+4\alpha^2)(1-8\alpha+4\alpha^2), \\
 & 1+24\alpha-72\alpha^2+96\alpha^3-48\alpha^4);
\end{align*}
etc. \\

\noindent
The only resulting intersection of the form we seek arising from non-vanishing denominators
is the following. The surface
\[ V_3: \; x^4+y^4 - (2u^2-v^2)^2 z^4 + (2u^2-v^2) k^2 w^4 = 0 \]
intersected with the quadric surface
\[ (2u-v) k w^2 = (x+y)^2 + 2(u-v) (x+y) z - (2u^2-v^2) z^2 \]
results in the identity
\begin{align*}
& 2\left( (3u-2v)(x^2+y^2) + (4u-3v)x y - (2u^2-v^2)(x+y+(u-v)z)z \right) \times \\
& \left( u(x^2+y^2) + v x y + (2u^2-v^2)(x+y+(u-v)z)z \right)- \\
&  (2u^2-v^2) \left( (x+y)^2 + 2(u-v)(x+y)z - (2u^2-v^2)z^2 - (2u-v)k w^2 \right) \times \\
& \left( (x+y)^2 + 2(u-v)(x+y)z - (2u^2-v^2)z^2 + (2u-v)k w^2 \right) \\
& = (2u-v)^2 \left( x^4+y^4 - (2u^2-v^2)^2 z^4 + (2u^2-v^2) k^2 w^4 \right).
\end{align*}
The surface $V_3$ has generic Picard rank equal to 3.
To find surfaces $V_3$ with infinitely many rational points,
it suffices for example to find infinitely many rational points on the intersection
\begin{align*}
F_0: & \; (x+y)^2 + 2(u-v) (x+y) z - (2u^2-v^2) z^2 = (2u-v) k w^2, \\
F_1: & \; u (x^2+y^2) + v x y + (2u^2-v^2)(x+y+(u-v)z)z = 0.
\end{align*}
We set ($x,y,z)=(p,q,r)$ and regard the equation of $F_1$ as defining a cubic curve
$C_1$ in the $u,v$-plane over ${\mathbb{Q}}(p,q,r)$, containing the rational point $(u,v)=(0,0)$.
As such, it is an elliptic curve with cubic model:
\[ V^2 = U(U^2 -(p^4+q^4)). \]
We seek points $(u,v)$ on $C_1$ such that $2u^2-v^2 \neq \pm \square$, in order to
keep small the Picard rank of the surface. Now the points $P_1=(-p^2,-p q^2)$,
$P_2=(-q^2,-p^2 q)$ on the cubic model are independent, and $P_1+P_2$ pulls back
to the point on $C_1$
\[ P(u,v) = (-p q(p+q)/((p^2+p q+q^2)r), (p+q)(p^2+q^2)/((p^2+p q+q^2)r)), \]
with
\[ 2u^2-v^2=-(p+q)^2(p^4+q^4)/((p^2+p q+q^2)^2r^2). \]
For this point $P$ to determine a point on the quadric $F_0$, we require
\[ k=p q r(2 p^2+3 p q+2 q^2)/((p+q)(p^2+p q+q^2)).  \]
On setting $z=(p^2+p q+q^2)r Z /(p+q)$, $w=(p^2+p q+q^2)W$, the surface now takes the form
\begin{equation}
\label{surface1}
V: x^4+y^4 - (p^4+q^4)^2 Z^4 - p^2 q^2 (2 p^2+3 p q+2 q^2)^2(p^4+q^4)W^4 = 0;
\end{equation}
the generic Picard rank is 3.  The intersecting quadric becomes
\[ F_0: \; p q(p+q)^2 (2p^2+3 p q+2 q^2) W^2 = -(x+y)^2 +2(p^2+p q+q^2)(x+y)Z-(p^4+q^4)Z^2; \]
and the intersection comprises the two quadrics
\begin{align*}
F_1: \; & -p q(x^2+y^2) + (p^2+q^2)x y - (p^4+q^4)(x+y) Z + (p^2+p q+q^2)(p^4+q^4) Z^2 = 0, \\
F_2: \; & (2p^2+3p q+2q^2)(x^2+y^2) + (3p^2+4p q+3q^2)x y - (p^4+q^4)(x+y)Z \\
     & +(p^2+p q+q^2)(p^4+q^4)Z^2 = 0.
\end{align*}
We have arranged that $F_1$ contain the point
$(x,y,Z)=(p,q,(p+q)/(p^2+p q+q^2))$, which allows parameterization of all points
on $F_1$ by:
\begin{align*}
x:y:z = & p q^2(p^2+q^2) a^2 + (p+q)(p^4+q^4) a b +p(p^4+q^4) b^2: \\
& p^2 q(p^2+q^2) a^2 + (p+q)(p^4+q^4) a b +q(p^4+q^4) b^2: \\
& (p-q)^2(p+q) a b.
\end{align*}
The inverse is given by
\[ a/b = -((p-q)q x - (p-q)p y + (p^4+q^4) z)/(p q (p^2+q^2) z). \]
Such a point gives rise to a rational point on the intersecting quadric $F_0$
precisely when
\begin{align*}
& p q(2p^2+3p q+2q^2) W^2 = -p^2 q^2(p^2+q^2)^2a^4-2p q(p+q)^2(p^2+q^2)(p^2-p q+q^2)a^3b\\
& -(p^4+q^4)(p^4+2p^3q+6p^2q^2+2p q^3+q^4)a^2b^2-2(p+q)^2(p^2-p q+q^2)(p^4+q^4)a b^3 \\
& -(p^4+q^4)^2b^4
\end{align*}
and we have arranged that there is a point on this elliptic quartic at
\[ (a,b,W)=(p^4+q^4, -p q(p^2+q^2), p q(p-q)^2(p^2+q^2)(p^4+q^4)). \]
This latter point is of infinite order, and hence its multiples
pull back to infinitely many rational points on (\ref{surface1}).
Using the same type of reasoning as in the proof of Theorem \ref{pic}, we get the following.

\begin{thm}\label{pic3}
 The set of diagonal quartic surfaces defined over $\mathbb{Q}$ with Picard number 3 and Zariski
dense set of rational points is infinite.
\end{thm}
\noindent

\begin{rem} {\rm The essence of the method described in this section was used in \cite{BreUl1}
to find infinitely many integral points on the surface $w^2=x^6+y^6+z^6$.}
\end{rem}
\begin{rem}{\rm The approach of this section cannot find diagonal quartic surfaces with Picard number
one, for the construction immediately implies the existence of two elliptic fibrations on
$\mathcal{S}$ (defined over $\mathbb{Q}$) and thus the Picard number is at least two (page 32 of Bright \cite{Bri}, Corollary 2.27).}
\end{rem}

\begin{rem}
{\rm A nice illustration of Van Luijk's theorem in practice can be also found in the Master's thesis of
D. Festi \cite{Festi}.}
\end{rem}

\section{Second construction} \label{sec3}

The second method we describe to construct diagonal quartics with infinitely many rational points
is even more direct than the previous one. Consider the genus three curve $C:\;Px^4+Qy^4+Rz^4=0$
and suppose that $(a,b,c)$, with $abc\neq 0$ is a rational point lying on $C$.  Substituting
\begin{equation*}
x=a T, \quad y=-\frac{c^3R}{b^3Q}+b T, \quad z=1+c T,
\end{equation*}
there results
\begin{equation*}
G(T)=\frac{-a^4 P R}{b^{12}Q^3}(b^8Q^2-b^4c^4Q R+c^8R^2 + 4b^4c Q(b^4Q-c^4R)T + 6b^8c^2Q^2T^2)=0.
\end{equation*}
The polynomial $G$ is of degree two in $T$ and thus if a rational number $S$ is chosen so that the
quartic curve
\begin{equation*}
\mathcal{C}:\;-Sw^4=G(T)
\end{equation*}
has infinitely many rational points, then there will be infinitely many rational points on the surface
$Px^4+Qy^4+Rz^4+Sw^4=0$.

To construct some examples, let $a, b, c$ be integer parameters and
let $R=-(Pa^4+Qb^4)/c^4$ be a non-zero rational number. Then the point $(a,b,c)$ lies on $C$ and
we seek $S$ such that there is a point of infinite order on the curve
\begin {align*}
\mathcal{C}: \; & -SW^4 = H(T) \\
          = & P Q(a^4P+b^4Q)(a^8P^2+3a^4b^4P Q+3b^8Q^2 + 4b^4c Q(a^4P+2b^4Q)T + 6b^8c^2Q^2T^2),
\end{align*}
where $W=\frac{b^3c Q}{a}w$. In order to guarantee the existence of a point on $\mathcal{C}$
we put $S=-H(t)$, so that $\mathcal{C}$ contains the point $(W,T)=(1,t)$.

\begin{exam}{\rm
To produce an explicit example, put $R=-P-Q$, $a=b=c=1$. Thus we consider the curve
$-S w^4= P Q(P+Q)(P^2+3P Q+3Q^2 + 4 Q(P+2Q)T + 6Q^2T^2)$ with $S=-P Q(P+Q)(P^2+3P Q+3Q^2+4Q(P+2Q)t+6Q^2t^2)$.
That is, the curve
\begin{align*}
\mathcal{C}: \;  & P Q(P+Q)(P^2+3P Q+3Q^2+4Q(P+2Q)t+6Q^2t^2) W^4 = \\
             & P Q(P+Q)(P^2+3P Q+3Q^2 + 4 Q(P+2Q)T + 6Q^2T^2).
\end{align*}
Using the point $(W,T)=(1,t)$ as a point at infinity, we find that $\mathcal{C}$ is birationally
equivalent to the elliptic curve defined over $\mathbb{Q}(P,Q)(t)$ with cubic model:
\[ \mathcal{E}: \; Y^2 = X(X^2 + 3(P^2+P Q+Q^2)(P^2+3P Q+3Q^2+4P Q t+8Q^2 t+6Q^2t^2)).  \]
From a geometric point of view,  $\mathcal{E}$ for fixed rational $P,Q$ represents a rational elliptic surface
in the plane $(X,Y,t)$.  Thus by results of Shioda, the generators of $\mathcal{E}(\mathbb{C}(t))$
lie among polynomial solutions $(X,Y)$ of the equation of $\mathcal{E}$ which satisfy the
conditions $\operatorname{deg}_{t}X\leq 2,\;\operatorname{deg}_{t}Y\leq 3$.  It is straightforward in any given instance
to compute all such polynomials defined over $\mathbb{Q}$.

