# Difference between revisions of "2011 USAJMO Problems/Problem 1"

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We will first take the expression modulo <math>3</math>. We get <math>2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3</math>. | We will first take the expression modulo <math>3</math>. We get <math>2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3</math>. | ||

− | + | Lemma 1: All perfect squares are equal to <math>0</math> or <math>1</math> modulo <math>3</math>. | |

We can prove this by testing the residues modulo <math>3</math>. We have <math>0^2 \equiv 0 \pmod 3</math>, <math>1^2 \equiv 1 \pmod 3</math>, and <math>2^2 \equiv 1 \pmod 3</math>, so the lemma is true. | We can prove this by testing the residues modulo <math>3</math>. We have <math>0^2 \equiv 0 \pmod 3</math>, <math>1^2 \equiv 1 \pmod 3</math>, and <math>2^2 \equiv 1 \pmod 3</math>, so the lemma is true. | ||

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Now, we take the original expression modulo <math>4</math>. For right now, we will assume that <math>n>1</math>, and test <math>n=1</math> later. For <math>n>1</math>, <math>2^n \equiv 0 \pmod 4</math>, so <math>2^n+12^n+2011^n=-1^n \pmod 4</math>. | Now, we take the original expression modulo <math>4</math>. For right now, we will assume that <math>n>1</math>, and test <math>n=1</math> later. For <math>n>1</math>, <math>2^n \equiv 0 \pmod 4</math>, so <math>2^n+12^n+2011^n=-1^n \pmod 4</math>. | ||

− | + | Lemma 2: All perfect squares are equal to <math>0</math> or <math>1</math> modulo <math>3</math>. | |

We can prove this by testing the residues modulo <math>4</math>. We have <math>0^2 \equiv 0 \pmod 4</math>, <math>1^2 \equiv 1 \pmod 4</math>, <math>2^2 \equiv 0 \pmod 4</math>, and <math>3^2 \equiv 1 \pmod 4</math>, so the lemma is true. | We can prove this by testing the residues modulo <math>4</math>. We have <math>0^2 \equiv 0 \pmod 4</math>, <math>1^2 \equiv 1 \pmod 4</math>, <math>2^2 \equiv 0 \pmod 4</math>, and <math>3^2 \equiv 1 \pmod 4</math>, so the lemma is true. | ||

## Revision as of 13:38, 26 August 2020

Find, with proof, all positive integers for which is a perfect square.

## Solution 1

Let . Then . Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd. Proof by Contradiction: We wish to show that the only value of that satisfies is . Assume that . Then consider the equation . From modulo 2, we easily know x is odd. Let , where a is an integer. . Dividing by 4, . Since , , so similarly, the entire LHS is an integer, and so are and . Thus, must be an integer. Let . Then we have . . . Thus, n is even. However, it has already been shown that must be odd. This is a contradiction. Therefore, is not greater than or equal to 2, and must hence be less than 2. The only positive integer less than 2 is 1.

## Solution 2

If , then , a perfect square.

If is odd, then .

Since all perfect squares are congruent to , we have that is not a perfect square for odd .

If is even, then .

Since , we have that is not a perfect square for even .

Thus, is the only positive integer for which is a perfect square.

## Solution 3

Looking at residues mod 3, we see that must be odd, since even values of leads to . Also as shown in solution 2, for , must be even. Hence, for , can neither be odd nor even. The only possible solution is then , which indeed works.

## Solution 4

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 since the problem is asking for positive integers, is always divisible by 12, so this will be disregarded in this process. If is even, then and . Therefore, the sum in the problem is congruent to , which cannot be a perfect square. Now we check the case for which is an odd number greater than 1. Then and . Therefore, this sum would be congruent to , which cannot be a perfect square. The only case we have not checked is . If , then the sum in the problem is equal to . Therefore the only possible value of such that is a perfect square is .

## Solution 5

We will first take the expression modulo . We get .

Lemma 1: All perfect squares are equal to or modulo . We can prove this by testing the residues modulo . We have , , and , so the lemma is true.

We know that if is odd, , which satisfies the lemma's conditions. However, if is even, we get , which does not satisfy the lemma's conditions. So, we can conclude that is odd.

Now, we take the original expression modulo . For right now, we will assume that , and test later. For , , so .

Lemma 2: All perfect squares are equal to or modulo . We can prove this by testing the residues modulo . We have , , , and , so the lemma is true.

We know that if is even, , which satisfies the lemma's conditions. However, if is odd, , which does not satisfy the lemma's conditions. Therefore, must be even.

However, a number cannot be even and odd at the same time, so this is impossible. Now, we only have to test . We know that , so the only integer is .

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