Perfect squares: 2011 USAJMO #1

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6870 posts

#1 h Apr 28, 2011, 9:55 PM • 41 Y

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Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.

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hrithikguy

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hrithikguy

#2 h Apr 28, 2011, 10:03 PM • 29 Y

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Let $2^n + 12^n + 2011^n = x^2$
$(-1)^n + 1 \equiv x^2 \pmod {3}$ .
Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd.
Proof by Contradiction:
I will show that the only value of $n$ that satisfies is $n = 1$ .
Assume that $n \ge 2$ .
Then consider the equation
$2^n + 12^n = x^2 - 2011^n$ .
From modulo 2, we easily that x is odd. Let $x = 2a + 1$ , where a is an integer.
$2^n + 12^n = 4a^2 + 4a + 1 - 2011^n$ .
Dividing by 4,
${2^{n-2} + 3^n \cdot 4^{n-1} = a^2 + a + \dfrac {1}{4} (1 - 2011^n})$ .
Since $n \ge 2$ , $n-2 \ge 0$ , so $2^{n-2}$ similarly, the entire LHS is an integer, and so are $a^2$ and $a$ . Thus, ${ \dfrac {1}{4} (1 - 2011^n})$ must be an integer.
Let ${ \dfrac {1}{4} (1 - 2011^n}) = k$ . Then we have $1- 2011^n = 4k$ .
$1- 2011^n \equiv 0 \pmod {4}$
$(-1)^n \equiv 1 \pmod {4}$ .
Thus, n is even.
However, I have already shown that $n$ must be odd. This is a contradiction. Therefore, $n$ is not greater than or equal to 2, and must hence be less than 2. The only positive integer less than 2 is 1.

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flare

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flare

#3 h Apr 28, 2011, 10:14 PM • 7 Y

Y by myh2910, samrocksnature, megarnie, son7, Adventure10, Mango247, and 1 other user

Or you could show that $2011^n$ must be congruent to 1 mod 3, and then $12^n$ is congruent to 0 mod 3, and using the possible residues for squares, $2^n$ must be 2 mod 3. This can only happen, though, if $n=1$ . Basically, the crucial step is showing that if a number $n$ is congruent to 1 mod 3, then $n^k$ must be as well (where $k$ is a positive integer).

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BOGTRO

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BOGTRO

#4 h Apr 28, 2011, 11:07 PM • 5 Y

Y by samrocksnature, megarnie, hairtail, Adventure10, Mango247

I took this mod 3, showing that n is odd, then expressed n=2k+1, took mod 8, and showed that the number could never be a perfect square if k was positive, leaving n=1. I'll give a more complete solution later.

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Pina1234

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Pina1234

#5 h Apr 28, 2011, 11:14 PM • 4 Y

Y by samrocksnature, megarnie, Adventure10, Mango247

I said n could not be even because then 2011^n would be a perfect square, and then you would have to add 2*2011^(n/2)+1, which would then have to be 12^n+2^n, which is obviously too small. Then, I used mod 4 to show that it couldn't be any odd number over 1 because any perfect square is 0 or 1 mod 4, so then n=1.

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profmusic

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profmusic

#6 h Apr 28, 2011, 11:14 PM • 4 Y

Y by samrocksnature, megarnie, Adventure10, Mango247

flare wrote:

Or you could show that $2011^n$ must be congruent to 1 mod 3, and then $12^n$ is congruent to 0 mod 3, and using the possible residues for squares, $2^n$ must be 2 mod 3. This can only happen, though, if $n=1$ .

n = 3?

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abcak

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abcak

#7 h Apr 28, 2011, 11:17 PM • 34 Y

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Can't you just go mod 3 then mod 4 done?

Mod 3 implies odd
Mod 4 implies even or 1.

There is approximately 1 number that satisfies both these, which is 1.

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professordad

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professordad

#8 h Apr 28, 2011, 11:17 PM • 3 Y

Y by samrocksnature, megarnie, Adventure10

Pina1234 wrote:

I used mod 4 to show that it couldn't be any odd number over 1.

