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2007 USAMO Problems/Problem 5: Difference between revisions

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{{USAMO newbox|year=2007|num-b=4|num-a=6}}
{{USAMO newbox|year=2007|num-b=4|num-a=6}}
Let <math>\displaystyle{a_{n}}</math> be <math>7^{7^{n}}+1</math>. We prove the result by induction.
The result is true for <math>\displaystyle{n=0}</math> because <math>\displaystyle{a_0 = 2^3}</math> which is the product of <math>3</math> primes. Now we assume the result hold for <math>\displaystyle{n}</math>. We note that the sequence of <math>\displaystyle{a_{n}}</math> is defined by the recursion
<math>\displaystyle{a_{n+1}= (a_{n}-1)^{7}+1}</math>
<math>= a_{n}^{7}-7a_{n}^{6}+21a_{n}^{5}-35a_{n}^{4}+35a_{n}^{3}-21a_{n}^{2}+7a_{n}</math>
<math>= a_{n}^{7}-7a_{n}(a_{n}^{5}-3a_{n}^{4}+5a_{n}^{3}-5a_{n}^{2}+3a_{n}-1)</math>
<math>= a_{n}^{7}-7a_{n}(a_{n}-1)(a_{n}^{4}-2a_{n}^{3}+3a_{n}^{2}-2a_{n}+1)</math>
<math>= a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{4}-2a_{n}^{3}+3a_{n}^{2}-2a_{n}+1)\right)</math>
<math>= a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)</math>.
Since <math>\displaystyle{a_n - 1}</math> is an odd power of <math>7</math>, <math>\displaystyle{7a_n}</math> is a perfect square.
By assumption, <math>\displaystyle{a_n}</math> is divisible by <math>2n + 3</math> primes and, since the second term of the last expression above is a difference of squares and is composite, it is divisible by <math>2</math> primes. Thus <math>\displaystyle{a_{n+1}}</math> is divisible by <math>\displaystyle{(2n + 3) + 2 = 2(n+1) + 3}</math> primes as desired.

Revision as of 21:23, 25 April 2007

Problem

Prove that for every nonnegative integer $n$, the number $7^{7^n}+1$ is the product of at least $2n+3$ (not necessarily distinct) primes.

Solution

2007 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

Let $\displaystyle{a_{n}}$ be $7^{7^{n}}+1$. We prove the result by induction.

The result is true for $\displaystyle{n=0}$ because $\displaystyle{a_0 = 2^3}$ which is the product of $3$ primes. Now we assume the result hold for $\displaystyle{n}$. We note that the sequence of $\displaystyle{a_{n}}$ is defined by the recursion

$\displaystyle{a_{n+1}= (a_{n}-1)^{7}+1}$

$= a_{n}^{7}-7a_{n}^{6}+21a_{n}^{5}-35a_{n}^{4}+35a_{n}^{3}-21a_{n}^{2}+7a_{n}$

$= a_{n}^{7}-7a_{n}(a_{n}^{5}-3a_{n}^{4}+5a_{n}^{3}-5a_{n}^{2}+3a_{n}-1)$

$= a_{n}^{7}-7a_{n}(a_{n}-1)(a_{n}^{4}-2a_{n}^{3}+3a_{n}^{2}-2a_{n}+1)$

$= a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{4}-2a_{n}^{3}+3a_{n}^{2}-2a_{n}+1)\right)$

$= a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)$.

Since $\displaystyle{a_n - 1}$ is an odd power of $7$, $\displaystyle{7a_n}$ is a perfect square. By assumption, $\displaystyle{a_n}$ is divisible by $2n + 3$ primes and, since the second term of the last expression above is a difference of squares and is composite, it is divisible by $2$ primes. Thus $\displaystyle{a_{n+1}}$ is divisible by $\displaystyle{(2n + 3) + 2 = 2(n+1) + 3}$ primes as desired.

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