Difference between revisions of "2011 USAJMO Problems/Problem 1"
Hrithikguy (talk | contribs) (Created page with 'Let <math>2^n + 12^n + 2011^n = x^2</math> <math>(-1)^n + 1 \equiv x^2 \pmod {3}</math>. Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd…') |
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+ | Find, with proof, all positive integers <math>n</math> for which <math>2^n + 12^n + 2011^n</math> is a perfect square. | ||
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Let <math>2^n + 12^n + 2011^n = x^2</math> | Let <math>2^n + 12^n + 2011^n = x^2</math> | ||
<math>(-1)^n + 1 \equiv x^2 \pmod {3}</math>. | <math>(-1)^n + 1 \equiv x^2 \pmod {3}</math>. |
Revision as of 20:43, 28 April 2011
Find, with proof, all positive integers for which
is a perfect square.
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