For generic $P,Q$, the point $(W,T)=(-1,t)$ is of finite order on $\mathcal{C}$ precisely when $P+2Q+3 Q t=0$.
When $t \neq -(P+2Q)/(3Q)$ therefore, this point is of infinite order, and its multiples pull
back to infinitely many rational points on the diagonal quartic surface
\[ P x^4 + Q y^4 -(P+Q) z^4 -P Q(P+Q)(P^2+3P Q+3Q^2+4Q(P+2Q)t+6Q^2t^2) w^4 = 0. \]
The generic Picard rank is in this instance equal to 1. \\
To refine the example further, consider the specific case $P=Q=1$, and set $t=-1+u$.
Computation shows that for $u \neq 0$,
the point $M(-1,-1+u)$ is a generator for $\mathcal{E}(\mathbb{Q}(u))$. Its multiples $kM$, $k=1,2,3...$, pull back to
infinitely many rational points $(x,y,z,w)$ on the diagonal quartic surface
\begin{equation}
\label{St}
\mathcal{S}_{u}: \; X^4+Y^4-2Z^4-2(1+6u^2)W^4=0.
\end{equation}
The point $M$ pulls back to $(X,Y,Z,W)=(-1+u,1+u,u,-1)$; and $2M$ pulls back to
\begin{align*}
(X,Y,Z,W) =  & (-1 - 3u - 36u^3 + 54u^4 - 162u^5 - 324u^7 - 729u^8 + 81u^9, \\
& 1 - 3u - 36u^3 - 54u^4 - 162u^5 - 324u^7 + 729u^8 + 81u^9, \\
& 3u(-1 - 12u^2 - 54u^4 - 108u^6 + 27u^8), \\
& -(1 + 12u^2 + 9u^4)(-1 + 27u^4)),
\end{align*}
etc.
}
\end{exam}
\noindent
To our knowledge this is the first explicit example of a family of diagonal quartics each with
Picard number equal to 1 and possessing infinitely many rational points. We thus get:
\begin{thm}\label{pic4}
The set of diagonal quartic surfaces defined over $\mathbb{Q}$ with Picard number 1 and infinitely many
rational points is infinite.
\end{thm}
\section{Third construction } \label{sec4}
In this section we propose a different idea, which we name the polynomial method. More precisely,
we are interested in finding polynomial curves of low degree which lie on the affine variety
$F(x,y,z,w)=0$, where
\begin{equation*}
F(X,Y,Z,W)=PX^4+QY^4+RZ^2+SW^2.
\end{equation*}
Throughout, we suppose $PQRS \neq 0$, and that $PQRS$ is not a perfect square.
Essentially, we seek polynomial curves, say in the variable $T$, which satisfy the conditions
$\operatorname{deg}X\leq 1,\operatorname{deg}Y\leq 1$ and $\operatorname{deg}Z\leq2, \operatorname{deg}W\leq 2$.
It is clear that if such an identity exists, and the curve of genus one given by the intersection
\begin{equation*}
\alpha U^2 =Z(T),\quad \beta V^2=W(T)
\end{equation*}
has infinitely many rational points, then there are infinitely many rational points on the diagonal
quartic surface
$PX^4+QY^4+R\alpha^2 U^4+S\beta ^2 V^4=0$. \\ \\
If $\operatorname{deg}X=0$, we have one of two identities:
\[ (P,Q,R,S)=(a c(a d-b c), -b d(a d-b c), -c d, a b), \quad (X,Y,Z,W)=(1,t,a+b t^2, c+d t^2), \]
where the product of coefficients is a square; and
\begin{align}
\label{ident0}
(P,Q,R,S)=((2a c-b^2)a^2, (2a c-b^2)c^2, 1, -2a c), & \\
(X,Y,Z,W)=(1,t, a b +2 a c t+b c t^2, a+b t+c t^2). \nonumber
\end{align}
If  $\operatorname{deg}X=1$, then by homogeneity, by multiplying $T$ by a scalar, and by absorption of fourth powers
into $P,Q,R,S$, it suffices only to consider polynomial solutions of $F(X,Y,Z,W)=0$ of the form
\begin{equation*}
X=T, \quad Y=1+T, \quad Z=r+q T+T^2, \quad W=w+v T+T^2.
\end{equation*}
Equating the coefficients of the resulting expression in $T$ to zero, there arises:
\begin{equation*}
\begin{cases}
Q + r^2R + w^2 S=0,\\
2Q + q r R + 2v w S=0, \\
6Q + q^2R + 2r R + (v^2 + 2w)S=0,\\
2Q + q R + v S=0, \\
P + Q + R + S =0.
\end{cases}
\end{equation*}
Provided $r(q-2r) \neq 0$, then the first, second, and fifth equation imply that
\begin{align*}
P= & (q r - 2r^2 - v w + r^2v w + 2w^2 - q r w^2)S/(r(-q + 2r)), \\
Q= & (r v - q w)w S/(q - 2r), \\
R= & (v - 2w)w S/(r(-q + 2r))
\end{align*}
Provided $w(2r^2-q)+r(q-2r) \neq 0$, then the third and fourth equations give
\begin{align*}
v = & 2q(r-1)w^2/(w(2r^2-q)+r(q-2r)), \\
 & (r-w)(w-1)w(q r-2r^2-q w+q^2w-2q r w+2r^2w)=0.
\end{align*}
The instances $(r-w)(w-1)w=0$ lead to $PQ=0$; so necessarily
\[ w(-q+q^2-2q r+2r^2)+r(q-2r) = 0. \]
Now $-q+q^2-2q r+2r^2 \neq 0$, otherwise $r(q-2r)=0$; so that
\[ w=r(2r-q)/(-q+q^2-2q r+2r^2). \]
Together, we obtain the identity $P X^4+Q Y^4+R Z^2+S W^2=0$, where
\begin{align}
\label{ident1}
(P,Q,R,S) = & ( -(q-r-1)^2(q^2-2q r+2r^2-2r), \; -r^2(q^2-2q r+2r^2-2r), \\
            & 2(q-r-1)r, \; 1 ) \nonumber
\end{align}
and $(X,Y,Z,W)=$
\[ \left( T, \; 1+T, \; r+q T+T^2, \; -(q-2r)r+2(r-1)r T+(q^2-2q r-q+2r^2)T^2 \right). \]
The exceptions above can easily be dealt with:
if $r(q-2r) \neq 0$ and $w(q-2r^2)-r(q-2r)=0$, then $w=r(q-2r)/(q-2r^2)$, leading to
$q(r-1)=0$.  Now $r \neq 1$, otherwise $P=0$; so necessarily $q=0$ and $(v-2)(r v-r+1)=0$.
The case $v=2$ leads to $R=0$; and thus $v=(r-1)/r$. This gives the identity
$P X^4+Q Y^4+R Z^2+S W^2=0$,
where
\begin{equation}
\label{ident2}
(P,Q,R,S)=\left( (1-r)(1+r)^2, \; (1-r)r^2, \; -1-r, \; 2r \right)
\end{equation}
and
\[ (X,Y,Z,W) = \left( T, \; 1+T, \;r+T^2, \; r+(-1+r)T+r T^2 \right). \]
Finally, if $r(q-2r)=0$, then the only non-degenerate identity that results is
$P X^4+Q Y^4+R Z^2+S W^2=0$, where
\begin{equation}
\label{ident3}
(P,Q,R,S)=\left( (1-w)^2, \; w^2, \; 2w(1-w), \; -1 \right),
\end{equation}
and
\[ (X,Y,Z,W) = \left( T, \; 1+T, \; T+T^2, \; w+2w T+T^2 \right). \]
In the identities at (\ref{ident0}), (\ref{ident1}), (\ref{ident2}), (\ref{ident3}), it is straightforward
by choice of parameter to force rational points on to the intersection of the two quadrics $Z,W$
in the variable $T$,
thereby providing examples of diagonal quartic surfaces with infinitely many points.
The generic Picard rank in each case is equal to two. But the examples are interesting in their own right,
and we give a few more details.
\begin{exam}{\rm
At (\ref{ident0}), we demand
\[  a b +2a c t+b c t^2 = \alpha u^2, \quad a+b t+c t^2 = \beta v^2. \]
If we set, for example, $(\alpha,\beta)=(a b,a)$, then the intersection
represents an elliptic curve with origin at $(t,u,v)=(0,1,1)$. The point
$Q(t,u,v)=(0,1,-1)$ is of infinite order, and pullbacks of its
multiples $mQ$ provide infinitely many points on the surface of generic rank 2
\[ (2a c-b^2)a^2 X^4+(2 a c-b^2)c^2 Y^4+ a^2 b^2 Z^4 = 2 a^3 c W^4. \]
When $m=2$ for instance,
\begin{align*}
(X,Y,Z,W)= & (b^4 - 20a b^2c + 4a^2c^2, \; 8a b(b^2 + 2a c),  \\
& -b^4 - 20a b^2c + 12a^2c^2, -3b^4 + 20a b^2c + 4a^2c^2).
\end{align*}
}
\end{exam}
\begin{exam}{\rm
On homogenizing (\ref{ident1}), we have the example
\[ (p-q+r)^2(2p r-q^2+2q r-2r^2)X^4 + r^2(2p r-q^2+2q r-2r^2)Y^4 - 2r(p-q+r)s^2 Z^4 + t^2 W^4 = 0, \]
with $X=T$, $Y=1+T$, and where we demand
\[ r+ q T+p T^2 = s Z^2, \qquad r(2r-q) - 2r(p-r)T + (-p q+q^2-2q r+2r^2)T^2 = t W^2. \]
It is only necessary to specialize $p,q,r,s,t$ so that this latter intersection
represents an elliptic curve of positive rank. This occurs for example when
$s=r$, $t=r(2r-q)$, where we take $(T,Z,W)=(0,1,1)$ as zero of the corresponding
elliptic curve. The point $Q(T,Z,W)=(0,1,-1)$ is of infinite order, and pullbacks of its
multiples $mQ$ provide infinitely many points on the surface
\begin{align}
\label{surf1}
(p-q+r)^2(q^2-2 p r-2 q r+2r^2)X^4 & + r^2(q^2-2 p r-2 q r+2r^2)Y^4+ \\
2r^3(p-q+r)Z^4 & - (q-2r)^2r^2W^4 = 0. \nonumber
\end{align}
For example when $m=2$:
\begin{align*}
(X,Y, & Z,W) = (8(q-2r)r(q^2+2p r-6q r+6r^2), \\
& q^4-20p q^2r+12q^3r+4p^2r^2+72p q r^2-72q^2r^2-72p r^3+120q r^3-60r^4, \\
&-3q^4+20p q^2 r+4q^3r+4p^2r^2-88p q r^2+32q^2r^2+88p r^3-72q r^3+36r^4, \\
& q^4+20p q^2r-28q^3r-12p^2r^2-56p q r^2+112q^2r^2+56p r^3-168q r^3+84r^4).
\end{align*}
}
\end{exam}
\begin{exam}{\rm
On homogenizing (\ref{ident2}), we have the example
\[ -(r-s)(r+s)^2 X^4 - r^2 (r-s) Y^4 - (r+s) p^2 Z^4 + 2r q^2 W^4 = 0, \]
with $X=T$, $Y=1+T$, and where we demand
\[ r+s T^2 = p Z^2, \qquad  r + (r-s)T + r T^2 = q W^2. \]
It suffices to specialize $p,q,r,s$ so that this latter intersection
represents an elliptic curve of positive rank. This occurs for example when $p=q=r$,
with $(T,Z,W)=(0,1,1)$ as origin of the elliptic curve. The point $Q(T,Z,W)=(0,1,-1)$
is of infinite order, and pullbacks of its multiples $mQ$ provide infinitely many
points on the surface
\begin{align}
\label{surf2}
-(r-s)(r+s)^2 X^4-r^2(r-s) Y^4-r^2(r+s) Z^4+2r^3 W^4 = 0.
\end{align}
For example when $m=2$:
\[ (X,Y,Z,W)=(-8r(3r+s), \; -15r^2-18r s+s^2, \; 9r^2+22r s+s^2, \; 21r^2+14r s-3s^2 ); \]
etc.
}
\end{exam}
\begin{exam}{\rm
On homogenizing (\ref{ident3}), we have the example
\[ -(v - w)^2 X^4 - w^2 Y^4 - 2(v-w)w t^2 Z^4 + u^2 W^4 = 0, \]
with $X=T$, $Y=1+T$, and where we demand
\[ T+T^2 = t Z^2, \qquad  w + 2 w T + v T^2 = u W^2. \]
It is only necessary to specialize $t,u,v,w$ so that this latter intersection
represents an elliptic curve of positive rank. As illustration, take $v/w=2a$, $t=4(1-a)$, $u=w$,
with $(T,Z,W)=(0,0,1)$ as origin for the elliptic curve. The point $Q(T,Z,W)=(-1/a,1/(2a),1)$
is of infinite order, and pullbacks of its multiples $mQ$ provide infinitely many
points on the surface
\begin{align}
\label{surf3}
(1-2a)^2 X^4 + Y^4 - 2(1-a)^2(1-2a) Z^4 - W^4 = 0.
\end{align}
For example when $m=2$:
\[ (X,Y,Z,W) = (4(1-a)a^2, \; (2-4a+a^2)^2, \; 2a(2-4a+a^2), \; -4+16a-20a^2+8a^3+a^4), \]
etc.
}
\end{exam}
Finally, we observe that it is possible, given $m \in \mathbb{Z}$, to obtain surfaces of this
type with $PQRS \equiv m$ modulo squares. For set $r=1$, $q=2+2m$ in (\ref{ident1}).
There results
\[ 4m^2(1+m) T^4+(1+m) (1+T)^4-m \left( 1-(1+2m)T^2 \right)^2-\left( 1+2(1+m)T+T^2 \right)^2=0, \]
and the product of the coefficients is $m$ modulo squares.
We now demand that
\[ 1-(1+2m)T^2 = \square, \qquad 1+2(1+m)T+T^2 = \square. \]
This intersection contains a point at $T=0$, and the curve is of positive rank.
Some points are given by $T=$
\[ \frac{4(-1+m)}{(1+m)(5+m)}, \quad \frac{4(-1+m) (-3-10m+m^2)(-1+10m+3m^2)}{(1+m)(13+22m+m^2)(1-5m+27m^2+m^3)}, \dots \]
These give the surfaces
\[ 4m^2(1+m)X^4 + (1+m)Y^4 - m Z^4 - W^4 = 0, \]
with infinitely many points given by
\[ (X,Y,Z,W)= \left( 4(-1+m), \; 1+10m+m^2, \; -3-10m+m^2, \; -1+10m+3m^2 \right); \]
\begin{align*}
(X,Y,Z,W) = & (4(-1+m)(3+10m-m^2)(1-10m-3m^2), \\
& (1+10m+m^2)(1+52m+38m^2+52m^3+m^4), \\
& 5+114m-233m^2-1140m^3-381m^4-94m^5+m^6, \\
& 1-94m-381m^2-1140m^3-233m^4+114m^5+5m^6);
\end{align*}
etc. We thus proved:

