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2007 USAMO Problems/Problem 5: Difference between revisions

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Now we assume the result holds for <math>\displaystyle{n}</math>. Note that <math>\displaystyle{a_{n}}</math> satisfies the [[recursion]]
Now we assume the result holds for <math>\displaystyle{n}</math>. Note that <math>\displaystyle{a_{n}}</math> satisfies the [[recursion]]


=== Solution 1 ===
<div style="text-align:center;"><math>\displaystyle{a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)</math></div>
<div style="text-align:center;"><math>\displaystyle{a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)</math></div>


Revision as of 17:19, 2 May 2007

Problem

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

Solution

We prove the result by induction.

Let $\displaystyle{a_{n}}$ be $7^{7^{n}}+1$. The result holds for $\displaystyle{n=0}$ because $\displaystyle{a_0 = 2^3}$ is the product of $\displaystyle{3}$ primes.

Now we assume the result holds for $\displaystyle{n}$. Note that $\displaystyle{a_{n}}$ satisfies the recursion

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

Since $\displaystyle{a_n - 1}$ is an odd power of $\displaystyle{7}$, $\displaystyle{7(a_n-1)}$ is a perfect square. Therefore $\displaystyle{a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}$ is a difference of squares and thus composite, i.e. it is divisible by $\displaystyle{2}$ primes. By assumption, $\displaystyle{a_n}$ is divisible by $\displaystyle{2n + 3}$ primes. Thus $\displaystyle{a_{n+1}}$ is divisible by $\displaystyle{2+ (2n + 3) = 2(n+1) + 3}$ primes as desired.

See also

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