Difference between revisions of "2017 USAJMO Problems/Problem 2"

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==Solution==
 
==Solution==
Let <math>a = 2^n - 1</math> and <math>b = 2^n + 1</math>. We see that <math>a</math> and <math>b</math> are relatively prime (they are consecutive positive odd integers).
 
  
Lemma: <math>(2^k + 1)^{-1} \equiv 2^k + 1 \pmod{2^{k+1}}</math>.
 
 
Since every number has a unique modular inverse, the lemma is equivalent to proving that <math>(2^k+1)^2 \equiv 1 \pmod{2^{k+1}}</math>. Expanding, we have the result.
 
 
Substituting for <math>a</math> and <math>b</math>, we have
 
<cmath>(2^k+1)^{2^k-1} + (2^k-1)^{2^k+1} \equiv 2^k - 1 + 2^k + 1 \equiv 0 \pmod{2^{k+1}},</cmath>
 
where we use our lemma and the Euler totient theorem: <math>a^\phi{n} \equiv 1 \pmod{n}</math> when <math>a</math> and <math>n</math> are relatively prime.
 
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:11, 19 April 2017

Problem:

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

Solution

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See also

2017 USAJMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
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All USAJMO Problems and Solutions
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