I found this problem in a Putnam guide, and the topic it was listed under was Fermat's infinite descent. I'm assuming we need to show that the only solutions are $(1,1)$ and $(0,y)$ for all $y$, but I'm unsure of how to use infinite descent to show this. All I have gotten is that both $x$ and $y$ must be positive and odd for $x\neq0$. Help would be greatly appreciated!
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4This question is similar to: Divisibility of $2^x-1 $. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. Found using an Approach0 search. Note the duplicate question says the problem came from a Korean Math Olympiad. The AoPS thread Number theory is also in the results. ... – John Omielan Dec 21 '24 at 05:50
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(cont.) To show that $x\nmid 2^x-1$, there's the AoPS thread n| 2^{n} -1 is not possible for n>1, and the related post here of Find all the primes that satisfy $p \mid 2^p - 1$. Finally, welcome to Math SE. – John Omielan Dec 21 '24 at 05:54
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In addition, for showing that for $x \gt 1, x\nmid 2^x-1$, as stated in the answer here and also basically in several answers among the duplicates to your question, there's also the more specific For $n \geq 2$, show that $n \nmid 2^{n}-1$ and Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$?. – John Omielan Dec 21 '24 at 06:04
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1"A5. Show that $n$ does not divide $2^n - 1$ for $n > 1$." was on the 1972 Putnam exam. – Gerry Myerson Dec 21 '24 at 19:11
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Note, that for every positive integer $x$ such that $2\not\mid x$ we have $$ 2^{x} \equiv \phi(x) \bmod x$$
, where $\phi(x)$ is number of positive integers $t\in[1;x]$, coprime with $x$.
Since, we want $\frac{2^x-1}{x}=y$ to be an integer, we must have $ 2^x \equiv 1\bmod x$ and therefore $\phi(x)\equiv 1\bmod x$. Finally, since $\phi(x) \le x$, we find that $ \phi(x) = 1$.
But, since $ 2\not\mid x$ (because if $2\mid x$ and $x\ge 1$, then the LHS of our first equation will be odd, and the RHS will be even), both 1 and 2 are coprime to $x$. Then, we have a contradiction: $\phi(x) \ge 2$ and $\phi(x) = 1$ in the same time.
Therefore, the only possibilites are $x=1$ (because when we count $\phi(1)$, we don't include 2), and $x=0$ (because when $x=0$, $2^x-1$ is even and so does $xy$). These two $x$ give us two solutions $(x_1,y_1)=(1,1)$ and $(x_2,y_2)=(0,\mathbb{Z})$.
Ivan Borisyuk
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