2006 IMO Problems/Problem 4

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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2006 IMO Problems/Problem 4

Problem

Solution