14. How many different pairs of integers $x, y$, lying between 1 and 1000, are there such that $x^{2}+y^{2}$ is divisible by 49?
14. If $x^{2}+y^{2}$ is divisible by 49, then $x^{2}+y^{2}$ is also divisible by 7. But $x^{2}$, when divided by 7, can only give remainders of $0, 1, 4$, or 2 (see the solution to problem 13). The remainder from dividing $x^{2}+y^{2}$ by 7 is equal to the sum of the remainders from dividing the numbers $x^{2}$ and $y^{2}$ by 7. But it is easy to check that out of all the sums of any two (the same or different) of the numbers $0, 1, 4$, and 2, only the sum $0+0=0$ is divisible by 7. Therefore, $x^{2}+y^{2}$ is divisible by 7 only when both $x^{2}$ and $y^{2}$ are divisible by 7, i.e., when both $x$ and $y$ are divisible by 7. On the other hand, if $x$ and $y$ are two numbers divisible by 7, then the sum $x^{2}+y^{2}$ is divisible by 49. Thus, the required number is the number of different pairs of positive integers $x$ and $y$ less than 1000 that are divisible by 7.

But $1000=7 \cdot 142+6$, so there are 142 numbers less than 1000 that are multiples of 7. Combining each of the 142 numbers $x$ with each of the 142 numbers $y$, we get a total of $142^{2}$ pairs of $x, y$. 142 of these pairs consist of the same numbers; all the other pairs we counted twice: once as the pair $x, y$ and
the second time as the pair $y, x$. Therefore, the number of different pairs $x, y$ is

$$
\frac{142^{2}-142}{2}+142=\frac{142 \cdot 143}{2}=10153
$$