Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution)
Line 15: Line 15:
  
  
−
Since <math>\displaystyle{a_n - 1}</math> is an odd power of <math>\displaystyle{7}</math>, <math>\displaystyle{7(a_n-1)}</math> is a perfect square. Therefore <math>\displaystyle{(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}</math> is a difference of squares and thus composite, i.e. it is divisible by <math>\displaystyle{2}</math> primes. By assumption, <math>\displaystyle{a_n}</math> is divisible by <math>\displaystyle{2n + 3}</math> primes. Thus <math>\displaystyle{a_{n+1}}</math> is divisible by <math>\displaystyle{2+ (2n + 3) = 2(n+1) + 3}</math> primes as desired.
+
Since <math>\displaystyle{a_n - 1}</math> is an odd power of <math>\displaystyle{7}</math>, <math>\displaystyle{7(a_n-1)}</math> is a perfect square. Therefore <math>\displaystyle{a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}</math> is a difference of squares and thus composite, i.e. it is divisible by <math>\displaystyle{2}</math> primes. By assumption, <math>\displaystyle{a_n}</math> is divisible by <math>\displaystyle{2n + 3}</math> primes. Thus <math>\displaystyle{a_{n+1}}</math> is divisible by <math>\displaystyle{2+ (2n + 3) = 2(n+1) + 3}</math> primes as desired.

Revision as of 21:34, 25 April 2007

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.

Solution

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

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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_USAMO_Problems/Problem_5&oldid=14727"