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Difference between revisions of "2006 IMO Problems/Problem 4"

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===Solution===
 
===Solution===
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Bogus 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>.
 
<math>x < 0</math>: LHS integer iff <math>x =-1</math>, but then <math>LHS = 2 \neq y^{2}</math>.
 
<math>(x,y) = (0,2)</math> is a solution.
 
<math>(x,y) = (0,2)</math> is a solution.

Revision as of 11:42, 7 September 2018

Problem

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


Solution

Bogus 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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