arithmetic progression of triangle sides Let $gcd(a,b,c)=1$ such that $a^2, b^2, c^2$ are in arithmetic progression.  Show they can be written in the form
$a=-p^2+2pq+q^2$
$b=p^2+q^2$
$c=p^2+2pq-q^2$
for relatively prime integers $p,q$ of different parities.
so if $a^2, b^2, c^2$ are in arithmetic progression, then 
$a^2=x, b^2=x+d, c^2=x+2d$, for $x,d\in\mathbb{Z}$
That means $x+x+d=x+d+d \Rightarrow 2x+d=x+2d \Rightarrow x=d$.
So then $a^2=x, b^2=2x, c^2=3x$
Now with primitive pythagorean triples, we have $(2pq, p^2-q^2, p^2+q^2)$.  I just don't see how to go from here....i'm sure it's easy, but I'm not seeing something.

$a^2, b^2,$ and $c^2$ are in AP if and only if $a^2+c^2=2b^2$.
Looking at $a,b,$ and $c$ modulo $4$, we see that they must all be odd integers. 
Then $u=\frac 12(c-a)$ and $w=\frac 12(c+a)$ are also integers. 
In fact $a=w-u$ and $c=w+u$. 
So $2b^2=a^2+c^2=(w-u)^2+(w+u)^2=2(u^2+w^2)$.
Hence $u^2+w^2 = b^2$.
So there exists coprime $p$ and $q$ such that 
If $2pq < q^2-p^2$, then
\begin{align}
   u &= 2pq \\
   w &= q^2-p^2 \\
   b &= q^2+p^2 \\
\hline
   a &= q^2-p^2-2pq \\
   c &= q^2-p^2+2pq \\
   b &= q^2+p^2 \\
\end{align}
If $2pq > q^2-p^2$, then
\begin{align}
   u &= q^2-p^2 \\
   w &= 2pq \\
   b &= q^2+p^2 \\
\hline
   a &= 2pq-(q^2-p^2) \\
   c &= 2pq+(q^2-p^2) \\
   b &= q^2+p^2 \\
\end{align}
