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2024 USAJMO Problems/Problem 3

Problem

Let $a(n)$ be the sequence defined by $a(1)=2$ and $a(n+1)=(a(n))^{n+1}-1$ for each integer $n\geq1$. Suppose that $p>2$ is prime and $k$ is a positive integer. Prove that some term of the sequence $a(n)$ is divisible by $p^k$.

Solution 1

Lemma $1$:

Given a prime $p$, a positive integer $k$, and an even $m$ such that $p^k|a(m)$, we must have that $p^{k+1}|a(m+2)$.


Proof of Lemma $1$:

$a(m+1)\equiv (a(m))^{m+1}-1\equiv -1 \mod p^{k+1}$

Then, $a(m+2)\equiv (a(m+1))^{m+2}-1\equiv (-1)^{m+2}-1\equiv 0 \mod p^{k+1}$


Therefore, by induction, if there exists an even integer $m$ such that $p|a(m)$, then for all integers $k$, $p^k|a(m+2k-2)$, so we are done if there exists an even $m$ such that $p|a(m)$.


Now, consider the case where there is some prime $p>2$ such that there are no even integers $m$ such that $p|a(m)$.

Lemma $2$:

In this case, we must have that $p|a(m)$ if $m\equiv -1 \mod p-1$ for all integers $m$.

Proof of Lemma $2$:

Suppose for the sake of contradiction that there exists some $m$ such that $m\equiv -1\mod p-1$ and $p$ does not divide $a(m)$. Then, we have $a(m+1)\equiv (a(m))^{m+1}-1\equiv 1-1\equiv 0\mod p$, by Fermat's Little Theorem. Since for all $p>2$, $p-1$ is even, then $m+1$ would be even. However this results in a contradiction.

Then, we get that if $m\equiv -1\mod p-1$, then $0\equiv a(m)\equiv (a(m-1))^m-1\mod p\implies (a(m-1))^{m+1}\equiv 1\equiv a(m-1)\mod p$.

Then, by LTE, $v_p(a(m))=v_p((a(m-1))^m-1)=v_p(a(m-1)-1)+v_p(m)>v_p(m)$. Since $\gcd(p-1,p)=1$, then $\gcd(p-1,p^k)=1$ for all positive integers $k$, so then by Chinese Remainder Theorem there exists integers $m$ such that $m\equiv -1\mod p-1$ and $m\equiv 0\mod p^k$, so we are done $\square$


Remark: I think this is a very cool NT problem.


-bronzetruck2016

See Also

2024 USAJMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

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