Forgive my stupidity, but how could you show using $\pmod{4}$ that it can't be any odd number over $1$ ? If we have the equation $2^n + 12^n + 2011^n \pmod{4} \equiv a^2 \pmod{4}$ and $n \geq 2$ , then we are left with $2011^n \equiv a^2 \pmod{4}$ , but isn't it possible to have $2011^n \equiv 1 \pmod{4}$ ?

EDIT: If it was like that then I would probably get a 2 or 3 now yay

This post has been edited 1 time. Last edited by professordad, Apr 28, 2011, 11:22 PM

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hrithikguy

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hrithikguy

#9 h Apr 28, 2011, 11:20 PM • 3 Y

Y by samrocksnature, megarnie, Adventure10

Mod 4, you get
$2^n + (-1)^n \equiv x^2 \pmod {4}$ .
The possible square residues modulo 4 are only 0 and 1.
Thus, n is either 1 or an even number.
But it can't be even if you look modulo 3 at the original equation.
So the answer is n=1

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BarbieRocks

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BarbieRocks

#10 h Apr 28, 2011, 11:22 PM • 4 Y

Y by samrocksnature, megarnie, CircleInvert, Adventure10

$2011^n \cong 3^n (\mod 4)$ but only when n is even this is congruent to 1 mod 4.

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Pina1234

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Pina1234

#11 h Apr 28, 2011, 11:24 PM • 4 Y

Y by samrocksnature, megarnie, Adventure10, Mango247

Well, I already showed that I couldn't be any even number because if it was 2011^n and n was even and you let n=2k, then it would be (2011^k)^2. Then to reach the next perfect square up, you would have to add 2*2011^k, which would then have to be comprised by 144^k+4^k, which is obviously too small.

Thus, n would have to be odd, so it would not be able to be 1 (mod 4) if n was anything other than 1.

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varunrocks

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varunrocks

#12 h Apr 28, 2011, 11:41 PM • 7 Y

Y by mathwiz0803, jchang0313, samrocksnature, megarnie, Adventure10, Mango247, and 1 other user

Long Horrible Solution

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tekgeek

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tekgeek

#13 h Apr 29, 2011, 1:11 AM • 12 Y

Y by Dmute, franchester, Williamgolly, anonman, myh2910, samrocksnature, megarnie, son7, Adventure10, and 3 other users

Let $s^2=2^n+12^n+2011^n$ .

In mod 3, this becomes

$s^2 \equiv 2^n+1 \equiv (-1)^n+1 \; (mod \; 3)$

Perfect squares in mod $3$ are congruent to $0$ or $1$ . If $n$ is even, then $s^2 \equiv 2 \; (mod \; 3)$ , which is not possible. So $n$ must be odd.

Now we take mod 4,

$s^2 \equiv 2^n+3^n \equiv 2^n+(-1)^n \; (mod \; 4)$ . For odd $n\geq 2, \; 4 \: | \: 2^n$ so $s^2 \equiv (-1)^n\equiv -1\equiv 3 \; (mod \; 4)$ . But squares in mod $4$ are $0$ or $1$ .

Thus the only possible solution is $\boxed{n=1}$ . Substituting gives

$2^1+12^1+2011^1=2025={45}^2$

$\boxed{}$

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anchenyao

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anchenyao

#14 h Apr 29, 2011, 10:59 PM • 4 Y

Y by Imayormaynotknowcalculus, samrocksnature, Adventure10, Mango247

To me, this problem seemed like a basic number theory exercise. Just take mod 4 and mod 3, finding a contradiction for n>1, so n=1 is the only solution.

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superpi83

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superpi83

#15 h Apr 29, 2011, 11:24 PM • 4 Y

Y by samrocksnature, megarnie, Adventure10, Mango247

even longer, even more horrible
Gak, why didn't I think of mod 3?!

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gracemoon124

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gracemoon124

#229 h Feb 25, 2024, 9:20 PM

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Taking $n=1$ , the value of the expression is $2025=45^2$ . We claim that this is the only solution.

Suppose that there’s another solution, for $n=k >1$ . Taking modulo $3$ , $(-1)^n+1\equiv 0\implies n$ is odd. Taking modulo $4$ , it’s $(-1)^n$ . Since $n$ is odd, the expression is $3\pmod 4$ , which is impossible (the only quadratic residues mod $4$ are $0$ and $1$ ). We are done. $\square$

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K124659

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K124659

#230 h Feb 25, 2024, 10:04 PM

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Would Fermat's last theorem be any help here?

