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Difference between revisions of "2006 IMO Problems/Problem 4"
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===Problem===
 
===Problem===
 
Determine all pairs <math>(x, y)</math> of integers such that <cmath>1+2^{x}+2^{2x+1}= y^{2}.</cmath>
 
Determine all pairs <math>(x, y)</math> of integers such that <cmath>1+2^{x}+2^{2x+1}= y^{2}.</cmath>
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 +
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===Solution===
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<math>x < 0</math>: LHS integer iff <math>x =-1</math>, but then <math>LHS = 2 \neq y^{2}</math>.
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<math>(x,y) = (0,2)</math> is a solution.
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for <math>x = 1,2</math> no solution.
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so assume <math>x > 2</math>. LHS is odd, so writing <math>y = 2n+1</math> gives us <math>2^{x-2}(1+2^{x+1}) = n(n+1)</math>. <math>n,n+1</math> are coprime, and so are <math>2^{x-2}, 1+2^{x+1}</math>. so <math>n = 2^{x-2}, n+1=1+2^{x+1}</math> or vice versa, but both lead to a contradiction

Revision as of 13:28, 2 October 2012

Problem

Determine all pairs $(x, y)$ of integers such that \[1+2^{x}+2^{2x+1}= y^{2}.\]


Solution

$x < 0$: LHS integer iff $x =-1$, but then $LHS = 2 \neq y^{2}$. $(x,y) = (0,2)$ is a solution. for $x = 1,2$ no solution. so assume $x > 2$. LHS is odd, so writing $y = 2n+1$ gives us $2^{x-2}(1+2^{x+1}) = n(n+1)$. $n,n+1$ are coprime, and so are $2^{x-2}, 1+2^{x+1}$. so $n = 2^{x-2}, n+1=1+2^{x+1}$ or vice versa, but both lead to a contradiction

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