Problem 532

Since $\sqrt{2}$ is irrational, neither $m = \sqrt{2} n$ nor $2m = \sqrt{2} n$ can occur for a pair $(m, n)$ of positive integers. Therefore, the pairs $(m, n)$ of positive integers not exceeding 100 split into the following three types depending on the values of $\sqrt{2} n$, $m$ and $2m$:

* Type (1): satisfies $\sqrt{2} n < m$,
* Type (2): satisfies $m < \sqrt{2} n < 2m$,
* Type (3): satisfies $2m < \sqrt{2} n$.

The condition $2m < \sqrt{2} n$ can be restated as $\sqrt{2} m < n$. Therefore, the fact that a pair $(a, b)$ is of type (1) is equivalent to the fact that the pair $(b, a)$ is of type (3). Consequently, the number of pairs belonging to type (1) is the same as the number of pairs belonging to (3).

There are altogether $100^2 = 10000$ pairs $(m, n)$ of positive integers not exceeding 100, and it is given that there are 2958 pairs belonging to type (2). Therefore, the number of pairs belonging to type (1) is
$$\frac{10000 - 2958}{2} = 3521.$$