\begin{thm}
{For any $m\in\mathbb{Z}$ there is a diagonal quartic surface
$ax^4+bY^4+cZ^4+dT^4=0$ containing infinitely many (non-trivial)
rational points and satisfying the property $abcd\equiv m\pmod{\mathbb{Q}^{2}}$.}
\end{thm}

\section{Identities for sixth powers} \label{sec5}
In this section we apply the polynomial method to find low degree curves on the two types of surface
that are of interest to us:
\begin{equation}\label{quadcube}
PX^6+QY^6+RZ^3+SW^2=0
\end{equation}
and
\begin{equation}\label{quadquad}
PX^6+QY^6+RZ^3+SW^3=0.
\end{equation}

First, consider the equation (\ref{quadcube}). We seek polynomial solutions of (\ref{quadcube}) of the form
\begin{equation*}
X=T,\quad Y=1+T,\quad Z=T^2+aT+b,\quad W=T^3+cT^2+dT+e.
\end{equation*}
Let $F(X,Y,Z,W)$ denote the left hand side of (\ref{quadquad}). Then after the above substitution
we obtain a degree six polynomial in $T$, say $F(X,Y,Z,W)=\sum_{i=0}^{6}c_{i}T^i$.
We are interested in solving in rationals the system
of equations $c_{i}=0$ for $i=0,\ldots,6$, with the assumption $PQRS\neq 0$. First, solve the
system
\begin{equation*}
c_{4}=c_{5}=c_{6}=c_{7}=0
\end{equation*}
with respect to $P, Q, d, e$. There results:
\begin{align*}
P&=\frac{1}{6}(3(a-2) R + 2(c-3)S),\\
Q&=-\frac{3aR+2cS}{6},\\
d&=\frac{3 (2a^2-5a + 2 b)R+2(c-5)cS}{4S},\\
e&=\frac{-3(2a^3-20a + 12ab+15ac-6a^2c-6bc)R+2c(3c^2-15c+20)}{12S}.
\end{align*}
Substituting into the first three equations and clearing the common denominator $144S$, we are left
with the system
\begin{equation*}
c_{1}(a,b,c,R,S)=c_{2}(a,b,c,R,S)=c_{3}(a,b,c,R,S)=0.
\end{equation*}
These equations are of degree $3, 2, 2$ respectively in the variable $b$. The polynomial
$M_{1}=\operatorname{Res}_{b}(c_{1},c_{2})=20736 R^2 S \times W_{1}(a,c,R,S)$, with $\operatorname{deg}_{R}W_{1}=7$;
and $M_{2}=\operatorname{Res}_{b}(c_{2},c_{3})=5184 R^2 S\times W_{2}(a,c,R,S)$, with $\operatorname{deg}_{R}W_{2}=5$.
Finally, $W(a,c,S)=\operatorname{Res}_{R}(W_{1},W_{2})$ has the following factorization:
\begin{align*}
W(a,c,S)=-2^{41}&3^{36}5^{2}S^{35}(a-2)^{11}(a-1)^{3}a^{13}(3a-2c)^{4}(c-3)^{7}c^{8} \times \\
& (4-4c+a(c+1))(2c+a(c-4))f_{10}(a,c)f_{11}(a,c)f_{12}(a,c)f_{13}(a,c),
\end{align*}
where
\begin{align*}
&f_{10}(a,c)=(a^3-7a^2+23a+1)c^2+(a+1)(a^2-16a-27)c+2a(2a^2+14a+27),\\
&f_{11}(a,c)=(a^3+a^2+7a-27)c^2-(7 a^3+15a^2-49a+3)c+4a (2a-1)^2,\\
&f_{12}(a,c)=\sum_{i=0}^{6}p_{i}(a)c^{i},\\
\end{align*}
with
\begin{align*}
&p_{0}=8 a^3 (1+18 a+2058 a^2+1584 a^3+444 a^4+72 a^5+8 a^6),\\
&p_{1}=-12 a^2 (1+49 a+1601 a^2+1609 a^3+1592 a^4+616 a^5+100 a^6+12 a^7),\\
&p_{2}=6 a (3+134 a+5587 a^2-20016 a^3+28417 a^4-9500 a^5+2771 a^6-54 a^7+98 a^8),\\
&p_{3}=-9-927 a-10296 a^2+49352 a^3-18054 a^4-38322 a^5+8348 a^6+1596 a^7-1221 a^8-387 a^9,\\
&p_{4}=3 (-1+a) (-135-1350 a-1570 a^2+9688 a^3-2738 a^4+1838 a^5-1170 a^6+368 a^7+29 a^8),\\
&p_{5}=-3 (-1+a)^2 (2025-1251 a+1825 a^2+169 a^3-329 a^4-21 a^5+59 a^6+3 a^7),\\
&p_{6}=(-1+a)^3 (729-486 a+783 a^2-548 a^3+147 a^4-6 a^5+a^6).
\end{align*}

The last factor $f_{13}$ is of degree $57$ in $a$ and $59$ in $c$.  It is clear that the only potentially
awkward factors for a case-by-case analysis are $f_{i}$ for $i=10,11,12,13$. The curve defined by
$f_{10}(a,c)=0$ in the $ac$ plane is rational, with parametrization:
\begin{equation*}
a=-\frac{6(t-5)}{t^2+15},\quad c=-\frac{12 \left(t^2-2 t+21\right)}{t^3-3 t^2+3 t-81}.
\end{equation*}
Similarly, the curve defined by $f_{11}(a,c)=0$ is rational with parameterization:
\begin{equation*}
a=-\frac{6 (t-5)}{t^2+15},\quad c=-\frac{8 (t-3)^2}{t^3+t^2+27 t+3}.
\end{equation*}
Surprisingly, the curve defined by $f_{12}(a,c)=0$ is also of genus 0, with nontrivial rational points
when $a c(3a-2c)=0$. However, these are the only rational points. For
\[ f_{12}(a,c) = \mbox{Norm}_{\mathbb{Q}(\theta)/\mathbb{Q}} (g_0(a,c)+g_1(a,c)\theta+g_2(a,c)\theta^2), \]
where $\theta^3=2$, and
\begin{align*}
g_0(a,c)= & 562a + 60a^2 + 20a^3 - 373c + 19a c - 71a^2c - 3a^3c - 45c^2 + 47a c^2 - 3a^2c^2 + a^3c^2, \\
g_1(a,c)= & 2(223a + 24a^2 + 8a^3 - 148c + 10a c - 32a^2c - 18c^2 + 18a c^2), \\
g_2(a,c) = & 354a + 36a^2 + 12a^3 - 235c + 15a c - 48a^2c - 27c^2 + 27a c^2.
\end{align*}
Thus $f_{12}(a,c)=0$ implies $g_0=g_1=g_2=0$, and it follows easily that either $(a,c)$ is the point
at infinity $(0/0,1/0)$, or $(a,c)=(0,0),(1,3/2),(2,3)$. No new identities result. \\
We are unable to say much about the zeros of $f_{13}(a,c)=0$, a curve of degree 70. By reducing
modulo a large number of primes, we believe the curve to be absolutely irreducible. A small search
produces zeros at $(a,c)=(0,0)$, $(1,0)$, $(2,0)$, $(1/2,0)$, $(2/9,0)$, $(0,2)$, $(0,3)$, $(0,-1/23)$,
$(1,2)$, $(2,3)$, $(2,4)$, $(1,3/2)$, $(1,3/2)$, $(-2,1/3)$, $(3/2,3)$, $(8,3)$, $(1,7/3)$, but no
new identities result. \\
Case-by-case analysis now results in the following table. Solutions occur in pairs
corresponding to the transformation $T \rightarrow -T-1$, and without loss of generality
we list only one of each pair.
\begin{equation*}
\begin{array}{|c|c|c|c|}
\hline
P & Q & R & S \\
\hline
x & y & z & w \\
\hline \hline
t^2 & 1 & -2t & -1 \\
\hline
T & 1+T & T+T^2 & (1+T)^3 - t T^3 \\
\hline
& & & \\
\hline
16t^3 & 8 & -1 & 3t \\
\hline
T & 1+T & 2((1+T)^2+2t T^2) & 4T((1+T)^2+t T^2) \\
\hline
& & & \\
\hline
1 & -108 & -4 & 3 \\
\hline
1 & T & 1-3T+6T^2 & -1+6T-12T^2+18T^3 \\
\hline
& & & \\
\hline
1 & -432 & 8 & -9 \\
\hline
1 & T & 1-3T+6T^2 & -1+4T-12T^2+12T^3 \\
\hline \hline
\end{array}
\end{equation*}

