Hello there, I have a problem and would like to know do anyone have an elementary proof of it and it goes like this:

Show that $ a^n + 1\neq m^2$ when $a=4,7,10$ for every $n$ and $m$ ($n$ and $m$ are natural numbers).

EDIT: I succeded in finding an elementary proof that $ a^n + 1\neq m^2$ when $a$ is of the form $a=3k+4$ , $k\geq0$ and I did it using properties of the "digital root" function

http://en.wikipedia.org/wiki/Digital_root

First, suppose that $a^n=m^2-1=(m-1)(m+1)$

We have the following cases:

1) $dr(m-1)=1$ imply $dr(m+1)=3$ which taken together imply $dr((m-1)(m+1))=3$

2) $dr(m-1)=2$ imply $dr(m+1)=4$ which taken together imply $dr((m-1)(m+1))=8$

3) $dr(m-1)=3$ imply $dr(m+1)=5$ which taken together imply $dr((m-1)(m+1))=6$

4) $dr(m-1)=4$ imply $dr(m+1)=6$ which taken together imply $dr((m-1)(m+1))=6$

5) $dr(m-1)=5$ imply $dr(m+1)=7$ which taken together imply $dr((m-1)(m+1))=8$

6) $dr(m-1)=6$ imply $dr(m+1)=8$ which taken together imply $dr((m-1)(m+1))=3$

7) $dr(m-1)=7$ imply $dr(m+1)=9$ which taken together imply $dr((m-1)(m+1))=9$

8) $dr(m-1)=8$ imply $dr(m+1)=1$ which taken together imply $dr((m-1)(m+1))=8$

9) $dr(m-1)=9$ imply $dr(m+1)=2$ which taken together imply $dr((m-1)(m+1))=9$

So we have $dr((m-1)(m+1))\in\lbrace 3,6,8,9 \rbrace$ but since $dr(3k+4)\in\lbrace 1,4,7 \rbrace$ implies that $dr((3k+4)^n)\in\lbrace 1,4,7 \rbrace$ the result follows because $\lbrace 3,6,8,9 \rbrace \cap \lbrace 1,4,7 \rbrace = \emptyset$