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

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==Solution 5==
 
==Solution 5==
I claim <math>(a,b) = (2n-1,2n+1)</math> always satisfies the conditions, n is integer, <math>n \geq 2</math>,
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I claim <math>(a,b) = (2n-1,2n+1)</math>, <math>n (\in \mathbb{N}) \geq 2</math> always satisfies above conditions.
  
Proof:
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Proof:
  
We can expand <math>(2n-1)^{2n+1} + (2n+1)^{2n-1}</math> using binomial theorem. However, since <math>2n-1 + 2n+1 = 4n</math>, all the terms with power <math>2n</math> larger than <math>2</math> modulo <math>4n</math> evaluate to 0, and thus can be omitted. We are left with the terms: <math>(2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n</math>, which is divisible by <math>4n</math>.
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Since there are infinitely many integers larger than or equal to 2, there are infinitely many distinct pairs <math>(a,b)</math>.  
  
Since there are infinitely many integers larger than or equal to 2, the proof is complete .<math>\blacksquare</math>
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We only need to prove:
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<math>a^b+b^a \equiv 0 \pmod{a+b}</math>
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We can expand <math>a^b + b^a = (2n-1)^{2n+1} + (2n+1)^{2n-1}</math> using binomial theorem. However, since <math>a + b = 2n-1 + 2n+1 = 4n</math>, all the <math>2n</math> terms (with more than 2 powers of) when evaluated modulo <math>4n</math> equal to 0, and thus can be omitted. We are left with the terms: <math>(2n+1)(2n)^1-1+(2n-1)(2n)^1+1 = 4n \cdot 2n</math>, which is divisible by <math>4n</math>.
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<math>(2n-1)^{2n+1} + (2n+1)^{2n-1} \equiv (2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n \equiv 0 \pmod{4n}</math>
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The proof is complete. <math>\blacksquare</math>
  
 
-AlexLikeMath
 
-AlexLikeMath

Revision as of 00:25, 10 June 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

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 $1,2$. $\boxed{}$

~AopsUser101

Solution 5

I claim $(a,b) = (2n-1,2n+1)$, $n (\in \mathbb{N}) \geq 2$ always satisfies above conditions.

Proof:

Since there are infinitely many integers larger than or equal to 2, there are infinitely many distinct pairs $(a,b)$.

We only need to prove:

$a^b+b^a \equiv 0 \pmod{a+b}$

We can expand $a^b + b^a = (2n-1)^{2n+1} + (2n+1)^{2n-1}$ using binomial theorem. However, since $a + b = 2n-1 + 2n+1 = 4n$, all the $2n$ terms (with more than 2 powers of) when evaluated modulo $4n$ equal to 0, and thus can be omitted. We are left with the terms: $(2n+1)(2n)^1-1+(2n-1)(2n)^1+1 = 4n \cdot 2n$, which is divisible by $4n$.

$(2n-1)^{2n+1} + (2n+1)^{2n-1} \equiv (2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n \equiv 0 \pmod{4n}$

The proof is complete. $\blacksquare$

-AlexLikeMath


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

See also

2017 USAJMO (ProblemsResources)
First Problem Followed by
Problem 2
1 2 3 4 5 6
All USAJMO Problems and Solutions