# 2007 USAMO Problems/Problem 5

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## Problem

(Titu Andreescu) 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.

## Solutions

### Solution 1

The proof is by induction. The base is provided by the $n = 0$ case, where $7^{7^0} + 1 = 7^1 + 1 = 2^3$. To prove the inductive step, it suffices to show that if $x = 7^{2m - 1}$ for some positive integer $m$ then $(x^7 + 1)/(x + 1)$ is composite. As a consequence, $x^7 + 1$ has at least two more prime factors than does $x + 1$. To confirm that $(x^7 + 1)/(x + 1)$ is composite, observe that \begin{align*} \frac{x^7 + 1}{x + 1} &= \frac{(x + 1)^7 - ((x + 1)^7 - (x^7 + 1))}{x + 1} \\ &= (x + 1)^6 - \frac{7x(x^5 + 3x^4 + 5x^3 + 5x^2 + 3x + 1)}{x + 1} \\ &= (x + 1)^6 - 7x(x^4 + 2x^3 + 3x^2 + 2x + 1) \\ &= (x + 1)^6 - 7^{2m}(x^2 + x + 1)^2 \\ &= \{(x + 1)^3 - 7^m(x^2 + x + 1)\}\{(x + 1)^3 + 7^m(x^2 + x + 1)\}. \end{align*} Also each factor exceeds 1. It suffices to check the smaller one; $\sqrt{7x}\leq x$ gives \begin{align*} (x + 1)^3 - 7^m(x^2 + x + 1) &= (x + 1)^3 - \sqrt{7x}(x^2 + x + 1) \\ &\geq x^3 + 3x^2 + 3x + 1 - x(x^2 + x + 1) \\ &= 2x^2 + 2x + 1\geq 113 > 1. \end{align*} Hence $(x^7 + 1)/(x + 1)$ is composite and the proof is complete.