\noindent
We obtain points on $P x^6+Q y^6+R z^6+S w^2=0$ by demanding $Z=z^2$. As an example, consider the
first entry in the above table, with demand $T+T^2=z^2$. The latter is parameterized by
\[ (T,z)=\left( a^2/(-a^2+b^2), \; a b/(-a^2+b^2) \right) \]
resulting in the (rather dull) identity $t^2 x^6 + y^6 -2t z^6 = w^2$ with
\[ (x,y,z,w)=(a^2, \; b^2, \; a b, \; a^6t-b^6). \]

\noindent
As second example, take the second entry in the table, where we demand
\[ (1+T)^2+2t T^2 = z^2. \]
This is parameterized by
\[ (T,z)=\left( \frac{2a b}{2t a^2-2a b-b^2}, \; \frac{2t a^2+b^2}{2t a^2-2 a b-b^2} \right) \]
resulting in the identity $2t^3 x^6+y^6-z^6+6t w^2=0$ with
\[ (x,y,z,w)=(2a b, 2t a^2-b^2, 2t a^2+b^2, 2a b(4t^2 a^4+b^4) ). \]

\noindent
The third entry (with $PQRS=\square$) demands
\[ 1-3T+6T^2 = z^2, \]
which is parameterized by
\[ (T,z)=\left( \frac{a(3a+2b)}{6a^2-b^2}, \; \frac{6a^2+3a b+b^2}{6a^2-b^2} \right), \]
resulting in the identity $x^6-108y^6-4z^6+3w^2=0$
where
\begin{align}\label{specsol}
(x,y,z,w)= & (6a^2-b^2, \; a(3a+2b), \; 6a^2+3a b+b^2, \\
& 270a^6+540a^5b+360a^4b^2+144a^3b^3+48a^2b^4+12a b^5+b^6).\notag
\end{align}
Similarly, the fourth entry leads to the identity $x^6-432y^6+8z^6=9w^2$
with
\begin{align*}
(x,y,z,w)= & (6a^2-b^2, \; a(3a+2b), \; 6a^2+3a b+b^2, \\
& 108a^6-72a^5b-216a^4b^2-144a^3b^3-42a^2b^4-8a b^5-b^6).
\end{align*}

\noindent
In the same way, we can obtain points on $P x^6+Q y^6+S m^2 z^6+R w^3=0$ by setting $(Z,W)=(w,m z^3)$,
with $m$ chosen so that the appropriate cubic curve is elliptic with positive rational rank.
As example, we consider just the first entry from the table, which leads to the cubic curve
\[ (1+T)^3 -t \; T^3 = m z^3. \]
This latter is elliptic with positive rational rank for $(t,m)=(1,6)$.
A generator is $Q(z,T)=(7/18, -37/54)$, and the resulting identity is
$x^6+y^6-36 z^6 + 2 w^3 =0$. Multiples $kQ$ correspond to solutions:
\begin{equation*}
\begin{array}{c|cccc}
k & x & y & z & w \\ \hline
1 & 37 & 17 & 21 & 629 \\ \hline
2 & 1805723 & 2237723 & 960540 & 4040707888729 \\ \hline
3 & 209143555850753 & 84691068680987 & 112490043311709 & 17712591252741962842340733211 \\ \hline
.. & .. & .. & .. & .. \\ \hline
\end{array}
\end{equation*}

\noindent
Second, we consider the equation (\ref{quadquad}) and look for polynomial solutions of the form
\begin{equation*}
X=T, \quad Y=1+T, \quad Z=b+a T+T^2, \quad W=d+c T+T^2.
\end{equation*}
If $F(X,Y,Z,W)$ denotes the left hand side in (\ref{quadquad}) then the above substitution
results in a degree six polynomial in $T$, say $F(X,Y,Z,W)=\sum_{i=0}^{6}c_{i}T^i$.
Consider the system of equations $c_{i}=0$ for $i=0,\ldots,6$.
Eliminating $a,b,c,d,S$, there results $P Q (P+R) (Q+R) H(P,Q,R)=0$, where
\begin{align*}
H(P,Q,R)=64&R^6+192(P+Q)R^5+48(5 P^2+53 P Q+5 Q^2)R^4+\\
           &32(P+Q)(5 P^2+139 P Q+5 Q^2)R^3+\\
           &12(5P^4+8P^3Q+1414P^2Q^2+8P Q^3+5 Q^4)R^2+\\
           &12(P+Q)(P^2-174 P Q+Q^2)(P^2-6 P Q+Q^2)R+\\
           &(P^2+18 P Q+Q^2)^3.
\end{align*}
Solutions with $P Q=0$ are not of interest, and rational solutions resulting from $(P+R)(Q+R)=0$
are trivial.  Consider finally $H(P,Q,R)=0$. Surprisingly, the equation defines a curve $C$ of genus
zero, with parametrization:
\begin{equation*}
P=2 v^6, \quad Q=-2,\quad R=(1-v-v^2)^3.
\end{equation*}
Consequently, we obtain just one rational solution of the system. After the substitution $T \rightarrow 1/(v T-1)$, there results the identity
\begin{equation*}
(1-T-T^2)^3+(1+T-T^2)^3=2-2T^6,
\end{equation*}
discovered by Elkies and mentioned in \cite{Elk2}.

\begin{rem}
{\rm Note that for the equation $x^6-108y^6-4z^6+3w^2=0$ we can prove much more. Indeed,
rewrite the equation in the form $(x^3-2z^3)(x^3+2z^3)=3(6y^3-w)(6y^3+w)$ and make the substitution
$w=t(x^3+2z^3)-6y^3$, leading to the equation of a genus 1 curve over $\mathbb{Q}(t)$:
\begin{equation}\label{speccurve}
\mathcal{C}:\;\left(3 t^2+1\right) x^3+\left(6 t^2-2\right) z^3-36 t y^3=0.
\end{equation}
Using the solution given by (\ref{specsol}) we can define a base change
$t=\phi(a)=(1+12 a^2+12 a^3)/(1+6 a+18 a^3)$ which give us a $\mathbb{Q}(a)$- rational point
on $\mathcal{C}_{\phi}$. This point can be used to produce infinitely many rational parametric solutions
of the equation defining $\mathcal{C}_{\phi}$, and thus leads to infinitely many rational curves on the
surface, say $\mathcal{S}$, defined by the equation $x^6-108y^6-4z^6+3w^2=0$ with $w=t(x^3+2z^3)-6y^3$.
As an immediate consequence of the existence of infinitely many rational curves on $\mathcal{S}$, we
get that the set $\mathcal{S}(\mathbb{Q})$ is Zariski dense in $\mathcal{S}$.}
\end{rem}

Finally, we adapt the third method above to construct some surfaces of the form
\begin{equation*}
Px^6+Qy^6+Rz^6+Sw^2=0
\end{equation*}
with infinitely many rational points. Consider the genus ten curve $C:\;Px^6+Qy^6+Rz^6=0$ and suppose
that $(a,b,c)$, with $abc\neq 0$ is a rational point lying on $C$. Put
\begin{equation*}
x=aT,\quad y=bT-\frac{c^3R}{b^3Q},\quad z=cT+1,
\end{equation*}
 Then, with $x, y, z$ defined above we get
\begin{equation*}
G(T)=\frac{R(b^6Q+c^6R)}{b^{30}Q^5}\sum_{i=0}^{3}C_{i}T^{i},
\end{equation*}
where
\begin{align*}
&C_{0}=b^{24} Q^4 - b^{18} c^6 Q^3 R + b^{12} c^{12} Q^2 R^2 - b^6 c^{18} Q R^3+c^{24}R^4,\\
&C_{1}=-6 b^6 c Q (-b^6 Q + c^6 R) (b^{12} Q^2 + c^{12} R^2),\\
&C_{2}=15 b^{12} c^2 Q^2 (b^{12} Q^2 - b^6 c^6 Q R + c^{12} R^2),\\
&C_{3}=-20 b^{18}c^3 Q^3 (-b^6 Q + c^6 R),\\
&C_{4}=15 b^{24}c^4Q^4.
\end{align*}
Now $G$ is of degree four in $T$ and thus if a rational number $S$ is chosen so that the quartic curve
\begin{equation*}
\mathcal{C}:\;-S w^2=G(T)
\end{equation*}
has infinitely many rational points, then there will automatically be infinitely many rational points
on the surface $Px^6+Qy^6+Rz^6+Sw^2=0$. As before, we can guarantee positive rank by taking $S=-G(t)$,
for a parameter $t$.\\
As an example, take $(a,b,c)=(1,1,1)$, $(P,Q,R)=(p,q,-p-q)$. It is straightforward to determine that
in consequence there are infinitely many points on either of the two surfaces
\begin{align*}
p x^6+q y^6 -(p+q) z^6 -15 p q(p+q)w^2 & = 0, \\
p x^6+q y^6 -(p+q) z^6 -p q(p^5+q^5) w^2 & = 0.
\end{align*}
If we specialize further to $(P,Q,R)=(1,1,-2)$, then we get the quartic curve
$-S w^2=2(31+90T+105T^2+60T^3+15T^4)$. Taking now $S=-2(31+90t+105t^2+60t^3+15t^4)$ we consider the curve
\begin{equation*}
\mathcal{C}: \; (31+90t+105t^2+60t^3+15t^4)w^2=31+90T+105T^2+60T^3+15T^4.
\end{equation*}
Taking the point $(T,w)=(t,1)$ as a point at infinity, then the point $Q(T,w)=(t,-1)$
is of infinite order, and its multiples pull back to infinitely many points on the surface.
For example, $2Q$ delivers the point, on simplifying by replacing $t$ by $-1-t$:
\begin{align*}
(x,y,z,w) = (& -1 + 3t + 60t^3 + 90t^4 + 90t^5 + 900t^6 + 675t^8 - 225t^9, \\
             & -1 - 3t - 60t^3 + 90t^4 - 90t^5 + 900t^6 + 675t^8 + 225t^9, \\
             & 3t(1 + 20t^2 + 30t^4 - 75t^8), \\
&  (-1 + 90t^4 + 900t^6 + 675t^8)(1 + 60t^2 + 420t^4 + 6300t^6 + 55350t^8 + \\
& 94500t^{10} + 94500t^{12} + 202500t^{14} + 50625t^{16}))
\end{align*}
on the surface
\[ x^6+y^6-2 z^6 = 2(1+15 t^2+15 t^4)w^2, \]
etc.

