1999 IMO Problems/Problem 4

Problem

Determine all pairs $(n,p)$ of positive integers such that

$p$ is a prime, $n$ not exceeded $2p$ , and $(p-1)^{n}+1$ is divisible by $n^{p-1}$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. official solution Clearly we have the solutions $(1,p)$ and $(2,2)$ , and for every other solution $p \geq 3$ . It remains to find the solutions $(x,p)$ with $x \geq 2$ and $p \geq 3$ . We claim that in this case $x$ is divisible by p and $x y 2p$ , whence $x = p$ . This will lead to

$\[p^{p-1}|(p-1)^{p}+1=p^{2}\left(p^{p-2}-{p \choose 1}p^{p-3}+ \cdots + -{p \choose p-3}p-{p \choose p-2}+1\right)\]$ (Error compiling LaTeX. Unknown error_msg)

therefore, because all the terms in the brackets excepting the last one is divisible by $p$ , $p-1 \leq 2$ . This leaves only $p=3$ and $x=3$ . Let us prove now the claim. Since $(p-1)^{x}+1$ is odd, so is $x$ (therefore $x < 2p$ ). Denote by q the smallest prime divisor of $x$ . $q|(p-1)^{x}+1$ we get $(p-1)^{x} \equiv -1 \pmod {q}$ and $(q,p-q)=1$ . But $(x,p-q)=1$ (from the choice of q) leads to the existence of integers $u,v$ such that $ux+v(q-1)=1$ , whence $p-1 \equiv (p-1)^{ux}$ . $(p-1)^{v(q-1)} \equiv (-1)^{u} \cdot 1^{v} \equiv -1 \pmod {q}$ , because $u$ must be odd. This shows that $q|p$ , therefore $q=p$ . In conclusion the required solutions are $(2,2), (3,3)$ and $(1,p)$ , where $p$ is an arbitrary prime.

1999 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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1999 IMO Problems/Problem 4

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