Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 USAJMO Problems/Problem 1"
(→‎Solution 3)
Line 17: Line 17:
  
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
==Solution 4==
 +
Let <math>a = 2x + 1</math> and <math>b = 2x-1</math>, where <math>x</math> leaves a remainder of <math>1</math> when divided by <math>4</math>.We seek to show that <math>(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x</math> because that will show that there are infinitely many distinct pairs <math>(a,b)</math> of relatively prime 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>.
 +
 +
Claim 1: <math>(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4</math>.
 +
We have that the remainder when <math>2x+1</math> is divided by <math>4</math> is <math>3</math> and the remainder when <math>2x-1</math> is divided by <math>4</math> is always <math>1</math>. Therefore, the remainder when <math>(2x+1)^{2x-1} + (2x-1)^{2x+1}</math> is divided by <math>4</math> is always going to be <math>(-1)^{2x-1} + 1^{2x+1} = 0</math>.
 +
 +
Claim 2: <math>(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x</math>
 +
We know that <math>(2x+1) \mod x \equiv 1</math> and <math>(2x-1) \mod x \equiv 3</math>, so the remainder when <math>(2x+1)^{2x-1} + (2x-1)^{2x+1}</math> is divided by <math>4</math> is always going to be <math>(-1)^{2x-1} + 1^{2x+1} = 0</math>.
 +
 +
Claim 3: <math>(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x</math>
 +
Trivial given claim <math>1,2</math>. <math>\boxed{}</math>
  
 
==See also==
 
==See also==
 
{{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}
 
{{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}

Revision as of 14:16, 19 February 2020

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Solution 4

Let $a = 2x + 1$ and $b = 2x-1$, where $x$ leaves a remainder of $1$ when divided by $4$.We seek to show that $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x$ because that will show 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 $a+b$.

Claim 1: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4$. We have that the remainder when $2x+1$ is divided by $4$ is $3$ and the remainder when $2x-1$ is divided by $4$ is always $1$. Therefore, the remainder when $(2x+1)^{2x-1} + (2x-1)^{2x+1}$ is divided by $4$ is always going to be $(-1)^{2x-1} + 1^{2x+1} = 0$.

Claim 2: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x$ We know that $(2x+1) \mod x \equiv 1$ and $(2x-1) \mod x \equiv 3$, so the remainder when $(2x+1)^{2x-1} + (2x-1)^{2x+1}$ is divided by $4$ is always going to be $(-1)^{2x-1} + 1^{2x+1} = 0$.

Claim 3: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x$ Trivial given claim $1,2$. $\boxed{}$

See also

2017 USAJMO (Problems • Resources)
First Problem Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&oldid=118114"