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"

m (wik)
(I removed the "Solution 1" heading from the middle of the solution.~~~~)
Line 11: Line 11:
 
Now we assume the result holds for <math>\displaystyle{n}</math>. Note that <math>\displaystyle{a_{n}}</math> satisfies the [[recursion]]
 
Now we assume the result holds for <math>\displaystyle{n}</math>. Note that <math>\displaystyle{a_{n}}</math> satisfies the [[recursion]]
  
−
=== Solution 1 ===
 
 
<div style="text-align:center;"><math>\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)</math></div>
 
<div style="text-align:center;"><math>\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)</math></div>
  

Revision as of 18:19, 2 May 2007

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

We prove the result 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.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_USAMO_Problems/Problem_5&oldid=14797"