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

(Added second solution)
m
(8 intermediate revisions by 6 users not shown)
Line 19: Line 19:
 
Note the condition that <math>p\equiv 1\mod{4}</math> guarantees that <math>a</math> is odd, since <math>3p-1 \equiv 2\mod{4}</math>
 
Note the condition that <math>p\equiv 1\mod{4}</math> guarantees that <math>a</math> is odd, since <math>3p-1 \equiv 2\mod{4}</math>
  
This makes <math>b = \frac{p+1}{2}</math>. Now we need to show that <math>a</math> and <math>b</math> are relatively prime. If <math>a</math> and <math>b</math> have a common factor <math>k</math>, then we know <math>k|a+b=2p</math>. Therefore <math>k=1,2,p,</math> or <math>2p</math>. We know <math>b=\frac{p+1}{2}<p</math>, so <math>p \nmid b</math> and <math>2p\nmid b</math>. We also know <math>a</math> is odd, so <math>2\nmid a</math>. This means the only common factor <math>a</math> and <math>b</math> have is <math>1</math>, so <math>a</math> and <math>b</math> are relatively prime.
+
This makes <math>b = \frac{p+1}{2}</math>. Now we need to show that <math>a</math> and <math>b</math> are relatively prime. We see that
 +
<cmath>gcd\left(\frac{3p-1}{2},\frac{p+1}2\right)=\frac{gcd(3p-1,p+1)}{2}</cmath>
 +
<cmath>=\frac{gcd(p-3,4)}{2}=\frac22=1</cmath>
 +
By the Euclidean Algorithm.
  
 
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.
Line 35: Line 38:
 
So
 
So
  
<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>
+
<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.
 +
 +
 +
== 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.
 +
 +
==Solution 4==
 +
I claim that the ordered pair <math>(2^{n} - 1, 2^{n} + 1)</math> satisfies the criteria for all <math>n \geq 2.</math>
 +
Proof:
 +
It is easy to see that the order modulo <math>2^{n+1}</math> of <math>(2^{n} - 1)^k</math> is <math>2, </math> since <math>(2^{n} - 1)^2 = 2^{2n} - 2 \cdot 2^{n} + 1 \equiv 1 \mod 2^{n+1}. </math>, and we can assert and prove similarly for the order modulo <math>2^{n+1}</math> of <math>(2^{n} + 1)^k.</math> Thus, the remainders modulo <math>2^{n+1}</math> that the sequences of powers of <math>2^{n} - 1</math> and <math>2^{n} + 1</math> generate are <math>1</math> for even powers and <math>2^{n} - 1</math> and <math>2^{n} + 1</math> for odd powers, respectively. Since <math>2^{n} + 1</math> and <math>2^{n} - 1</math> are both odd for <math>n \geq 2, </math> <cmath>(2^n + 1)^{2^n - 1} + (2^n - 1)^{2^n + 1} \equiv 0 \mod 2^{n+1}. </cmath> Since there are infinitely many powers of <math>2</math> and since all ordered pairs <math>(2^n - 1, 2^n+1)</math> contain relatively prime integers, we are done.
 +
<math>\boxed{}</math>
 +
 +
-fidgetboss_4000
 +
 +
==See Also==
 +
{{USAMO newbox|year= 2017|before=First Problem|num-a=2}}

Revision as of 20:25, 18 June 2020

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.

Solution 4

I claim that the ordered pair $(2^{n} - 1, 2^{n} + 1)$ satisfies the criteria for all $n \geq 2.$ Proof: It is easy to see that the order modulo $2^{n+1}$ of $(2^{n} - 1)^k$ is $2,$ since $(2^{n} - 1)^2 = 2^{2n} - 2 \cdot 2^{n} + 1 \equiv 1 \mod 2^{n+1}.$, and we can assert and prove similarly for the order modulo $2^{n+1}$ of $(2^{n} + 1)^k.$ Thus, the remainders modulo $2^{n+1}$ that the sequences of powers of $2^{n} - 1$ and $2^{n} + 1$ generate are $1$ for even powers and $2^{n} - 1$ and $2^{n} + 1$ for odd powers, respectively. Since $2^{n} + 1$ and $2^{n} - 1$ are both odd for $n \geq 2,$ \[(2^n + 1)^{2^n - 1} + (2^n - 1)^{2^n + 1} \equiv 0 \mod 2^{n+1}.\] Since there are infinitely many powers of $2$ and since all ordered pairs $(2^n - 1, 2^n+1)$ contain relatively prime integers, we are done. $\boxed{}$

-fidgetboss_4000

See Also

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