Is it distinct? (Are the Solutions to this Problem Distinct?)

Number Theory Level 3

Let $\overline{XY}$ be a two-digit number such that:

$X > Y$
$\left(\overline{XY}\right)^2 - \left(\overline{YX}\right)^2$ is a perfect square.

How many possible two-digit numbers $\overline{XY}$ are there?

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5 4 2 1 3

1 solution

Christian Daang
Apr 3, 2017

Based from the conditions, $(10X + Y)^2 - (10Y+X)^2 = a^2$ for some a.

$\implies (10X+Y + 10Y+X)(10X+Y - 10Y-X) = a^2 \\ (11)(9)(X+Y)(X-Y) = a^2$

To become the LHS be a perfect square, one should be 11. $\implies X+Y = 11$ since $\max(X-Y) = 9$ .

Since $X > Y$ , $\implies (X, Y) \in \big( (6, 5) , (7, 4) , (8, 3) , (9, 2) \big)$

Since $11(X+Y)$ and 9 are already a perfect square, then, $X-Y$ should also be a perfect square. Base from the set of solution for $(X , Y )$ , there is only one solution that is accepted, and that is $(X, Y) = (6 , 5)$

$\therefore$ there is only $\boxed{one}$ possible two digit number $\overline{XY}$ that exist and that is $65$ .

Is it distinct? (Are the Solutions to this Problem Distinct?)

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