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). 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>. | ||
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{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:58, 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
.
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