Find all natural solutions $(a,b,c) $ such that $a^2−b, b^2−c, $ and $c^2−a $ are all perfect squares

Question

Find all natural solutions $(a,b,c)$ such that

$a^2-b$,

$b^2-c$,

$c^2-a$

are all perfect squares.

Answer 1

Say $a>b$

Since $b<a\leq 2a-1$ we have: $$(a-1)^2<\underbrace{a^2-b}_{=x^2}<a^2\implies a-1 <x<a$$ which is impossible.

So $a\leq b$. With the same procedure we see that $b\leq c$ and $c\leq a$, so $a=b=c$

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So we have $$(a-1)^2\leq a^2-a =x^2<a^2$$ and thus $x=a-1$ so $a^2-2a+1= a^2-a \implies a=1$