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

Revision as of 10:42, 27 June 2019

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 $gcd\left(\frac{3p-1}{2},\frac{p+1}2\right)=\frac{gcd(3p-1,p+1)}{2}$ $=\frac{gcd(p-3,4)}{2}=\frac22=1$ 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.

Solution 3

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.

@@ Line 41: / Line 41: @@
 Since <math>a+b=4n</math>, all such pairs work, and we are done.
+== Solution 3 ==
+Let <math>x</math> be odd where <math>x>1</math>. We have <math>x^2-1=(x-1)(x+1),</math> so <math>x^2-1 \equiv 0 \pmod{2x+2}.</math> This means that  <math>x^{x+2}-x^x \equiv 0 \pmod{2x+2},</math> and since x is odd, <math>x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},</math> or <math>x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},</math> as desired.
 ==See Also==
 {{USAMO newbox|year= 2017 |before=[[2016 USAMO]]|after=[[2018 USAMO]]}}

2017 USAMO (Problems • Resources)
Preceded by 2016 USAMO	Followed by 2018 USAMO
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