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 "2017 USAJMO Problems/Problem 1"

m (→‎Solution 2)
(→‎Solution 2)
Line 9: Line 9:
  
 
==Solution 2==
 
==Solution 2==
−
Let <math>x</math> be any odd number above 1. We have <math>x^2-1=(x-1)(x+1).</math> Since <math>x-1</math> is even, <math>x^2-1 \equiv 0 \pmod{2x+2}.</math> This means that  <math>x^{x+2}-x^x \equiv 0 \pmod{2x+2},</math> and since x is odd, <math>x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},</math> or <math>x^{x+2}+x+2^x \equiv 0 \pmod{2x+2}.</math> This means for any odd x, the ordered triple <math>(x,x+2)</math> satisfies the condition. Since there are infinitely many values of <math>x</math> possible, there are infinitely many ordered triples, as desired.  
+
Let <math>x</math> be any odd number above 1. We have <math>x^2-1=(x-1)(x+1),</math> so <math>x^2-1 \equiv 0 \pmod{2x+2}.</math> This means that  <math>x^{x+2}-x^x \equiv 0 \pmod{2x+2},</math> and since x is odd, <math>x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},</math> or <math>x^{x+2}+x+2^x \equiv 0 \pmod{2x+2},</math> as desired.
−
 
 
−
{{CURRENTTIME}}
 
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:21, 20 April 2017

Problem

Prove that there are infinitely many distinct pairs $(a,b)$ of relatively prime integers $a>1$ and $b>1$ such that $a^b+b^a$ is divisible by $a+b$.

Solution 1

Let $a = 2n-1$ and $b = 2n+1$. We see that $(2n \pm 1)^2 = 4n^2-4n+1 \equiv 1 \pmod{4n}$. Therefore, we have $(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}$, as desired.

(Credits to laegolas)

Solution 2

Let $x$ be any odd number above 1. We have $x^2-1=(x-1)(x+1),$ so $x^2-1 \equiv 0 \pmod{2x+2}.$ This means that $x^{x+2}-x^x \equiv 0 \pmod{2x+2},$ and since x is odd, $x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},$ or $x^{x+2}+x+2^x \equiv 0 \pmod{2x+2},$ as desired.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2017 USAJMO (Problems • Resources)
First Problem Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&oldid=85319"