Find two arithmetic progressions of three square numbers I want to know if it is possible to find two arithmetic progressions of three square numbers, with the same common difference:
\begin{align}
\ & a^2 +r = b^2 \\
  & b^2 +r = c^2  \\ 
& a^2 +c^2 = 2\,b^2  \\ 
\end{align}
and
\begin{align}
\ & d^2 +r = e^2 \\
  & e^2 +r = f^2  \\ 
& d^2 +f^2 = 2e^2  \\ 
\end{align}
where $a,b,c,d,r \in \Bbb N$.
Here is an example that almost works:
\begin{align}
\ & 23^2 +41496 = 205^2 \\
  & 205^2 + 41496 = 289^2  \\ 
& 23^2 +289^2 = 2\,(205)^2  \\ 
\end{align}
and
\begin{align}
\ & 373^2 + 41496 = 425^2 \\
  & 425^2 + 41496 = \color{#C00000}{222121}  \\ 
& 23^2 + \color{#C00000}{222121} = 2\,(205)^2  \\ 
\end{align}
where the difference is $41496$, but the last element isn't a square number.
I can't find an example of two progressions with three numbers and the same common difference. Could you demonstrate that such progressions are nonexistent using reductio ad absurdum to this statement?

There are infinitely many solutions to the system,
\begin{align}
\ & a^2 +r_1 = b^2 \\
  & b^2 +r_1 = c^2  \\ 
& a^2 +c^2 = 2b^2  \\ 
\hline
\ & d^2 +r_2 = e^2 \\
  & e^2 +r_2 = f^2  \\ 
& d^2 +f^2 = 2e^2  \\ 
\end{align}
with $\color{blue}{r_1=r_2}$. Eliminating $r_1$ between the first two equations (and similarly for $r_2$), one must solve the Pythagorean-like,
$$a^2+c^2=2b^2\\ d^2+f^2=2e^2$$
which has solution,
$$a,b,c = p^2 - 2q^2,\; p^2 + 2p q + 2q^2,\; p^2 + 4p q + 2q^2\\ d,e,f = r^2 - 2s^2,\; r^2 + 2r s + 2s^2,\; r^2 + 4r s + 2s^2$$
Hence,
$$r_1 = -a^2+b^2 = -b^2+c^2 = 4 p q (p + q) (p + 2 q)\\ r_2 = -d^2+e^2 = -e^2+f^2 = 4 r s (r + s) (r + 2 s)$$
Thus one must solve,
$$p q (p + q) (p + 2 q) =  r s (r + s) (r + 2 s)$$
This is essentially the same equation in this post, hence one solution (among many) is,
$$p,\;q = 2 n (m + 6 n),\; m (m + 4 n)\\
\;r,\;s = m (m + 2 n),\; 4 n (m + 3 n)$$
For example, let $m,n = 1,1$, then,
$$a,b,c = 146, 386, 526\\  d,e,f = 503, 617, 713\\ r_1=r_2= 127680$$
and so on.
