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

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==Solution 3==
 
==Solution 3==
 
Because problems such as this usually are related to expressions along the lines of <math>x\pm1</math>, it's tempting to try these. After a few cases, we see that <math>\left(a,b\right)=\left(2x-1,2x+1\right)</math> is convenient due to the repeated occurrence of <math>4x</math> when squared and added. We rewrite the given expressions as: <cmath>\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.</cmath> After repeatedly factoring the initial equation,we can get: <cmath>\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).</cmath> Expanding each of the squares, we can compute each product independently then sum them: <cmath>\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},</cmath> <cmath>\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.</cmath> Now we place the values back into the expression: <cmath>\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.</cmath> Plugging any positive integer value for <math>x</math> into <math>\left(a,b\right)=\left(2x-1,2x+1\right)</math> yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs <math>\left(a,b\right)</math>. <math>\square</math>
 
Because problems such as this usually are related to expressions along the lines of <math>x\pm1</math>, it's tempting to try these. After a few cases, we see that <math>\left(a,b\right)=\left(2x-1,2x+1\right)</math> is convenient due to the repeated occurrence of <math>4x</math> when squared and added. We rewrite the given expressions as: <cmath>\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.</cmath> After repeatedly factoring the initial equation,we can get: <cmath>\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).</cmath> Expanding each of the squares, we can compute each product independently then sum them: <cmath>\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},</cmath> <cmath>\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.</cmath> Now we place the values back into the expression: <cmath>\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.</cmath> Plugging any positive integer value for <math>x</math> into <math>\left(a,b\right)=\left(2x-1,2x+1\right)</math> yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs <math>\left(a,b\right)</math>. <math>\square</math>
 +
-fatant
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:19, 20 October 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.

Solution 3

Because problems such as this usually are related to expressions along the lines of $x\pm1$, it's tempting to try these. After a few cases, we see that $\left(a,b\right)=\left(2x-1,2x+1\right)$ is convenient due to the repeated occurrence of $4x$ when squared and added. We rewrite the given expressions as: \[\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.\] After repeatedly factoring the initial equation,we can get: \[\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).\] Expanding each of the squares, we can compute each product independently then sum them: \[\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},\] \[\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.\] Now we place the values back into the expression: \[\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.\] Plugging any positive integer value for $x$ into $\left(a,b\right)=\left(2x-1,2x+1\right)$ yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs $\left(a,b\right)$. $\square$ -fatant

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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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