2007 USAMO Problems/Problem 5

Revision as of 21:44, 25 April 2007 by Hamster1800 (talk | contribs) (Moved the solution to above the box)


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.


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

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.

2007 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
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