A Diophantine Problem For Nitin Kumar

We want $x,y \ge 1$ . There is one obvious solution for $(x,y)$ , namely $(1,1)$ .

For $x \ge 2$ , we want $2^x$ to divide $3^x-1 = 2(1 + 3 + 3^2 + \cdots + 3^{x-1})$ . Thus we must have $1 + 3 + 3^2 + \cdots + 3^{x-1}$ even, and hence $x = 2z$ must be even, where $z \ge 1$ .

It is a simple induction to prove that $2^{2z-1} > 3^z +1$ for all $z \ge 3$ . Thus $2^{2z-1}$ can divide neither $3^z-1$ nor $3^z + 1$ when $z\ge 3$ . But $2^{2z}$ must divide $3^{2z}-1 = (3^z+1)(3^z-1)$ and since $3^z+1$ and $3^z-1$ differ by $2$ , one of these two terms is divisible by $2$ , but not by $4$ . Thus it follows that $2^{2z-1}$ must divide one of $3^z+1$ and $3^z-1$ , so we deduce that $z \le 2$ .

• If $z=1$ then $x=2$ and $2^x=4$ divides $3^x-1=8$ , giving $y=2$ .
• If $z=2$ then $x=4$ and $2^x=16$ divides $3^x-1=80$ , giving $y=5$ .

Thus the only integer solutions are $(x,y) = (1,1)\,,\,(2,2)\,,\,(4,5)$ . There are $\boxed{3}$ solutions.