Problem 4

Determine all pairs

(x,y)

of integers such that

1+2^{x}+2^{2x+1}=y^{2}

Step 6 of 7: Rule out one sign and bound the other

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\varepsilon=1\Longrightarrow m=1\text{ (impossible)},\qquad \varepsilon=-1\Longrightarrow 1+m=2^{x-2}(m^2-8)

Detailed analysis

If $\varepsilon=1$ , then $1-m=2^{x-2}(m^2-8)$ forces $m^2<8$ , hence $m=1$ , but then the equation is false. Thus $\varepsilon=-1$ , and $1+m=2^{x-2}(m^2-8)$ . Since $x\ge3$ , $1+m\ge2(m^2-8)$ , so $2m^2-m-17\le0$ and the positive odd integer $m$ satisfies $m\le3$ . It cannot be $1$ , hence $m=3$ .