Difference between revisions of "2017 USAJMO Problems/Problem 1"

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==Solution 1==
 
==Solution 1==
Let <math>a = 2n-1</math> and <math>b = 2n+1</math>. We see that <math>(2n \pm 1)^2 = 4n^2-4n+1 \equiv 1 \pmod{4n}</math>. Therefore, we have <math>(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1  = 4n \equiv 0 \pmod{4n}</math>, as desired.  
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Let <math>a = 2n-1</math> and <math>b = 2n+1</math>. We see that <math>(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}</math>. Therefore, we have <math>(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1  = 4n \equiv 0 \pmod{4n}</math>, as desired.  
  
 
(Credits to mathmaster2012)
 
(Credits to mathmaster2012)

Revision as of 22:28, 24 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 $(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}$. 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 mathmaster2012)

Solution 2

Let $x$ be odd where $x>1$. 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.

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See also

2017 USAJMO (ProblemsResources)
First Problem Followed by
Problem 2
1 2 3 4 5 6
All USAJMO Problems and Solutions
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