So $b^2=a(a^2+4)$. If $d=(a,a^2+4)$, then $a=dt^2$ for some integer $t$. Clearly, $d \in \{1,2,4\}$. So it suffices to exclude the cases $d=1$ and $d=4$. But if this is the case, then $a$ and $a^2+4=c^2$ are both perfect squares. So we have $c^2-a^2=4$ which is clearly not possible. Hence the claim.

Can someone pls check if this is correct: $$b=\sqrt {a (a^2+4)} \implies a|4$$Hence, $ a=1,2,4$ from which $1,4$ don't work Hence, $a=2$ and which proves the statement $a=2t^2$

AlastorMoody wrote: Can someone pls check if this is correct: $$b=\sqrt {a (a^2+4)} \implies a|4$$Hence, $ a=1,2,4$ from which $1,4$ don't work Hence, $a=2$ and which proves the statement $a=2t^2$ How does $b=\sqrt {a (a^2+4)}$ imply $a|4$?

enthusiast101 wrote: AlastorMoody wrote: Can someone pls check if this is correct: $$b=\sqrt {a (a^2+4)} \implies a|4$$Hence, $ a=1,2,4$ from which $1,4$ don't work Hence, $a=2$ and which proves the statement $a=2t^2$ How does $b=\sqrt {a (a^2+4)}$ imply $a|4$? Since, $b $ is an integer, $a (a^2+4) $ is a perfect square $ \implies (a^2+4)=ka \implies a^2+4 \equiv 4 \equiv 0 (\text {mod a }) $

AlastorMoody wrote: $a (a^2+4) $ is a perfect square $ \implies (a^2+4)=ka $ How do you deduce this?