Difference between revisions of "2011 USAJMO Problems/Problem 1"
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However, I have already shown that <math>n</math> must be odd. This is a contradiction. Therefore, <math>n</math> is not greater than or equal to 2, and must hence be less than 2. The only positive integer less than 2 is 1. | However, I have already shown that <math>n</math> must be odd. This is a contradiction. Therefore, <math>n</math> is not greater than or equal to 2, and must hence be less than 2. The only positive integer less than 2 is 1. | ||
-hrithikguy | -hrithikguy | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | If <math>n = 1</math>, then <math>2^n + 12^n + 2011^n = 2025 = 45^2</math>, a perfect square. | ||
+ | |||
+ | If <math>n > 1</math> is odd, then <math>2^n + 12^n + 2011^n \equiv 0 + 0 + (-1)^n \equiv 3 \pmod{4}</math>. | ||
+ | |||
+ | Since all perfect squares are congruent to <math>0,1 \pmod{4}</math>, we have that <math>2^n+12^n+2011^n</math> is not a perfect square for odd <math>n > 1</math>. | ||
+ | |||
+ | If <math>n = 2k</math> is even, then <math>(2011^k)^2 < 2^{2k}+12^{2k}+2011^{2k} </math> <math>= 4^k + 144^k + 2011^{2k} <</math> <math> 2011^k + 2011^k + 2011^{2k} < (2011^k+1)^2</math>. | ||
+ | |||
+ | Since <math>(2011^k)^2 < 2^n+12^n+2011^n < (2011^k+1)^2</math>, we have that <math>2^n+12^n+2011^n</math> is not a perfect square for even <math>n</math>. | ||
+ | |||
+ | Thus, <math>n = 1</math> is the only positive integer for which <math>2^n + 12^n + 2011^n</math> is a perfect square. |
Revision as of 01:08, 17 May 2011
Find, with proof, all positive integers for which
is a perfect square.
Solution
Let
.
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
that satisfies is
.
Assume that
.
Then consider the equation
.
From modulo 2, we easily that x is odd. Let
, where a is an integer.
.
Dividing by 4,
$2^{n-2} + 3^n \cdot 4^{n-1} = a^2 + a + \dfrac {1}{4} (1 - 2011^n})$ (Error compiling LaTeX. Unknown error_msg).
Since
,
, so
similarly, the entire LHS is an integer, and so are
and
. Thus, $\dfrac {1}{4} (1 - 2011^n})$ (Error compiling LaTeX. Unknown error_msg) must be an integer.
Let $\dfrac {1}{4} (1 - 2011^n}) = k$ (Error compiling LaTeX. Unknown error_msg). Then we have
.
.
Thus, n is even.
However, I have already 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.
-hrithikguy
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.