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

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Prove that there are infinitely many distinct pairs <math>(a,b)</math> of relatively prime positive integers <math>a>1</math> and <math>b>1</math> such that <math>a^b+b^a</math> is divisible by <math>a+b</math>.
 
Prove that there are infinitely many distinct pairs <math>(a,b)</math> of relatively prime positive integers <math>a>1</math> and <math>b>1</math> such that <math>a^b+b^a</math> is divisible by <math>a+b</math>.
  
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== Solution ==
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== Solution 1 ==
 
Let <math>n=a+b</math>. Since <math>gcd(a,b)=1</math>, we know <math>gcd(a,n)=1</math>. We can rewrite the condition as  
 
Let <math>n=a+b</math>. Since <math>gcd(a,b)=1</math>, we know <math>gcd(a,n)=1</math>. We can rewrite the condition as  
  
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Therefore, for all primes <math>p \equiv 1\mod{4}</math>, the pair <math>\left(\frac{3p-1}{2},\frac{p+1}{2}\right)</math> satisfies the criteria, so infinitely many such pairs exist.
 
Therefore, for all primes <math>p \equiv 1\mod{4}</math>, the pair <math>\left(\frac{3p-1}{2},\frac{p+1}{2}\right)</math> satisfies the criteria, so infinitely many such pairs exist.
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== Solution 2 ==
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Take <math>a=2n-1, b=2n+1, n\geq 2</math>. It is obvious (use the Euclidean Algorithm, if you like), that <math>\gcd(a,b)=1</math>, and that <math>a,b>1</math>.
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Note that
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<cmath>a^2 = 4n^2-4n+1 \equiv 1 (\bmod 4n)</cmath>
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<cmath>b^2 = 4n^2+4n+1 \equiv 1 (\bmod 4n)</cmath>
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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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Since <math>a+b=4n</math>, all such pairs work, and we are done.

Revision as of 22:05, 20 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 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. If $a$ and $b$ have a common factor $k$, then we know $k|a+b=2p$. Therefore $k=1,2,p,$ or $2p$. We know $b=\frac{p+1}{2}<p$, so $p \nmid b$ and $2p\nmid b$. We also know $a$ is odd, so $2\nmid a$. This means the only common factor $a$ and $b$ have is $1$, so $a$ and $b$ are relatively prime.

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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