Difference between revisions of "2024 USAJMO Problems/Problem 3"

Revision as of 08:31, 28 March 2024

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 $1$ :

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

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

@@ Line 6: / Line 6: @@
 == Solution 1 ==
+Lemma <math>1</math>:
+Given a prime <math>p</math>, a positive integer <math>k</math>, and an even <math>m</math> such that <math>p^k|a(m)</math>, we must have that <math>p^{k+1}|a(m+2)</math>.
+Proof of Lemma <math>1</math>:
+<math>a(m+1)\equiv (a(m))^{m+1}-1\equiv -1 \mod p^{k+1}</math>
+Therefore, <math>a(m+2)\equiv (a(m+1))^{m+2}-1\equiv (-1)^{m+2}-1\equiv 0 \mod p^{k+1}</math>
 ==See Also==
 {{USAJMO newbox|year=2024|num-b=2|num-a=4}}
 {{MAA Notice}}

2024 USAJMO (Problems • Resources)
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Difference between revisions of "2024 USAJMO Problems/Problem 3"

Revision as of 08:31, 28 March 2024

Contents

Problem

Solution 1

See Also