I am currently investigating the specific square number $a^n+1$ and whether it can become a square. I know that $a^n+1$ cannot be a square if n is even because then I can write n=2x, and so $(a^n)^2$+1 is always smaller than $(a^n+1)^2$.
But what about odd powers of n? Can they allow $a^n+1$ to become a square? Or a more general case, can $a^n+1$ ever be a square number?
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Can $a^n+1$ ever be a square? PARTIAL ANSWER
If $a^n+1=m^2$ then $a^n=(m^2-1)=(m+1)(m-1)$. If $a$ is odd, then both $(m+1)$ and $(m-1)$ are odd, and moreover, two sequential odd numbers have no factors in common. Thus $(m+1)=r^n$ and $(m-1)=s^n$, $\gcd{r,s}=1$, because the product of $(m+1)$ and $(m-1)$ must be an $n^{th}$ power. $r^n-s^n=2$ has no integer solutions for $n>1$, so $a^n+1\neq m^2$ if $a$ is odd.
To answer the question in the title:
The next square after $n^2$ is $(n+1)^2=n^2+2n+1 > n^2+1$ if $n>0$.
Therefore, $n^2+1$ is never a square, unless $n=0$.