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 $\lbrace 3,6,8,9 \rbrace \cap \lbrace 1,4,7 \rbrace = \emptyset$



Numbers of a certain form not expressible as squares
I recommend this [survey](http://www.ams.org/journals/bull/2004-41-01/S0273-0979-03-00993-5/S0273-0979-03-00993-5.pdf) about the solution of Catalan's conjecture. We learn from here that the special case you are considering, namely $x^p-y^q=1$ for $p=2$, was solved by Chao Ko in 1964. In 1976 Chein published a simpler proof, see [here](http://www.ams.org/journals/proc/1976-056-01/S0002-9939-1976-0404133-1/S0002-9939-1976-0404133-1.pdf).


