USAJMO 2011/1

For discussing Olympiad Level Number Theory problems
tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

USAJMO 2011/1

Unread post by tanmoy » Mon Feb 23, 2015 8:32 pm

Find,with proof,all positive integers $n$ for which $2^{n}+12^{n}+2011^{n}$ is a perfect square.
"Questions we can't answer are far better than answers we can't question"

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

Re: USAJMO 2011/1

Unread post by Tahmid » Mon Feb 23, 2015 10:42 pm

if $n=1$ ; then $2^{n}+12^{n}+2011^{n}$ is a perfect square .

take $mod 3$ when $n=2m$ for all $m\in \mathbb{N}$
take $mod 4$ when $n=2m+1$ for all $m\in \mathbb{N}$
in each case a contradiction creates.

so $n=1$ is only the solution .

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

Re: USAJMO 2011/1

Unread post by tanmoy » Tue Feb 24, 2015 12:18 pm

$\text{My proof}$:
Take the whole expression mod $12$. Note that the perfect squares can only be of the form $0, 1, 4$ or $9 (mod 12)$.Note that,$12^{n}$ is always divisible by $12$, so this will be disregarded in this process.There are two cases to consider,either $n$ is odd or even.If $n$ is even, then $2^n \equiv 4 \pmod{12}$ and $2011^n \equiv 7^n \equiv 1 \pmod {12}$. Therefore, the sum in the problem is congruent to $5 \pmod {12},$ which cannot be a perfect square. Now we check the case for which n is an odd number.If $n=1$,then $2^{n} + 12^{n} + 2011^{n}=45^{2}$.Let $n>1$.Then $2^n \equiv 8 \pmod{12}$ and $2011^n \equiv 7^n \equiv 7 \pmod {12}$. Therefore, this sum would be congruent to $3 \pmod {12},$ which cannot be a perfect square.Therefore the only possible value of n such that $2^n+12^n+2011^n$ is a perfect square is $n=1$. :)
"Questions we can't answer are far better than answers we can't question"

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