In a golden pair, the sum of the outer terms is $a$ greater than the sum of the inner terms, the sum of the squares of the outer terms is $b$ greater than the sum of the squares of the inner terms, and finally, the sum of the fourth powers of the outer terms is $c$ greater than the sum of the fourth powers of the inner terms. Calculate the terms of the golden pair. - Let finally $a=7, b=21, c=2625$.
I. Let the questioned ratio pair be

$$
x: y=z: v, \quad \text { that is } \quad(1) x v=y z
$$

thus, according to the data, the following additional equations hold:

$$
\begin{aligned}
x+v & =y+z+a \\
x^{2}+v^{2} & =y^{2}+z^{2}+b \\
x^{4}+v^{4} & =y^{4}+z^{4}+c
\end{aligned}
$$

Square equation (2) and subtract from it twice equation (1) and equation (3)

$$
0=2 a(y+z)+a^{2}-b
$$

Square equation (3) and subtract from it twice the square of equation (1) and equation (4)

$$
0=2 b\left(y^{2}+z^{2}\right)+b^{2}-c
$$

Thus, we have obtained a solvable system of equations with two unknowns for the two internal terms. Denoting the right-hand sides with a single letter each

$$
\begin{aligned}
y+z & =\frac{b-a^{2}}{2 a}=d \\
y^{2}+z^{2} & =\frac{c-b^{2}}{2 b}=e
\end{aligned}
$$

assuming that $a \neq 0$, and $b \neq 0$, for now we focus on this case. Subtract (6) from the square of (5); by division

$$
y z=\frac{d^{2}-e}{2}=f
$$

Based on (5) and (7), $y$ and $z$ are the roots of the following equation:

$$
\begin{gathered}
u^{2}-d u+f=0, \quad \text { that is } \\
y, z=u_{1,2}=\frac{1}{2}\left(d \pm \sqrt{d^{2}-4 f}\right)
\end{gathered}
$$

Using the above, $x$ and $v$ can be determined as follows:

Based on (5) and (2)

$$
x+v=d+a=\frac{b+a^{2}}{2 a}=g
$$

Furthermore, based on (1) and (7)

$$
x v=f
$$

Thus, similarly to (8)

$$
x, v=\frac{1}{2}\left(g \pm \sqrt{g^{2}-4 f}\right)
$$

The result (8) means that one root of the equation concerning $u$ is taken as $y$, and the other as $z$. The selection can be done in two ways, but we do not get essentially different solutions by swapping $y$ and $z$, since our procedure would only apply the known interchangeability of the ratio pair. The same applies to (9); the selection of $x$ and $v$ is independent of the selection of $y$ and $z$, so generally there are 4 non-essentially different solutions to the system of equations (1)-(4).

Expressing the two discriminants in terms of the original parameters:

$$
\begin{aligned}
& D_{1}=d^{2}-4 f=2 e-d^{2}=\frac{c}{b}-b-\left(\frac{b-a^{2}}{2 a}\right)^{2} \\
& D_{2}=g^{2}-4 f=(d+a)^{2}-4 f=2 e-d^{2}+b=D_{1}+b
\end{aligned}
$$

thus

$$
\begin{aligned}
& y, z=\frac{1}{2}\left(\frac{b-a^{2}}{2 a} \pm \sqrt{\frac{c}{b}-b-\left(\frac{b-a^{2}}{2 a}\right)^{2}}\right) \\
& x, v=\frac{1}{2}\left(\frac{b+a^{2}}{2 a} \pm \sqrt{\frac{c}{b}-\left(\frac{b-a^{2}}{2 a}\right)^{2}}\right)
\end{aligned}
$$

$y$ and $z$, and $x$ and $v$ are real if $D_{1} \geqslant 0$, and $D_{2} \geqslant 0$. If $D_{l}$, or $D_{2} 0$ is equal, and the other is positive, then the two internal, or the two external terms are equal and form a geometric mean between the other two terms. $D_{1}=D_{2}=0$ according to the above would only be possible if $b=0$, which we excluded in our current calculation.

In the excluded case $a=0$, it follows from (1) and (2) that the terms of the $x, v$ pair are equal to $y$ and $z$ in some order, so (3) and (4) only hold if $b=0$, and $c=0$, the parameters are not independent of each other.

In the case $b=0$, similarly, it follows that $c=0$ (here we use the square of (1)), but adding twice (1) to (3)

$$
(x+v)^{2}=(y+z)^{2}, \quad x+v= \pm(y+z)
$$

thus either $a=0$, or $a=2(x+v)$. (Conversely, from $c=0$ similarly follows $b=0$, the other case - i.e., $x^{2}+v^{2}=-\left(y^{2}+z^{2}\right)$ is impossible.)

II. In the case of the parameter triplet $a=7, b=21, c=2625$, $d=-2, e=52, f=-24, g=5, D_{1}=10^{2}, D_{2}=11^{2}$, thus one of $y$ and $z$ is -6, the other is 4, and one of $x$ and $v$ is -3, the other is 8. One of the 4 ratio pairs is:

$$
(-3):(-6)=4: 8
$$

Mihály Szalay (Budapest, Vörösmarty M. st., III. grade)