In my previous question, link(Existence of Natural Numbers Satisfying Given Arithmetic Progression and Product Conditions) I asked about a specific type of AP and thanks to @ Misha Lavrov and @Simon Goater, I was able to find solutions.
After that I was interested in finding two AP of perfect squares, AP1 (a$^2$,b$^2$,c$^2$) and AP2(A$^2$,B$^2$,C$^2$) where AP1 and AP2 are not equal and both AP have the same common difference.
My question is: Can I find such two AP1 and AP2 satisfying the condition that $\sqrt{\frac{b^2+B^2}{2}}$ is an integers.
Do there exist distinct arithmetic progressions of perfect squares satisfying the above conditions? If so, how can such sequences be systematically found?
My Progress till now:
Theorem: If $a^2, b^2, c^2$ are in A.P. Then there exist m,n belongs to natural number such that $b^2 = m^2 + n^2 , a^2 = (m - n)^2$ and $c^2 = (m+n)^2$ with common difference $2mn$ and the converse is also true.
Since, we know $a^2, b^2, c^2$ and $A^2, B^2, C^2$ are in A.P and they have the same common difference.
Therefore, $b^2 = m^2 + n^2 , a^2 = (m - n)^2$ and $c^2 = (m+n)^2$ with common difference $2mn$ and $B^2 = r^2 + s^2 , A^2 = (r - s)^2$ and $c^2 = (r+s)^2$ with common difference $2rs$ and $2mn = 2rs$. This is same as finding two right angle triangle with same area L. Therefore, $n = 2L/m$ and $s = 2L/r$.
Such that $b^2 = m^2 + (2L/m)^2$ and $B^2 = r^2 + (2L/r)^2$.
Therefore, $b^2 + B^2 = m^2 + (2L/m)^2 + r^2 + (2L/r)^2 = (m^2 + r^2) + (2L)^2(r^2 + m^2)/(rm)^2 = (m^2 + r^2)(1 + 4L^2/(rm)^2) = (m^2 + r^2)((rm)^2 + 4L^2)/(rm)^2$
Now if we can show that $\sqrt{\frac{(m^2 + r^2)((rm)^2 + 4L^2)}{2(rm)^2}}$ is an integer or not. Then we are done.
Or if we put $z = L/rm$. Then we have to show it for $\sqrt{\frac{(m^2 + r^2)((1 + 4z^2)}{2}}$.
Notice $(1+4z^2)$ can never be a perfect square as its between two consecutive perfect square and is also odd. Therefore the only way $\frac{(m^2 + r^2)((1 + 4z^2)}{2}$ is a perfect square if $(m^2 + r^2) = 2^{2k+1}(1+4z^2)(d^2)$ where d is the odd terms in factorization of $(m^2 + r^2)$.
Therefore, $\frac{(m^2 + r^2)((1 + 4z^2)}{2} = 2^{2k}(1+4z^2)^2(d^2)$ or $(m_{1}^2 + r_{1}^2) = 2(1+4z^2)(d^2)$, where $m = 2^{k}m_{1}$ and $r = 2^{k}r_{1}$
Let $N = \sqrt{\frac{b^2+B^2}{2}} = 2^{k}(1+4z^2)(d)$.
Notice $N^2 = \frac{b^2+B^2}{2}$ implies $b^2, N^2, B^2 $ are also in A.P. using the above theorem and Pythagoras parametrization we get there exist x, y(natural number) such that $N^2 = (x^2 + y^2)^2$.
Therefore, $(2^{k}d)(1+4z^2) = x^2 + y^2$.
After that I am not able to proceed further.
