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This question has already answered here in other topics, but all answers that I read have some techniques like congruence and fermat numbers and the book which I am reading shows this question in chapter about divisibility. So I have been looking for a way of resolving this problem just using divisibility.

I found a document that brings an answer considering just divisibility, but there are two parts of the solution that I haven't already understood yet. I have read it many times but no success. Here they are

1) why can't $x$ and $y$ be both positive?

2) why did he consider the power $a^{-ny}$, instead of $a^{ny}$, since it is well-known that $t$ divides $a^{ny}-1$?

See remarks on the attached picture in

https://photos.app.goo.gl/RyjPbLkYzJiMpWoG9

Marcos Paulo
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1 Answers1

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  • Regarding your first part, $x,y$ cannot be both positive. If so, then $d\geq m+n$. But this is impossible, since $d|a \implies d \leq a$. Do not forget that $m,n$ are both positive by the hypothesis.

  • Regarding your second part, we have assumed, by the first part, Without Loss of Generality, that $y$ is negative. Now, the term $(a^{ny}-1)$ is not an integer since $a^{ny}$ is a fraction. Hence, we need to negate it to make the power positive. i.e. $(a^{-ny}-1)$.

I hope now it is clear.

Maged Saeed
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  • Very clear about part 2...but no so much about the first because in that solution it is written $d \leq m$ which I understand because $d$ divides $m$. Now you said that $d$ divides $a$ and I got more confused. Could you please give me more details about the fact that if $x, y >0$ implies $d\geq m+n$? – Marcos Paulo Dec 07 '18 at 12:49
  • Oh I see. I was trying to give the property that if a number$d$ divides another number $a$ then $d$ must be less than or equal to $a$. In our case, it follows that $d$ shall be less than or equal to $m$ and less than or equal to $n$ because it divides them. – Maged Saeed Dec 07 '18 at 13:03
  • Hope I was clear. – Maged Saeed Dec 07 '18 at 13:03
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    Now I understand your point, but $d\geq m+n$ is still freaking me out. Please – Marcos Paulo Dec 07 '18 at 14:14
  • @user621908, if $x$ and $y$ both positive, then what is the value of $mx+ny = d$? clearly it is greater than $m$ and $n$. Consider the case where $x,y = 1$ which is the least positive integer. now $d= 1\times m + 1 \times n$ is greater than $m$ and $n$. It is exactly equal to $m+n$. Do you agree? – Maged Saeed Dec 07 '18 at 14:23
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    It is ok now. This specific example was very helpful... I appreciate your time and patience – Marcos Paulo Dec 07 '18 at 15:24
  • Oh thanks, you are more than welcome. :) – Maged Saeed Dec 07 '18 at 15:26