Difference between revisions of "2011 USAJMO Problems/Problem 1"
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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>. | 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>. | ||
+ | ==Solution 5== | ||
+ | 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>. | ||
+ | [color=#f00][b]Lemma 1: All perfect squares are equal to <math>0</math> or <math>1</math> modulo <math>3</math>.[/b][/color] | ||
+ | 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 know that if <math>n</math> is odd, <math>-1^n+1^n \equiv 0 \pmod 3</math>, which satisfies the lemma's conditions. However, if <math>n</math> is even, we get <math>2 \pmod 3</math>, which does not satisfy the lemma's conditions. So, we can conclude that <math>n</math> is odd. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | [color=#f00][b]Lemma 2: All perfect squares are equal to <math>0</math> or <math>1</math> modulo <math>3</math>.[/b][/color] | ||
+ | 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 know that if <math>n</math> is even, <math>-1^n \equiv 0 \pmod 4</math>, which satisfies the lemma's conditions. However, if <math>n</math> is odd, <math>-1^n \equiv -1 \equiv 3 \pmod 4</math>, which does not satisfy the lemma's conditions. Therefore, <math>n</math> 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 <math>n=1</math>. We know that <math>2^1+12^2+2011^2=45^2</math>, so the only integer is <math>\boxed{n=1}</math>. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:37, 26 August 2020
Find, with proof, all positive integers for which is a perfect square.
Contents
[hide]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 .
[color=#f00][b]Lemma 1: All perfect squares are equal to or modulo .[/b][/color] 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 .
[color=#f00][b]Lemma 2: All perfect squares are equal to or modulo .[/b][/color] 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 .
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