IMO 2006 · Problem 4

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

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{(x,y)\in\{(0,2),(0,-2),(4,23),(4,-23)\}.} \]

Solution

If \(x<0\), then \[ 1<1+2^x+2^{2x+1}\le 2, \] so the expression cannot be an integer square.

For \(x=0\), the equation gives \(y^2=4\), hence \(y=\pm2\). For \(x=1,2\), it gives respectively \(y^2=11,37\), neither of which is a square.

Now suppose \(x\ge3\). Put \[ n=2^{x-1}\ge4,\qquad Y=|y|. \] The equation becomes \[ Y^2=8n^2+2n+1, \] so \(Y\) is odd and \[ (Y-1)(Y+1)=2n(4n+1). \] Of the two consecutive even integers \(Y-1,Y+1\), exactly one is divisible by \(4\); the other is twice an odd integer. Since \(n\) is a power of \(2\) and \(4n+1\) is odd, it follows that \[ n\mid Y-1\quad\text{or}\quad n\mid Y+1. \]

On the other hand, \[ Y^2-(2n+1)^2=2n(2n-1)>0 \] and \[ (3n)^2-Y^2=n^2-2n-1>0 \] because \(n\ge4\). Thus \[ 2n+1<Y<3n. \] As \(Y\) is an integer, this implies \[ 2n<Y-1<Y+1\le3n. \] The only multiple of \(n\) in \((2n,3n]\) is \(3n\). Therefore the divisibility above forces \[ Y+1=3n. \] Substituting \(Y=3n-1\) into the equation yields \[ 9n^2-6n+1=8n^2+2n+1, \] hence \(n(n-8)=0\). Since \(n>0\), we obtain \(n=8\), so \(x=4\) and \(y=\pm23\).

Finally, \[ 1+2^0+2^1=4,\qquad 1+2^4+2^9=529=23^2, \] so all four listed pairs satisfy the equation.

The solution was checked against the official IMO shortlist solutions.