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?
$$(a,b,c,d,e,f,r)=(1,29,41,23,37,47,840)$$ satisfies $$a^2 +r = b^2,\quad b^2 +r = c^2,\quad a^2 +c^2 = 2b^2$$ $$d^2 +r = e^2,\quad e^2 +r = f^2,\quad d^2 +f^2 = 2e^2$$