Fomin D:

How many pairs of natural numbers ( $m, n$ ), each not exceeding 1000, exist such that

$\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} ?$
Consider all pairs $(m, n)$ of natural numbers for which $m+n+1=s \geq 3$. We have $\frac{1}{s}2>\sqrt{2}>1 / 2>1 / s$, and since the number $\sqrt{2}$ is irrational, it falls into exactly one of the specified intervals. Thus, among the pairs $(m, n)$ with a fixed value of $s$, there will be exactly one for which the inequalities specified in the condition are satisfied. At the same time, the intersection point $\left(\frac{s}{1+\sqrt{2}}, \frac{\sqrt{2} s}{1+\sqrt{2}}\right)$ of the lines $y=\sqrt{2} x$ and $y+x=s$ falls into the square with vertices $(m, n),(m+1, n),(m, n+1),(m+1, n+1)$. We are interested in pairs $(m, n)$ where both numbers do not exceed 1000, so the inequalities

must be satisfied

$\frac{\sqrt{2} s}{1+\sqrt{2}}<1001$. From this, $3 \leq s \leq 1001 \cdot \frac{1+\sqrt{2}}{\sqrt{2}}=1708.8 \ldots$. In total,

## Answer

1706 pairs.