\section{Final remarks} \label{sec6}

We can try to extend the polynomial method of the last two sections in the following manner.
Seek $P,Q,R,S \in \mathbb{Q}$ and polynomials $G_{i}, F\in\mathbb{Q}[t]$, with
$\operatorname{deg}G_{i}\leq 2$ for $i=1,2,3$ and $\operatorname{deg}F\leq 4$, such that either
\begin{equation}
\label{quartsurf}
PG_{1}(t)^4+QG_{2}(t)^4+RG_{3}(t)^4+SF(t)^2=0,
\end{equation}
or
\begin{equation}
\label{sextsurf}
PG_{1}(t)^6+QG_{2}(t)^6+RG_{3}(t)^6+SF(t)^3=0.
\end{equation}
If we can find such polynomials and numbers $P,Q,R,S$, the existence
of a rational point, say $(t,w)$, on the quartic curve
\begin{equation*}
C:\;w^2=F(t),
\end{equation*}
then implies respectively the existence either of a rational point on the surface $Px^4+Qy^4+Rz^4+Sw^4=0$,
or a rational point on the surface $Px^6+Qy^6+Rz^6+Sw^6=0$, in each case with
$(x,y,z,w)=(G_{1}(t),G_{2}(t),G_{3}(t),w)$. \\ \\
The system of equations underlying (\ref{quartsurf}) is too large for complete analysis.
However, there seem to be numerous such identities, and we list just one of the simpler ones:
\[(P,Q,R,S) = (a^2+2a b+2b^2-2c, 4(a^2+2a b+2b^2 - 2c), (a^2-2a b+2b^2)c^2, -a^2+2a b-2b^2), \]
\begin{align*}
(G_1,G_2,G_3)= & (2a(2a b-c) - 4b(2a b-c)t + (4b^3+(a-2b)c)t^2, \\
& 2(a^3-(a-b)c) - 2a(2a b-c)t + b(2a b-c)t^2,  \\
& 2(2a b-c) - 2(a^2+2b^2-c)t + (2a b-c)t^2 ),
\end{align*}
with
\begin{align*}
F(t) & =4\left( 2a^4(a^2+2ab+2b^2) -2a^3(3a+2b)c + (5a^2-2ab+2b^2)c^2 - c^3 \right) - \\
& 8(2ab - c) \left( 2a^2(a^2+2ab+2b^2) - 2a(2a+b)c + c^2 \right)t + \\
& 8(2ab - c) \left( 3ab(a^2+2ab+2b^2) - (a^2+6ab+2b^2)c + c^2 \right)t^2 - \\
& 4(2ab - c) \left( 4b^2(a^2+2ab+2b^2) -2b(a+4b)c + c^2 \right)t^3 + \\
& \left( 8b^4(a^2+2ab+2b^2) -8b^3(a+3b)c + (a^2-2ab+10b^2)c^2 - c^3 \right)t^4.
\end{align*}
The generic Picard rank is 2.\\ \\
In the case of (\ref{sextsurf}), we offer the identity

\[\begin{array}{l}
2\left( (p+q)t^2+4qt+p-q \right)^6-2\left( 2qt^2+(2p-2q)t-p+3q \right)^6\\[0.1in]
\;\;-(p^2-5q^2)^3(t^2-t-1)^6-\left( (p^2+4pq-q^2)t^4+(2p^2+22q^2)t^3 \right. \\[0.1in]
\;\;-(3p^2-24pq+9q^2)t^2+
\left. (6p^2-16pq+18q^2)t+8pq-11q^2-p^2 \right)^3=0,\end{array} \]
which, on taking $p=1,\,q=1,$ yields the identity
\[ 2(t^2+2t)^6 - 2(t^2+1)^6 + (t^2-t-1)^6 = (t^4+6 t^3+3t^2+2t-1)^3. \]
For generic $t_0$, the elliptic quartic
\[ t^4+6t^3+3t^2+2t-1 = (t_0^4+6t_0^3+3t_0^2+2t_0-1)W^2 \]
with $(t,W)=(t_0,1)$ as origin, has point $Q(t,W)=(t_0,-1)$ of infinite order.
Pullbacks of $mQ$, $m=2,3,...$, now give infinitely many solutions of
\[ 2X^6-2Y^6+Z^6 = (t_0^4+6t_0^3+3t_0^2+2t_0-1)^3W^6. \]
For example, $2Q$ delivers $(X,Y,Z,W)=$
\begin{align*}
(t(2+t)(-1+2t+2t^2+ 14t^3+20t^4+32t^5+26t^6-4t^7+2t^8) \times & \\
(3+6t-6t^2-6t^3+24t^5+18t^6+12t^7+2t^8), & \\
(1+t^2)(1+8t^2-68t^4-192t^5+296t^6+1488t^7+2560t^8+2112t^9+656t^{10} & \\
-144t^{11}+292t^{12}+480t^{13}+248t^{14}+48t^{15}+4t^{16}), & \\
(1+t-t^2)(1-4t-10t^2-60t^3+50t^4+112t^5+412t^6+600t^7+1540t^8+2640t^9 & \\
+3640t^{10}+3440t^{11}+2240t^{12}+960t^{13}+320t^{14}+64t^{15}+4t^{16}), & \\
1+4t-6t^2+28t^3+126t^4+336t^5+420t^6+104t^7-732t^8-1104t^9-1176t^{10} & \\
-1008t^{11}-784t^{12}-448t^{13}-96t^{14}+4t^{16}). &
\end{align*}
Accordingly, observe that we deduce
infinitely many integer solutions of the Diophantine equation
\[ U_1^6+2 V_1^6+2 W_1^6 = U_2^6+2 V_2^6+2 W_2^6; \]
and more generally, integer solutions of the arbitrarily long chain
\[ 2 X_1^6-2 Y_1^6+Z_1^6 \; = \; 2 X_2^6-2 Y_2^6+Z_2^6 \; = \; 2 X_3^6-2 Y_3^6+Z_3^6 \; = \; \dots \]
No further solution to (\ref{sextsurf}) over $\mathbb{Q}$ was discovered. However,
the following curious identity came to light:
\begin{equation*}
125u^6 - 2(u^2+1)^6 + 2(\epsilon u^2-\epsilon^{-1})^6 = (2\epsilon u^4-3u^2-2\epsilon^{-1})^3,
\end{equation*}
where $\epsilon = (1+\sqrt{5})/2$. \\ \\

redpajama_set_name: RedPajamaArXiv


Eustress (or good stress) and bad stress (acute or chronic) cause the exact same reaction in the human body. Even during voluntary "stressful" activities such as sport or exercise or when we receive unexpected good news, the brain stem, the oldest and more primordial part of the human brain, immediately mobilizes the body's resources. The brain stem does not know, and one might say does not care, what triggers the sudden demand for additional physical activity. All the brain stem knows, prior to any higher brain intervention such as a decision to be afraid of something, or a decision to exercise, is that more blood is needed immediately to fulfill physical demands that may already be occurring (in the case of exercise or a real and impending threat) or that may be presumed to occur (in the case of perceived danger in a situation).
When the motor areas of the brain and the limbic system become activated by a positive but sudden event, most of the reticular activating system of the brain stem is also mobilized. This activation includes greatly increased stimulation of the vasoconstrictor and cardioacceleratory areas of the vasomotor center of the brain stem. Thus, the increase in arterial pressure permits to keep pace with an expected increase in muscle activity. A similar rise in pressure occurs during negatively stressful situations. The need to prepare to meet the danger posed by the stressor mobilizes the reticular activating system and the vasomotor center of the brain stem.
During dangerous situations (real or perceived), arterial pressure rises to as high as twice its normal value within a few seconds. This dramatic increase can immediately supply blood to any or all muscles of the body needed to respond. This translates into an enormously increased ability to fight against or to flee from the cause of danger. It is indeed a significant survival factor that no conscious decision is needed when this split-second mobilization is required.
Soft drinks, sweetened fruit juices, and sugar-loaded sports drinks raise blood pressure, according to a International Study of Macro/Micronutrients and Blood Pressure (INTERMAP). The researchers measured the consumption of sugar-sweetened drinks, sugars, and diet beverages (which contain high quantities of glucose and fructose) over the course of four days, administered two 24-hour urine collections and eight blood-pressure recordings, and asked questions about the patients' lifestyle and medical history. Results show that there is a direct correlation between fructose and glucose intake and increases in blood pressure and that sugar-sweetened beverages are associated with a 1.1-mm-Hg increase in systolic and 0.4-mm-Hg increase in diastolic blood pressure after adjustment for weight and height.