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ostriches88

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ostriches88

#231 h Mar 18, 2024, 4:58 PM

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sol

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Cusofay

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Cusofay

#232 h Mar 26, 2024, 5:38 PM

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First we can see that the sum is a perfect square for $n=1$ . Now if $n\geq 2$ , taking the sum modulo $3$ implies that $n$ is odd but mod $4$ implies $n$ even which is not possible.

$\mathbb{Q.E.D.}$

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chenghaohu

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chenghaohu

#233 h Mar 31, 2024, 2:51 AM

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I claim that the only integer which work is n =1. n =1 results in 2025 = 45 squared.

To show that it is the only such number, we show that for n>2, there are no solutions. Take mod 4, we find that the expression is equivalent to (-1)^n mod 4. The only QRs mod 4 are 1 and 0, so n is a multiple of 2.

Then we take mod 3. It becomes equivalent to (-1)^n + 1^n mod 3. Since n is even from the previous statement, (-1)^n + 1^n is congruent to 2 mod 3. But 2 is not a QR mod 3, therefore all n>2 doesn't work.

In conclusion we get that n = 1 is the only positive integer which makes 2^n + 12^n + 2011^n a perfect square.

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a_0a

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a_0a

#234 h Jul 2, 2024, 3:40 PM

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Solution

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gladIasked

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gladIasked

#235 h Aug 16, 2024, 2:44 AM

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Note that when $n=1$ , $2+12+2011=2025=45^2$ . Now assume $n\ge 2$ . Taking modulo $4$ on both sides tells us that $(-1)^n\equiv 1\pmod 4$ , implying that $n$ is even.
Taking modulo $3$ on both sides tells us that $(-1)^n + 1\equiv 0\pmod 3\implies (-1)^n\equiv -1\pmod 3$ . Thus, $n$ must be odd, a contradiction. Therefore, $\boxed{n=1}$ is the only solution. $\blacksquare$

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Eka01

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Eka01

#236 h Sep 13, 2024, 10:50 AM

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For $n=1$ , the expression is $2025$ which is $45^2$ .
We prove that no other $n$ work.
First notice that $2^n +12^n+2011^n \equiv 0,2(mod \ 3)$ when $n$ is odd and even respectively.
Now for $n \geq 2$ ,
$2^n+12^n+2011^n \equiv 3,1(mod \ 4)$ when $n$ is odd and even respectively.
For $n$ to be a square, it must be $0,1(mod \ 3)$ and $0,1(mod \ 4)$ but all of these conditions cannot be true simultaneously for an integer greater than $1$ . Hence $\boxed{n=1}$ is the only solution.

This post has been edited 2 times. Last edited by Eka01, Sep 13, 2024, 10:51 AM

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pie854

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pie854

#237 h Nov 16, 2024, 8:14 AM

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Note that $2+12+2011=2025=45^2$ , so $n=1$ works. Suppose $n\geq 2$ , considering mod 4 we can find that $n$ must be even. Say $n=2k$ then $2^{2k}+12^{2k}+2011^{2k}\equiv 4^k+1\pmod {12}$ . The quadratic residues mod 12 are $0,1,4,9$ . It's quite clear that none of them are possible.

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blueprimes

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blueprimes

#238 h Dec 18, 2024, 5:11 PM

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For $n \ge 2$ , note that $2^n + 12^n + 2011^n \equiv (-1)^n \pmod{4}$ , and since $-1$ is not a quadratic residue $\pmod{4}$ we have $n$ is even. But then $2^n + 12^n + 2011^n \equiv (-1)^n + 1 \equiv 2 \pmod{3}$ which fails. It remains to check $n = 0, 1$ , to which $n = 1$ is the only solution by inspection. We are done.

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Jupiterballs

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Jupiterballs

#239 h Dec 27, 2024, 1:27 PM

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Why is my solution different then the everyone?

