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

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So
  
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<cmath>a^b+b^a = a(a^2)^n+b(b^2)^{n-1} \equiv a\cdot 1^n + b\cdot 1^{n-1} \equiv a+b = 4n \equiv 0 (\bmod 4n)</cmath>
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<cmath>a^b+b^a = a(a^2)^n+b(b^2)^{n-1} \equiv </cmath> <cmath>a\cdot 1^n + b\cdot 1^{n-1} \equiv a+b = 4n \equiv 0 (\bmod 4n)</cmath>
  
 
Since <math>a+b=4n</math>, all such pairs work, and we are done.
 
Since <math>a+b=4n</math>, all such pairs work, and we are done.

Revision as of 20:08, 14 October 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 1

Let $n=a+b$. Since $gcd(a,b)=1$, we know $gcd(a,n)=1$. We can rewrite the condition as

\[a^{n-a}+(n-a)^a \equiv 0 \mod{n}\] \[a^{n-a}\equiv-(-a)^a \mod{n}\] Assume $a$ is odd. Since we need to prove an infinite number of pairs exist, it suffices to show that infinitely many pairs with $a$ odd exist.

Then we have \[a^{n-a}\equiv a^a \mod{n}\] \[1 \equiv a^{2a-n} \mod{n}\]

We know by Euler's theorem that $a^{\varphi(n)} \equiv 1 \mod{n}$, so if $2a-n=\varphi(n)$ we will have the required condition.

This means $a=\frac{n+\varphi(n)}{2}$. Let $n=2p$ where $p$ is a prime, $p\equiv 1\mod{4}$. Then $\varphi(n) = 2p*\left(1-\frac{1}{2}\right)\left(1-\frac{1}{p}\right) = p-1$, so \[a = \frac{2p+p-1}{2} = \frac{3p-1}{2}\] Note the condition that $p\equiv 1\mod{4}$ guarantees that $a$ is odd, since $3p-1 \equiv 2\mod{4}$

This makes $b = \frac{p+1}{2}$. Now we need to show that $a$ and $b$ are relatively prime. We see that \[gcd\left(\frac{3p-1}{2},\frac{p+1}2\right)=\frac{gcd(3p-1,p+1)}{2}\] \[=\frac{gcd(p+1,4)}{2}=\frac22=1\] By the Euclidean Algorithm.

Therefore, for all primes $p \equiv 1\mod{4}$, the pair $\left(\frac{3p-1}{2},\frac{p+1}{2}\right)$ satisfies the criteria, so infinitely many such pairs exist.

Solution 2

Take $a=2n-1, b=2n+1, n\geq 2$. It is obvious (use the Euclidean Algorithm, if you like), that $\gcd(a,b)=1$, and that $a,b>1$.

Note that

\[a^2 = 4n^2-4n+1 \equiv 1 (\bmod 4n)\]

\[b^2 = 4n^2+4n+1 \equiv 1 (\bmod 4n)\]

So

\[a^b+b^a = a(a^2)^n+b(b^2)^{n-1} \equiv\] \[a\cdot 1^n + b\cdot 1^{n-1} \equiv a+b = 4n \equiv 0 (\bmod 4n)\]

Since $a+b=4n$, all such pairs work, and we are done.

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