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Difference between revisions of "2006 IMO Problems/Problem 4"
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for <math>x = 1,2</math> no solution.
 
for <math>x = 1,2</math> no solution.
 
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
 
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
 +
 +
NOTE: This solution does not seem to be correct. Take e.g. <math>x=4</math>. Then <math>LHS=529</math> and <math>y=\pm23</math>. The argument of the solution is promising, but the following is ''not'' true: Let <math>a</math>, <math>b</math> be co-prime and let <math>c</math>, <math>d</math> be also coprime and additionally let <math>ab = cd</math>. Then it must either be <math>(a,b) = (c,d)</math> or <math>(b,a) = (c,d)</math>.

Revision as of 08:24, 15 August 2013

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

NOTE: This solution does not seem to be correct. Take e.g. $x=4$. Then $LHS=529$ and $y=\pm23$. The argument of the solution is promising, but the following is not true: Let $a$, $b$ be co-prime and let $c$, $d$ be also coprime and additionally let $ab = cd$. Then it must either be $(a,b) = (c,d)$ or $(b,a) = (c,d)$.

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