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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?

nabu1227
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  • There are no two consecutive positive integers both being a perfect square. $10^n+1$ has an algebraic factor when $n\ge 3$ is odd. I do not know a complete proof yet but it should be relatively easy to show that $10^n+1$ is never a square for positive integer $n$. – Peter Aug 16 '18 at 12:40
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    However, Catalan's conjecture (now proven) states that the only consecutive perfect powers are $8$ and $9$ answering your question. – Peter Aug 16 '18 at 12:42
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    @Peter $10^n+1=m^2\implies 10^n=(m+1)(m-1)$ but then one of the two factors must be divisible by $2^a\times 5^n$ and the other is $2^{n-a}$. but it is clear that those two expressions can not differ by $2$. – lulu Aug 16 '18 at 12:45
  • @lulu I just noticed that $10^n+1\equiv 2\mod 3$ for $n\ge 1$ showing that $10^n+1$ can never be a perfect square. – Peter Aug 16 '18 at 12:48
  • @Peter Ah, even easier. – lulu Aug 16 '18 at 12:48
  • hello, can I request that you all who commented here edit 10^n to a^n for the sake of relevance? Thank you very much! – nabu1227 Sep 09 '18 at 14:32

2 Answers2

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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 $n=0$.

lhf
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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.

Keith Backman
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