<s>[INST] 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?
 [/INST] 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.
</s>