On December 28, 2018 call for Matching Grants Program was closed with 135 applications received for current grants competition.
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text:
Drax boosts education and skills with work experience placements
Drax is offering young people a valuable insight into the world of work at its sites in the UK, including its power station near Selby in North Yorkshire, as part of the renewable energy company's commitment to supporting education and skills.
This is the first time Drax has run its in-person work experience programme since March 2020 when the company had to stop all in person educational activity, including visits and work experience, due to Covid restrictions.
The programme gives students the opportunity to work alongside highly trained staff, observe practical tasks and ask questions to improve understanding and help them decide if a Science, Technology, Engineering and Maths (STEM) career is right for them. Students who took part could choose from a variety of business areas including engineering, IT, finance, and facilities.
Abigail Cheney, who is in Year 10 at school, spent a week at Drax Power Station shadowing engineers and getting an insight into what working for the energy company is like.
"I'm really glad I took part in the Drax work experience programme. The work I did during the week was really interesting and gave me an insight into what a career in engineering could be like. The atmosphere in the department was really nice, everyone was so friendly and made me feel included. Overall, it was an amazing opportunity which expanded my knowledge and gave me a better idea of the sector I want to work in in the future."
Danny Adams, who is in Year 12 at school, also spent a week at Drax working with colleagues in the finance department.
"My work experience week at Drax was amazing. It was a great opportunity for me to explore the area of work I'm interested in which is finance. I worked with some fantastic people who made me feel comfortable and like part of the team. I especially enjoyed the site tour. The week has been great for helping me get to know new people and socialise whilst gaining new skills on the job. I would definitely look into applying for a role at Drax in the future."
Bruce Heppenstall, Plant Director, said:
"We work closely with schools in our communities to inspire young people from all backgrounds to study STEM subjects, so the next generation has the education and skills needed to support businesses like ours as we continue to develop and grow.
"We're pleased to be able to offer in-person work placements at our sites again and hope that this will give students a valuable insight into the world of work and spark an interest in STEM subjects."
Drax welcomed 26 work experience students across its sites in the UK, including Cruachan Power Station in Scotland. At the end of the week, each student was presented with a certificate of achievement and a £100 Amazon voucher to spend on books and educational resources.
Prior to attending the five day in-person work experience, all students completed a two-day virtual work experience, which introduced them to career opportunities at Drax and gave them the opportunity to improve their employability skills, through activities such as CV coaching and interview practice.
More information about work experience at Drax is available on the website.
Drax runs a variety of initiatives to support STEM education and skills including offering power station tours and running interactive workshops for students.
Photo caption: Work experience students Abigail Cheney, Victoria Adkin, Ethan Jarrett, Laura Hull and Danny Adams in front of the biomass domes at Drax Power Station
Megan Hopgood
T: 07936 350 175
Drax first introduced virtual work experience in June 2020 when it was unable to run its usual on-site work experience programmes due to Covid restrictions and decided to continue the programme after receiving a positive response from those who took part.
Students who took part in Drax's virtual work experience programme aged 14-18 could choose from a variety of different business areas and sites to do the in-person week of their placement. The programme provides an opportunity to learn about the renewable energy company, focusing on developing employability skills and learning about their business area through practical experience and conversations with employees.
Offering work experience virtually also enables more students to participate from across England and Scotland, who may previously have found it difficult to take part, due to barriers preventing their attendance, such as geography, opportunity, and economic factors.
About Drax
Drax Group's purpose is to enable a zero carbon, lower cost energy future and in 2019 announced a world-leading ambition to be carbon negative by 2030, using bioenergy with carbon capture and storage (BECCS) technology.
Drax's around 3,000 employees operate across three principal areas of activity – electricity generation, electricity sales to business customers and compressed wood pellet production and supply to third parties. For more information visit www.drax.com
Power generation:
Drax owns and operates a portfolio of renewable electricity generation assets in England and Scotland. The assets include the UK's largest power station, based at Selby, North Yorkshire, which supplies five percent of the country's electricity needs.
Having converted Drax Power Station to use sustainable biomass instead of coal it has become the UK's biggest renewable power generator and the largest decarbonisation project in Europe. It is also where Drax is piloting the groundbreaking negative emissions technology BECCS within its CCUS (Carbon Capture Utilisation and Storage) Incubation Area.
Its pumped storage, hydro and energy from waste assets in Scotland include Cruachan Power Station – a flexible pumped storage facility within the hollowed-out mountain Ben Cruachan.
The Group also aims to build on its BECCS innovation at Drax Power Station with a target to deliver 4 million tonnes of negative CO2 emissions each year from new-build BECCS outside of the UK by 2030 and is currently developing models for North American and European markets.
Pellet production and supply:
The Group has 17 operational pellet plants and developments with nameplate production capacity of around 5 million tonnes a year.
Drax is targeting 8 million tonnes of production capacity by 2030, which will require the development of over 3 million tonnes of new biomass pellet production capacity. The pellets are produced using materials sourced from sustainably managed working forests and are supplied to third party customers in Europe and Asia for the generation of renewable power.
Drax's pellet plants supply biomass used at its own power station in North Yorkshire, England to generate flexible, renewable power for the UK's homes and businesses, and also to customers in Europe and Asia.
Drax supplies renewable electricity to UK businesses, offering a range of energy-related services including energy optimisation, as well as electric vehicle strategy and management.
To find out more go to the website www.energy.drax.com
Families enjoy festive fun at Drax's Skylark Centre
Drax supports key workers at sea during the Christmas period
Glasgow apprentices win top prizes and Drax awards
Third generation Drax technician wins hat trick of awards
Drax invests in 30 new rail wagons supporting energy security and jobs whilst cutting supply chain emissions
UK's largest power station lights up for Poppy Appeal
Drax raises £9,000 for Humber charity CatZero
Families enjoy spooktacular fun at Drax's Skylark Centre

redpajama_set_name: RedPajamaCommonCrawl


MoneyWorks was built from the ground up for the needs of business, finance and accounting professionals. It isn't bloated with wider organisational "features" like HR & CRM, that are costly, complicated and need an IT consultancy to configure. Instead you get a powerful, feature rich platform that unlocks your company's accounting data to help you run a better business.
MoneyWorks is fully supported, regularly updated and flexible enough to fit within your workflow and integrate with your existing business software.
You need your accounting software to fit in with your business processes, not the other way around.
Designed for maximum flexibility and interoperability, MoneyWorks allows you to structure your accounts to match the needs of your business.
MoneyWorks can interface seamlessly with most leading payroll, CRM or POS systems, and generic systems like FileMaker Pro, MS Office or mySQL, making it possible to streamline your workflow.
Better information helps make better business decisions. Dashboard charts show you at a glance how your business is performing, while on screen lists give you quick access to your data.
For in-depth analysis and forecasting, MoneyWorks has the most powerful reporting of any accounting system in its class. If the comprehensive included reports and powerful subsummary features are still not enough, the MoneyWorks Gold report writer makes it possible to create the exact reports you need.
You need software that won't bog down, wasting valuable staff-time (and hence money) through slow operation. MoneyWorks is fast. People who have switched from other systems tell us that the speed of MoneyWorks is phenomenal, making their operations much more efficient.
Comprehensive documentation: facilitates staff self-training.
Not tying accounting software to payroll: upgrading accounting software to keep up with payroll rules should not be necessary. So you won't get stung for an annual upgrade that you don't really need.
A fully-compatible works-alike product line: no expensive data conversion or retraining if you need to upgrade.
Cross-platform support: no repurchase if you change platforms in the future.
Effective user interface: helps users get the job done quickly and accurately.
Performance: fast, efficient software means staff time is not wasted.
No computer degree needed. The friendly interface and graphic workflows make MoneyWorks fast, easy and efficient to operate.
Since 1992, MoneyWorks has helped tens of thousands of businesses around the world.


In the year of Canada's federal "gender budget," 2018 has seen many large lenders announce funding for women entrepreneurs. This is a strong signal that gender equality in entrepreneurship, lending and investing has become a priority for the Canadian economy.
At the same time, we need to change our mindset about financing female-led businesses, or the funding imbalance will continue despite these good intentions.
WEOC Organizing Member Women's Enterprise Centre in BC just published "Money and Mindset: Borrowing is more than dollars and cents for women entrepreneurs," an article outlining the effect of unconscious bias and other key factors on lending to women business owners.
The symbols, role models and language that lenders choose in their marketing contribute to unconscious bias and can be a significant barrier to diversity.
The article goes on to list nine other factors that affect women entrepreneurs' financing and growth, among them: women are less likely to recognize the need for financing, they experience higher documentation requirements and they borrow smaller amounts than their male counterparts. The latter can be a significant barrier to growth, as it leaves women strapped when new opportunities or circumstances arise.
Recent announcements like the partnership between WEOC and BDC and the women's entrepreneurship fund by BMO show a major shift, and a step in the right direction.
WEOC calls on other lending institutions and business service providers to recognize and respond to the different needs of female-led companies.
It is important to combine greater investment in women-owned businesses with access to comprehensive support, mentorship, guidance and training resources. This illustrates the need for partnerships between lenders and organizations that provide loans and other supports specifically for women business owners.
The Women's Enterprise Initiative, of which Women's Enterprise Centre is a part, has supported women business owners with these wraparound services since 1995. A 2016 review of outcomes found that, during the period 2007-2012, WEI-assisted firms grew more than unassisted firms in terms of employment and sales, they stayed in business longer and their labour productivity was higher.
Previous studies have also shown that WEI-assisted firms attribute up to 50% of their revenue growth to the assistance provided by Women's Enterprise organizations in the WEI, and they are 50% more likely to stay in business longer than 5 years compared to the national average.
Wouldn't it be nice if these wraparound services were available to all women business owners across Canada, not just in the Western provinces?!
WEOC, the national association of organizations providing services to women entrepreneurs, has been trying to assist the non-profit organizations in each region of the country to get funded to provide the full "wrap", including loans, which supplement the work which BDC and the traditional lenders provide.
Similar services are provided on a national basis by WEOC's sister organization in the US. When will Canada invest in this part of the infrastructure for women entrepreneurs?


