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2007 USAMO Problems/Problem 5

Revision as of 20:27, 7 May 2007 by Carpo (talk | contribs) (→‎Solution)

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 1

We proceed 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.

Solution 2

Notice that $7^{{7}^{k+1}}+1=(7+1) \frac{7^{7}+1}{7+1} \cdot \frac{7^{7^2} + 1}{7^{7^1} + 1} \cdots \frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$. Therefore it suffices to show that $\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$ is composite.

Let $x=7^{7^{k}}$. The expression becomes

$\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1} = \frac{x^7 + 1}{x + 1}$

which is the shortened form of the geometric series $\displaystyle x^{6}-x^{5}+x^{4}-x^3 + x^2 - x+1$. This can be factored as $\displaystyle (x^{3}+3x^{2}+3x+1)^{2}-7x(x^{2}+x+1)^{2}$

Since $x$ is an odd power of $7$, $\displaystyle 7x$ is a perfect square, and so we can factor this by difference of squares. Therefore, it is composite.

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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