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.

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).

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.

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.

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?

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.

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

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

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

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.

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

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.

Gak, why didn't I think of mod 3?!

abcak wrote: 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. Ya that's what I did. I first showed that n=1 worked. Then I said "assume for the sake of contradiction there is a solution greater than or equal to 2...." and continued. But, for some reason, instead of saying implies odd/even i said implies 1,3 mod 4 and 0,2 mod 4 =p

I'm just wondering, can they take off points for being convoluted even if the proof is complete?

Depends on just how convoluted. In general, a correct solution will get the full points.

as in, my super-convoluted proof for the even case. see above.

Again, if the solution is fully correct, in general it will get a perfect score, unless the write-up is particularly bad. The proof above looks fine.

Let $2^n + 12^n + 2011^n = a^2 (a \in \mathbb{Z})$. For all positive integers $n$, we have $2^n \equiv (-1)^n \;(\bmod\; 3)$, $12^n \equiv 0 \;(\bmod\; 3)$, and $2011^n \equiv 1 \;(\bmod\; 3)$. Since $a^2 \equiv 0 \;(\bmod\; 3)$ or $a^2 \equiv 1 \;(\bmod\; 3)$, we must have $(-1)^n \equiv -1 \;(\bmod\; 3)$. This means $n$ is an odd number, so let $n = 2k+1 \, (k \in \mathbb{N})$. We have \begin{align*} 2^{2k+1} + 12^{2k+1} + 2011^{2k+1} &= a^2 \\ \iff 2 \cdot 4^k + 12 \cdot 144^k + 2011 \cdot (2011^2)^k &= a^2 \end{align*}If $k>0$, then $2 \cdot 4^k \equiv 0 \;(\bmod\; 4)$, $12 \cdot 144^k \equiv 0 \;(\bmod\; 4)$, and $2011 \cdot (2011^2)^k \equiv -1 \cdot 1^k \equiv -1 \;(\bmod\; 4)$. However, $a^2 \equiv 0 \;(\bmod\; 4)$ or $a^2 \equiv 1 \;(\bmod\; n)$, so $k>0$ leads to a contradiction. Thus, $k=0$ (we can verify this is possible because $2 \cdot 4^0 + 12 \cdot 144^0 + 2011 \cdot (2011^2)^0 = 2025 = 45^2$), and $n = 2 \cdot 0 + 1 = \fbox{1}$ is the only solution.

I claim that the only possible solution is $n=1$. First, notice that for all odd positive integers $n$ greater than $1$, we have that \[2^n+12^n+2011^n \equiv 0+0+3^n \equiv 3 \mod 4,\]which can never be a perfect square. Similarly, note that for all even positive integers $n$, we have that \[2^n+12^n+2011^n \equiv 1^{\frac{n}{2}}+0+1^n \equiv 2 \mod 3,\]which can also never be a perfect square. Finally, note that $n=1$ yields $2025$, which is a perfect square. Therefore $n=1$ is the only possible solution, finishing the problem.

The only solution is $n=1$. If $n\ge 3$, then $2^n + 12^n + 2011^n \equiv 3^n \pmod{8}$. Now note that odd perfect squares must be $\equiv 1 \pmod{8}$ which forces $n\equiv 0 \pmod{2}$. This means that $2^{2n'} + 12^{2n'} + 2011^{2n'}$ is a perfect square. Now we have $2^{2n'} + 12^{2n'} + 2011^{2n'} \equiv 1 + 0 + 1 =2 \pmod{3}$ which is not possible. Thus $n\le 2$. Now check gives that $n = 2$ fails and that $n=1$ works. Thus it is the only solution.

We claim the answer is $n=1$ If $n \geq 2$ $$2^{n}+12^{n}+2011^{n} \equiv 3^{n} \pmod{4}$$So $n$ cannot be odd. $$2^{n}+12^{n}+2011^{n} \equiv 2^{n}+1 \pmod{3}$$So $n$ cannot be even. Plugging in $n=1$ we achieve $$2^{1}+12^{1}+2011^{1} = 2025 =45^{2}$$

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. $\square$

Would Fermat's last theorem be any help here?

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

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.

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. $\blacksquare$

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.

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.

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.

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,$\fbox{n = 4(0) + 1 = 1 is the only solution}$