Because of the way the world is trending with different mobile Internet technologies today, there are many website designers out there looking for work. It's an incredibly vibrant market, primarily because the number of people and businesses who need websites designed far outweigh the actual number of designers. However, it's difficult to really get your message across.
If you're a designer, how, exactly, do you go about informing potential clients of your talents and abilities to create for them a high-end website? This is where a request for proposal for website design comes into play. Throughout the article below, we will discuss different categories of information to include in your proposal, different tips on how to write a RFP, and even tips on creating a great call to action to get interested readers to follow through with contacting you.
There are many do-it-yourself site-building template services out there, but most lack the capacity to actually construct a vibrant, feature-rich site. Not to mention that most of these services will not even allow for a custom web address, unless the individual pays a large price tag – which, if someone is paying money for, they might as well get the most out of it. Explain in any request for proposal in web design that those DIY avenues are very short, inherently complex, and are not customizable.
When pitching a RFP for web design, two incredibly important categories—perhaps the most important categories—to touch on are that of overall budgeting and time. People want to know how much it is going to cost them, and subsequently how long it will take until their project is finished. You must be adept at the art of appraisals and quotes. For instance, if a person wants a website to sell eBooks, with a built in forum for customers to start a community, it's your job to assess everything you may need to complete this site and to quote the individual an amount of money for the job and the amount of time it will take.
Make it clear in your website request for proposal that a website is the number-one way to accomplish branding online. When you think about a huge name in the market like Apple or McDonald's or Coca-Cola, you think about their overall brand just as much as their products. It's their logo, their commitment to customers, the different forms of advertising they have, etc; they brand better than anyone else. Make the point that a professional website not only implements professional and attractive design elements that will come across as professional and user-friendly, but that a website also ties in to social communications online and allows a company to brand itself.
Most people who would be so inclined to consider having a website built for them do not understand much about web hosting. A great way to gain traction with your RFP for web design is to go into some detail about the importance of a quality web host. For instance, if the individual would like to have streaming videos on their site, or build a user community in a social network-style site, they will need a specific web host offering these types of features. As the designer, you can really show your expertise by explaining the intricacies in web hosting packages, explaining what uptime means, going over security measures, and other pieces of vital information that convey the importance of a premium web host to the individual.
How much does the average person know about website functionality? As the designer, you are tasked with relaying this information in a consumable, palatable way. For example, we live in an age of mobile computing, and it may only be a decade before the vast majority of all Internet communications happen via smartphones and tablets. In your request for proposal, address the emerging mobile market and make an effort to accurately explain why today's websites need responsive template designs to work across a range of platforms, and also why various design elements need to be multi-format (e.g. Flash isn't Apple-friendly, and not all script languages work the same) in order for a website to provide the same user experience regardless of from which device a person accesses the site with.
In your RFP for web design, one of the key aspects you can touch on and really flesh out is the category of search engine optimization, or SEO. The average person does have at least a rudimentary understanding of this concept, as they understand that the keywords they search via search engine retrieves specific results. However, what they may not realize is just how important keywords and proper optimization structures are to their site's success. Explain the importance of locally registering a site, providing fresh content, finding keywords that are competitive without being saturated, and how to tag and cluster accurately for search engine spiders to actually grab onto the website rather than reading it as spam and pushing it to the back of the pack. Don't worry about coming across as too technical sounding; it's okay if the proposal sounds complex in its explaining of things like SEO—and other site design elements—because this will drive the point home that the do-it-yourself route is perhaps not the best way to go for the individual.
Another way you can work to garner interest in your request for proposal is to address the issue of site maintenance. Think about selling someone a new car from a lot. Sure, you can pitch the car as a great piece of machinery and get them interested in driving it home. But what guarantees are you offering if the car happens to break down in a month? You will not have to offer a similar type of warranty per se, but you should make it clear that you are willing to perform basic site maintenance, and maybe even act as a web master (for an additional fee) so that the site runs smoothly and without issue.
Do not pressure anyone into a click or action; use a basic call to action at the end (example below) by piquing interest and curiosity.
Again, edit, edit, edit! Why would someone trust you to build a website if you can't even spell?
Have you ever seen one of those weight-loss programs online that have the incredibly long single page of text? You keep scrolling and scrolling looking for a price, but the sales pitch never ends! Every other line, you'll find words in bold, caps lock, red font, with a dozen exclamation points: "YOU NEED THIS PRODUCT!!!!" OK, simmer down! These types of proposal pages are incredibly common with weight-loss products, Forex, and a wide range of work-from-home scams. The problem—well, one among many—is that they seek to sell too hard!
If you write a proposal using the information provided here thus far, you don't have to scream at people. As long as you're addressing a problem, offering a solution, keeping things concise, palatable and readable, you will earn someone's attention if they want a website built. Then all you need to do is close with a semi-strong call to action.
Example: "If you would like to know more about how you can have a professional website built for you today, click here!" Any subtle variation of that, whether you want to change the site pitch or emphasize a particular company, is golden if the reader is already interested.
Keep things clean, tight, and professional, and those so inclined to have a website built will seek you out for more information. This is where the actual sales pitch will come out.


Club SciKidz Houston – Here the kids have the chance to reach their potential and have fun doing it. Join us at Club SciKidz Houston.
Here the kids have the chance to reach their potential and have fun doing it. Join us at Club SciKidz Houston.


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Landmarks Tours
London Landmarks Tour – West
London Landmarks Tour – East
Free London Landmarks Tour – South
Free London in a Day Tour
Free British Museum Tour
Free Harry Potter Tour*
Free Street Art & Graffiti Tour*
Free Soho Walking Tour*
Free London Food Tour*
Free Historical Pub Tour
Night Time Tours
Free Jack the Ripper Tour*
Free Ghost Tour*
London Pub Crawl
Shoreditch Pub Crawl
Camden Pub Crawl
The Ultimate Guide to visiting parks in London
London Nature & Parks
EXPLORE LONDON NATURE & PARKS WITH STRAWBERRY TOURS
It's no secret that London is a city known for its several parks, some of which are arguably the most famous and well-known parks in the world.
But if you think that all that there is to do in a park in London is to stroll around and snap photos of birds and flowers, then you are sorely mistaken. As a matter of fact, with so many different activities and events happening in London's parks throughout the year, one could easily spend an entire day in a London park without even venturing out of it. From landmarks, lakes, palaces and museums, to even ice rinks, zoos and open air theatres, there's no telling what you'll come across during a visit to a London park.
Where to find parks in London
The majority of the most popular parks in London tend to be centered around Westminster and Soho, such as Hyde Park, Green Park and St. James's Park; but the parks in London don't just stop there.
If you want to visit more than one park in one day, you could easily stroll around Green Park, Hyde Park, Holland Park and St. James's Park in seven to eight hours given that they are in close proximity to each other. The Regent's Park is also relatively close, although it's about a 20 to 25 minute walk from Hyde Park.
Other London parks may be trickier to get to because of their location, so you will probably need to take a tube, bus or car in order to visit them. Battersea Park, for example, may be located south of the River Thames, but it's also in the same area as other popular parks in London like Clapham Common and Greenwich Park.
Hampstead Heath, on the other hand, is about a 45 minute walk from Regent's Park so you will probably need to take pubic transportation in order to get to it, while Victoria Park is located in Stratford, northeast of Whitechapel.
What to do in parks in London
If you're hoping to do some sightseeing around a famous London park, or even have a quiet afternoon sitting on a deck chair reading a book, the sky is the limit when it comes to all the different activities you can do in London's many parks.
Regent's Park has a lake so you can hire a boat or pedalo, and it's also home to the ZSL London Zoo and the nearby Primrose Hill. Holland Park boasts a tennis court, golf practice nets, a netball and a cricket pitch, and even has an open air theatre during the summer months. The park also has a giant outdoor chess set, a merry-go-round, a zip wire, tire swing, and even a separate fenced-in area for young children.
But out of all the parks in London, none offer more activities than Hyde Park, which is hands-down the most visited park in London. Here you will find not only a ton of historic monuments (like the Marble Arch, Speaker's Corner, and Kensington Palace, just to name a few), there's also a massive 40-acre lake called The Serpentine where tourists can hire a boat or pedalo, or even feed the local ducks and wildlife. Visitors can also take a stroll along the seven-mile long Diana Princess of Wales Memorial Walk, dip their feet in the Diana, Princess of Wales Memorial Fountain, or even play a friendly game of football at The Hub, Hyde Park's main sports facility.
And let's not forget about Hyde Park's magical Winter Wonderland, during which the park is transformed into an ever-popular Christmas attraction which includes a skating rink, roller coasters, an ice bar, Bavarian-style food stalls, and much, much more.
When to visit Parks in London
Although you can visit any park in London any time of the year, some parks in London are perhaps best visited during specific seasons.
During the summer months, many parks in London sell deckchairs so you can sit and read a book or even just enjoy the tranquillity of being surrounded by nature. During spring you could stroll around and admire the picturesque trees, flowers and gardens as they come into bloom, or transform into a gorgeous golden paradise during the autumn months.
If you're visiting London during the Christmas holidays, Hyde Park is an obvious one as its annual Winter Wonderland event draws in thousands of tourists each year, while Holland Park and St. James's Parkare especially beautiful during the spring and autumn months. If you're visiting London during the summer months, watching an outdoor performance in The Regent's Park or Holland Park is a must.
Also be aware that all the parks in London have different opening and closing times depending on the season. During the summer months (between May and July) some parks may stay open as late as 9:30 p.m., while during the winter months (between November and February) some parks could close as early as 4:30 p.m. Holland Park is open from dawn until dusk every day throughout the year, while Hyde Park is open between the hours of 5 a.m. and midnight every day, all-year round.
If you're in London over the Christmas holidays, Winter Wonderland usually opens around the end of November until the beginning of January each year, from 10 a.m. to 10 p.m.
But regardless of when you plan on visiting a park in London, it won't take you long to discover that London's parks are beautiful any time of the year, come rain or come shine.
Email: london@strawberrytours.com


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If you understand this script then it says a lot about the year you were born
In the days before any kind of recording equipment, a very useful skill was being able to record what people were saying by hand.
This was difficult to do when most people speak at least twice the speed we write
But those that had this ability would use a set of symbols that would look completely alien to youngsters today. However, this skill was in demand and could land you a job.
The writing was a combination of dots, dashes, loops and swirls and for some people of a certain age would look all too familiar.
Today it's all too easy to record what people are saying in lectures or meetings; there are dozens of apps and recording devices so we can refer to what people have said.
But before technology people used a system called shorthand and for those that mastered it meant writing at a a speed of 280 words per minute.
Your average person speaks at a rate of 125 to 150 words per minute but most of us write at a much slower rate than this.
The most common shorthand system in the UK was Pitman shorthand developed by Englishman Sir Isaac Pitman and first presented in 1837.
Gregg shorthand was the most popular in the U.S. invented by John Robert Gregg in 1888 and was thought to be a lot easier to learn than Pitman.
Both methods involve symbols for the sounds of the words more than the letters. There are even symbols for words such as it, the, to and for so they could be written quickly.
Despite digital technology shorthand is still used for quick notetaking especially in the UK where recording equipment is not allowed in most law courts.
Teeline is the more common form of shorthand today used by the National Council of Journalists for its training program in the UK.
There is also a fun element to shorthand if you know it and those around you don't as the possibilities of writing secret notes in a journal or personal notebook are endless.
Find out how simple it is in the clip below:
Please share this story with all those people you think would recognize these weird and wonderful symbols.
The Highest Bill You've Probably Held Is A $100… But There Are Lots More And They're Totally Real.
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Country Girl Stuns The Judges With Her Gospel Performance
Crocodile And Snake Battle In The Wild
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Fascinating! I Had NO Idea Elephants Could Do This!


