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Difference between revisions of "2011 USAJMO Problems/Problem 1"
(→‎Solution)
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==Solution==
 
==Solution==
  
−
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>.
+
Then <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.
 
Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd.
 
Proof by Contradiction:
 
Proof by Contradiction:
Line 11: Line 11:
 
Then consider the equation
 
Then consider the equation
 
<math>2^n + 12^n = x^2 - 2011^n</math>.
 
<math>2^n + 12^n = x^2 - 2011^n</math>.
−
From modulo 2, we easily that x is odd. Let <math>x = 2a + 1</math>, where a is an integer.
+
From modulo 2, we easily know x is odd. Let <math>x = 2a + 1</math>, where a is an integer.
 
<math>2^n + 12^n = 4a^2 + 4a + 1 - 2011^n</math>.
 
<math>2^n + 12^n = 4a^2 + 4a + 1 - 2011^n</math>.
 
Dividing by 4,
 
Dividing by 4,
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Since <math>n \ge 2</math>, <math>n-2 \ge 0</math>, so <math>2^{n-2}</math> similarly, the entire LHS is an integer, and so are <math>a^2</math> and <math>a</math>. Thus, <math> \dfrac {1}{4} (1 - 2011^n})</math> must be an integer.
 
Since <math>n \ge 2</math>, <math>n-2 \ge 0</math>, so <math>2^{n-2}</math> similarly, the entire LHS is an integer, and so are <math>a^2</math> and <math>a</math>. Thus, <math> \dfrac {1}{4} (1 - 2011^n})</math> must be an integer.
 
Let <math> \dfrac {1}{4} (1 - 2011^n}) = k</math>. Then we have <math>1- 2011^n = 4k</math>.
 
Let <math> \dfrac {1}{4} (1 - 2011^n}) = k</math>. Then we have <math>1- 2011^n = 4k</math>.
−
<math>1- 2011^n \equiv 0 \pmod {4}</math>
+
<math>1- 2011^n \equiv 0 \pmod {4}</math>.
 
<math>(-1)^n \equiv 1 \pmod {4}</math>.
 
<math>(-1)^n \equiv 1 \pmod {4}</math>.
 
Thus, n is even.
 
Thus, n is even.
 
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==
 
==Solution 2==

Revision as of 16:13, 5 July 2014

Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.

Solution

Let $2^n + 12^n + 2011^n = x^2$. Then $(-1)^n + 1 \equiv x^2 \pmod {3}$. 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 $n$ that satisfies is $n = 1$. Assume that $n \ge 2$. Then consider the equation $2^n + 12^n = x^2 - 2011^n$. From modulo 2, we easily know x is odd. Let $x = 2a + 1$, where a is an integer. $2^n + 12^n = 4a^2 + 4a + 1 - 2011^n$. 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 $n \ge 2$, $n-2 \ge 0$, so $2^{n-2}$ similarly, the entire LHS is an integer, and so are $a^2$ and $a$. 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 $1- 2011^n = 4k$. $1- 2011^n \equiv 0 \pmod {4}$. $(-1)^n \equiv 1 \pmod {4}$. Thus, n is even. However, I have already shown that $n$ must be odd. This is a contradiction. Therefore, $n$ 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 $n = 1$, then $2^n + 12^n + 2011^n = 2025 = 45^2$, a perfect square.

If $n > 1$ is odd, then $2^n + 12^n + 2011^n \equiv 0 + 0 + (-1)^n \equiv 3 \pmod{4}$.

Since all perfect squares are congruent to $0,1 \pmod{4}$, we have that $2^n+12^n+2011^n$ is not a perfect square for odd $n > 1$.

If $n = 2k$ is even, then $(2011^k)^2 < 2^{2k}+12^{2k}+2011^{2k}$ $= 4^k + 144^k + 2011^{2k} <$ $2011^k + 2011^k + 2011^{2k} < (2011^k+1)^2$.

Since $(2011^k)^2 < 2^n+12^n+2011^n < (2011^k+1)^2$, we have that $2^n+12^n+2011^n$ is not a perfect square for even $n$.

Thus, $n = 1$ is the only positive integer for which $2^n + 12^n + 2011^n$ is a perfect square.


Solution 3

Looking at residues mod 3, we see that $n$ must be odd, since even values of $n$ leads to $2^n + 12^n + 2011^n = 2 \pmod{3}$. Also as shown in solution 2, for $n>1$, $n$ must be even. Hence, for $n>1$, $n$ can neither be odd nor even. The only possible solution is then $n=1$, 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). Use a chart if you believe this isn't true. Note that since the problem is asking for positive integers, $12^n$ is always divisible by 12, so this will be disregarded in this process. If $n$ is even, then $2^n \equiv 4 \pmod{12}$ and $2011^n \equiv 7^n \equiv 1 \pmod {12}$. Therefore, the sum in the problem is congruent to $5 \pmod {12}$, which cannot be a perfect square. Now we check the case for which $n$ is an odd number greater than 1. Then $2^n \equiv 8 \pmod{12}$ and $2011^n \equiv 7^n \equiv 7 \pmod {12}$. Therefore, this sum would be congruent to $3 \pmod {12}$, which cannot be a perfect square. The only case we have not checked is $n=1$. If $n=1$, then the sum in the problem is equal to $2+12+2011=2025=45^2$. Therefore the only possible value of $n$ such that $2^n+12^n+2011^n$ is a perfect square is $n=1$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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