By checking $\pmod{10}$ , we get the $n = 4k+1$ for some positive integer $k$
Now, by doing $16^k\cdot 2 + 12^{4k}\cdot 12 + 2011^{4k+1} \equiv 0+0+3 \equiv 3 \pmod{8}$
which is a contradiction as perfect squares are $\equiv 0,1,4 \pmod{8}$
This means that $k=0$ is the only solution.
therefore, $\fbox{n = 4(0) + 1 = 1 is the only solution}$

This post has been edited 1 time. Last edited by Jupiterballs, Dec 27, 2024, 1:27 PM

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Scilyse

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Scilyse

#240 h Jan 30, 2025, 6:24 PM

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The answer is $n = 1$ only, which works as $2^1 + 12^1 + 2011^1 = 2025 = 45^2$ . Indeed, for $n \geq 2$ even note that $2^n + 12^n + 2011^n \equiv 2^n + 1^n \equiv 2 \pmod{3}$ which is not a quadratic residue, and for $n \geq 3$ odd note that $2^n + 12^n + 2011^n \equiv 2^n + 3^n \equiv 3 \pmod{4}$ which is also not a quadratic residue.

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alex_xie

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alex_xie

#241 h Feb 20, 2025, 2:33 AM

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We claim that only $n=1$ works.
We can split this question into two cases:

Case 1: $n=1$
If $n=1$ , then $2^n+12^n+2011^n=2+12+2011=2025=45^2$ , which satisfies the condition.

Case 2: $n \ge 2$
If $n \ge 2$ , we can take our expressions mod 3 and 4.
Mod 3:
$2^n+12^n+2011^n \equiv (-1)^n+1^n \equiv 0$ or $1 \pmod {3}$
From this, we can see that n must be odd.
Mod 4:
$2^n+12^n+2011^n \equiv (-1)^n \equiv 0$ or $1 \pmod {3}$
From this, we can see that n must be even.
However, that two statements we have contradict, so no numbers greater than or equal to 2 make our expression a perfect square.

Therefore, only $\boxed{n=1}$ satisfies our condition.

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LeYohan

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LeYohan

#242 h Mar 9, 2025, 12:44 AM

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We claim that $n = 1$ is our only answer, as $2 + 12 + 2011 = 2025 = 45^2$ .

Assuming $n \geq 2$ , we have that $2^n + 12^n \equiv 0 \pmod {4}$ and $2011^n \equiv 3^n \pmod {4}$ .
For the expression to be square it would need to be congruent to $1$ or $0$ $\pmod {4}$ , so $2011^n \equiv 3^n \equiv 1 \pmod {4}$ , which only happens when $n$ is even.

Now using $\pmod {3}$ we have that $12^n \equiv 0 \pmod {3}$ and $2^n + 2011^n \equiv 2^n + 1^n \equiv 2^n + 1 \pmod {3}$ . Remembering that $n$ is even, we get that
$2^n + 1 \equiv 2^{2k} + 1 \equiv 1 + 1 \equiv 2 \pmod {3}$ , so the whole expression is congruent to $2 \pmod {3}$ , however $2$ is not a cuadratic residue $\pmod {3}$ , so we get a contradiction. $\square$

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de-Kirschbaum

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de-Kirschbaum

#243 h Mar 27, 2025, 2:27 AM

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First note that when $n=1$ we have $2+12+2011=2025=45^2$ so clearly $n=1$ works. Thus, we will consider only the cases where $n \geq 2$ . Under mod $4$ we have $2^n+12^n+2011^n \equiv 3^n \equiv x^2 \mod{4}$ where $x^2 \in \{0,1\}$ . If $n=2$ then $9 \equiv 1 \mod{4}$ so we must have $n \geq 3$ .

Note that $2^n+12^n+2011^n$ is odd, so $x^2 \equiv 1 \mod{8}$ . Thus $3^n \equiv 1 \mod{8} \implies n \equiv 0 \mod{2}$ thus $n=2k$ . Substitute and considering mod $3$ we get $4^k+12^{2k}+2011^{2k} \equiv 1+1 \equiv 2 \equiv x^2 \equiv \{0,1\} \mod{3}$ . Thus we have a contradiction, so $n=1$ is the only $n$ that works.

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