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

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==Solution 4== | ==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) | + | 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, <math>12^n</math> is always divisible by 12, so this will be disregarded in this process. If <math>n</math> is even, then <math>2^n \equiv 4 \pmod{12}</math> and <math>2011^n \equiv 7^n \equiv 1 \pmod {12}</math>. Therefore, the sum in the problem is congruent to <math>5 \pmod {12}</math>, which cannot be a perfect square. Now we check the case for which <math>n</math> is an odd number greater than 1. Then <math>2^n \equiv 8 \pmod{12}</math> and <math>2011^n \equiv 7^n \equiv 7 \pmod {12}</math>. Therefore, this sum would be congruent to <math>3 \pmod {12}</math>, which cannot be a perfect square. The only case we have not checked is <math>n=1</math>. If <math>n=1</math>, then the sum in the problem is equal to <math>2+12+2011=2025=45^2</math>. Therefore the only possible value of <math>n</math> such that <math>2^n+12^n+2011^n</math> is a perfect square is <math>n=1</math>. |

{{MAA Notice}} | {{MAA Notice}} |

## Revision as of 11:57, 5 June 2016

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

## Contents

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