Difference between revisions of "2007 USAMO Problems/Problem 5"

Revision as of 13:39, 7 June 2014

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.

Hint 1 of 3

You may be stuck in "factoring" $\frac{a^7 + 1}{a + 1} = a^6 - a^5 + a^4 - a^3 + a^2 - a + 1$ for $a = 7^{7^n}$ . Keep trying! This is a #5 on a USAMO!

Hint 2 of 3

Believe it or not, $\frac{a^7 + 1}{a + 1}$ is a difference of squares!

Final Hint

One of the perfect squares is $(x+1)^6$ .

Solution

Solution 1

We proceed by induction.

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

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

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

Since ${a_n - 1}$ is an odd power of ${7}$ , ${7(a_n-1)}$ is a perfect square. Therefore ${a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}$ is a difference of squares and thus composite, i.e. it is divisible by ${2}$ primes. By assumption, ${a_n}$ is divisible by ${2n + 3}$ primes. Thus ${a_{n+1}}$ is divisible by ${2+ (2n + 3) = 2(n+1) + 3}$ 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 $x^{6}-x^{5}+x^{4}-x^3 + x^2 - x+1$ . This can be factored as $(x^{3}+3x^{2}+3x+1)^{2}-7x(x^{2}+x+1)^{2}$ .

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

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2007 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
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