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

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For $n>1$, this is a special case of Catalan's conjecture (proved by Mihailescu). These special cases may well have elementary proofs, search the literature. The case $n=1$ you can do by inspection. –  Felipe Voloch Jan 28 '13 at 15:31
    
At least 4 and 7 seem simple, factor $m^2-1=(m+1)(m-1)$ and observe that not both can be a power of 2 (except for $m=3$) or a power of 7, resp. And, what is the motivation for this? Voting to close. –  quid Jan 28 '13 at 15:40
    
Thank you, I forgot that Catalan´s conjecture have status of a theorem, and sure there must be some elementary method for proving these three cases. –  Antisha Jan 28 '13 at 15:47
    
Motivation is to find elementary proof when $a$ is of the form $a=3k+4$ when k is natural number, or zero. –  Antisha Jan 28 '13 at 15:51
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Your proof can be simplified greatly. If $a=3k+4$, then $a\equiv 1\pmod{3}$, hence $a^n+1\equiv 2\pmod{3}$. On the other hand, it is easy to see that $m^2\equiv 2\pmod{3}$ is impossible. –  GH from MO Jan 28 '13 at 20:39

1 Answer 1

up vote 3 down vote accepted

I recommend this survey 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.

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I commented above (and then mistakenly deleted) that the specific result desired by the poster was realized through mod 3 considerations. I suspect that the poster is even more interested in when elementary considerations provide quick answers. For example, knowing when 2 is not a qth power mod a covers the cases mentioned and many more. Gerhard "Looks For Really Simple Answers" Paseman, 2013.01.28 –  Gerhard Paseman Jan 28 '13 at 18:57
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As a general aside: there is also a technically non-elementary (but in certain ways simpler) proof by Mignotte "A new proof of Ko Chao's theorem" (2004) –  quid Jan 28 '13 at 19:11
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Also, while I presently do not see how to improve on Chein's argument, I still think that something simpler and possibly less complete is wanted. Gerhard "AskMe About System Design" Paseman, 2013.01.28 –  Gerhard Paseman Jan 28 '13 at 19:13
    
I found an elementary proof that $a^n + 1\neq m^2$ when $a$ is of the form $a=3k+4$. In it I used the properties of this function: mathworld.wolfram.com/DigitalRoot.html Thank you all for your valuable information. –  Antisha Jan 28 '13 at 19:23
    
Perhaps it would be useful to include your elementary proof in your original post. –  GH from MO Jan 28 '13 at 19:27

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