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Meet J.L. Delozier, Runner Up in Spring 2016 Flash Fiction Contest
Dr. Jennifer ("J.L.") Delozier submitted her first short story, handwritten in pencil on lined school paper, to Isaac Asimov's magazine while still in junior high school. Several years later, she took a creative writing elective at Penn State University and was hooked. She received her BS and MD degrees in a compact six years, which was followed by the blur of internship, residency, and the launch of her medical career. But she never forgot her first love. When she decided to delve into longer fiction, she spent some time rediscovering her favorite physician writers. From the deductive reasoning of Sir Arthur Conan Doyle to the cutting edge science of Michael Crichton, she remains inspired by facts that lie on the edge of reality: bizarre medical anomalies, new genetic discoveries and so on. These are the foundation of her debut thriller, Type and Cross. Published April, 2016, it is available in print and on Kindle anywhere books are sold.
Dr. Delozier lives in Pennsylvania with her husband and four rescue cats. You can find her on Goodreads, her website www.jldelozier.com or on social media: Twitter.com/jldelozier and Facebook.com/jldeloziermd.
Read Jennifer's winning entry and then come back here to learn more about where she gets her ideas and tips for attending a writing conference!
WOW: "Justin's Room" is a chilling tale with a twist ending. How did you get the idea for this particular story?
Jennifer: All my ideas tend to arise from something I overhear and can't get out of my head. Justin's Room started when I overhead two people discussing a recently released book - I have no idea what book - which opens with a close-up view of someone sitting at a table calmly eating breakfast. As the scene develops, the reader becomes aware that this character is not in his own house and has, in fact, murdered all the occupants. I immediately thought to myself, "I need to use something like that." It's such a powerful image, I knew it would have even more impact in a shorter work, without a bunch of other concepts to dilute the punch in the gut. After all, "A novel is a war, but a short story is a landmine." That image just blows your legs off.
WOW: I love that analogy! Perfect. You mentioned you were a fan of Michael Crichton. What's your favorite novel by him and why?
Jennifer: The toughest question BY FAR. You may as well have asked me to pick my favorite pet. I can narrow it down to three. The Andromeda Strain, Sphere, and Jurassic Park. Why those? Hard for me to say. I thought the first was a brilliant debut. For the second, I loved the combo of sci-fi, mysticism and psychological thriller. And as for Jurassic Park, it had that magical quality, like the first Harry Potter. It took a dry scientific concept, explained it easily (cloning recovered DNA was a fresh idea then) and took it on a wild ride, all the way from "wow, gee wiz" to "holy shit, we're in trouble." I love the idea of taking real science and bending it to the extremes.
WOW: Having attended ThrillerFest this past summer, can you give our readers some tips on how to get the most out of writing conferences?
Jennifer: ThrillerFest was my first conference, so I'm hardly the one to be giving advice here. My best attempt - plan your schedule in advance. You can't attend every lecture, and many of the conference rooms fill up fast, so make sure you know where you want to go and what time you need to be there. On a lighter note, consider dressing better than I did. No ball gowns or anything, but it's nice to be prepared for possible photo ops with your heroes. I'm used to medical conferences, where I'd roll out of bed and stumble to the conference room in sweats. I learned after day one I wouldn't want to see Twitter posts of me and, say Lee Child, in which he looks like the dashing British gent he is, and I look like a frump.
WOW: You work as a federal physician. Can you tell us a bit about what sounds like a fascinating field?
Jennifer: When I first read this question, I thought it said a "feral" physician, which would have been much more interesting! I take care of Veterans at our local VA clinic, but I also participate in DEMPS - the disaster emergency medical personnel system, whereby I get deployed to federal medical shelters during large enough events ("large enough" means the President, not just the state governor, has declared a state of emergency and the local/state resources can't handle the scope of the situation). If you want to know what it's like, I'd encourage you to read my second book, Storm Shelter, which, though fiction, is loosely based on my previous deployments. Unfortunately, you'll have to wait about a year - I just turned the manuscript into my editor a few weeks ago!
WOW: We'd love to hear about your novel, TYPE AND CROSS, which was published this past spring, as well as the upcoming STORM SHELTER. Could you give us a synopsis and tell us a bit about your path to publication?
Jennifer: Here's a blurb for Type and Cross. I wanted it to appear on the cover, but alas - it didn't. I think my editor found it too cheesy!
A homemade virus with deadly global implications.
A disgruntled scientist with a coded plan.
And a nightmare-plagued, tequila-swilling psychologist-for-hire tasked with finding him before he succeeds.
Dr. Persephone Smith is gifted with enhanced empathy—the ability to get inside the heads of those criminals too twisted for other psychologists to unravel. But her success comes with a steep price. Scarred from too many descents into the dark minds of her subjects, Seph has lost her faith in God and Man. She lives only for her work—and for her sister, Grace.
When a bioterrorist triggers a pandemic, the government assigns Seph and a crack team of scientists to hunt him down and force him to divulge the cure. Dr. William Baine turns out to be easy to find but hard to hold, and Seph soon discovers he has a genetic secret of his own—a secret which holds the key to both his coded journal and the formula for the cure. His special abilities are a perfect foil to her own, and, as their psychological bond deepens, he leads her on a cat-and-mouse chase around the world, from the smoky tobacco shops of Bermuda to the underground tunnels of Geneva. There, with the pandemic threatening to exterminate most of humanity, Seph must confront her fears and use her empathetic connection to a killer to save the world, her sister and, ultimately, herself.
What's your type? Your survival depends on it.
Storm Shelter is actually a prequel and is set ten years prior to Type and Cross, when Seph is still a green clinical psychologist taking care of patients at the Philadelphia VA. Here's an excerpt from my query: "When Seph is deployed to an abandoned air hangar turned federal medical shelter during a massive hurricane, her worst nightmares come true. One by one, as the wind howls overhead, staff and evacuees disappear into the dark recesses of the vast space. The missing return as mutilated corpses. The living, trapped in the shrieking metal structure by the storm, descend into varying levels of paranoia and even madness. Seph must become both counselor and detective to determine who, or what, is calling them prey. Is the panic and mayhem just "shelter shock," as the head physician, Anne Parrish, insists? Or is everyone, Seph included, in danger of losing their minds—and their lives?" It has a lot more horror in it than Type and Cross did. It also has a limited, claustrophobic setting, as opposed to the global scope of Type and Cross. It's more Stephen King than Michael Crichton.
As far as my path to publication, I've floundered around like a true newbie and done everything "wrong" according to the rules, but so far, it's worked out ok! I had no social media platform to speak of until I started querying for Type and Cross. I built them as I was awaiting responses to my queries. My first mistake: I queried both agents and four or five small presses simultaneously. I later learned this is considered poor practice, but I had no idea of that at the time. I got several full requests from both agents and two of the small presses. While all the agents ultimately passed, both small presses offered me a contract. Then I had the difficult situation of choosing! I ultimately went with WiDo for many reasons, but the primary one is that they've been in existence for almost a decade, which is an eternity for a small press (since many of them fold so quickly.) They have the right of first refusal on Storm Shelter, and I'm hoping they'll like it enough to publish it.
I still desperately want an agent to be my partner on what I hope will be a long career, but my current plan is to write a third and final Persephone Smith book and then branch out with something new (I have a list of 28 undeveloped ideas) at which point I'll query again. For now, I'm taking a bit of a break and working on more short stories. I've submitted two and am awaiting responses (Alfred Hitchcock and Ellery Queen magazines - fingers crossed!) I also expanded Justin's Room to a longer length more appropriate for a print magazine. Wish me luck!
WOW: Best of luck, of course! You're determination and enthusiasm to the craft are a great inspiration for us all.
Labels: 2016 Spring Flash Fiction Contest, Contestant interview, J.L. Dozier, Renee Roberson, Runner up, Type and Cross
J.L., wow, thanks for sharing the inspiration behind Justin's Room! That is chilling. I love the quote, "A novel is a war, but a short story is a landmine."
Thanks for also sharing your path to publication. WiDo is a great publisher! I'm looking forward to Storm Shelter. :)
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TownePlace Suites by Marriott Miami Homestead is rated 4.6 out of 5 by 165.
Rated 5 out of 5 by LisaL from Weekend stay This stay came as a surprise. Everything was exceptionally maintained. Extremely quiet which is one issue we find at many hotels. We would definitely stay here again.
Rated 5 out of 5 by MDoty from One of the best recent hotel stays Everything about this hotel suited us perfectly. The one bedroom suite was large, attractive, and very comfortable. The bed was amazingly comfortable. Everything was spotlessly clean and the kitchen was well stocked. I really liked the separate thermostats in each room, and the two tv's. The breakfast was very nice and I really appreciated the plain oatmeal, which I don't usually find in hotel breakfasts. Staff was very helpful and cheerful.
Rated 5 out of 5 by PearlieMills from awesome I loved everything about my stay, the room was really up to date and very outstanding, fantastic breakfast and great pool. location is great.
Rated 4 out of 5 by Rut from Homestead Speedway exercise 2wk stay for a military exercise at the Homestead racetrack. Hot breakfast plus ample space in the lobby area made for a great start to the morning. Close to some good food joints and only 10min journey to the track.
Rated 4 out of 5 by MR from They enter your room without consent bypassing DND tag due to 3 day policy Ok so, I usually try to keep a tidy room and use my towels sparingly. Changing them out with the maids as I see fit. Most Marriott properties that I have stayed allow you to not take room cleaning service in exchange for points. So this brings me to my experience with this hotel. After 3 consecutive days with my do not disturb sign on the door I came back to my room to find a letter on my counter that management had entered my room due to policy and cleaned it without my consent. I do not with to have anyone in my room without me either being there or leaving the do not disturb sign off the door expecting the room will be cleaned.. This did not sit well with me and felt like an invasion of my privacy.


Hotel El Viandante is just outside downtown Santa Elena. Surrounded by wonderful nearby places to explore and visit. It is also very conveniently located at just a 15-minutes drive from the famous Monteverde Cloud Forest Reserve.
Hotel El Viandante is a clean and cheerful, bright and friendly place, where you will find the staff helpful and fun!.
Owners-operators, Renzo and Grace, take great pride in making sure that their guests are comfortable and feel welcome and at home at El Viandante.
One of the unique on-site services that El Viandante offers its guests is Mountain Bike Rentals, Bike Tours and Guides for those with an adventuresome spirit.
Renzo and Grace share a passion ...for each other, their little son, Paolo, and for the joy that comes from meeting the wonderful variety of guests who come to stay with them at Hotel El Viandante.
They are both well traveled and adventuresome, so they find meeting their guests and sharing stories an exhilarating experience.
Both Renzo and Grace longed for new adventures, the peacefulness of nature, its diverse beauty, and each was drawn to Costa Rica where they met. Combining their dreams, Renzo and Grace have created a special and unique destination for their guests to enjoy, and they are pleased to welcome you to their Hotel El Viandante.
Children 0 - 3 years old free.


Why Do Homebuyers Prefer a Fixed-Rate Mortgage?
A: Long-term, fixed-rate mortgages are preferred by most homebuyers because they offer security and stability. The interest rate does not fluctuate over the life of the loan, so the total amount of principal and interest always remains the same. The monthly payment can change, however, if local property taxes, which are normally part of the monthly mortgage payment, increase.