, 2002

The remainders that a square number can have upon division by 11 are 0, 1, 4, 9, 5 and 3. But since, apart from zero, no complementary remainders modulo 11 occur, both $x^{2}$ and $y^{2}$, and consequently also $x$ and $y$, must be divisible by 11.

Among the numbers from 1 to 1000 there are exactly $[1000/11] = 90$ multiples of 11. Accordingly there are at most $90 \cdot 89 / 2 = 4005$ numbers of the form $x^{2}+y^{2}$ with $x \neq y$ that are divisible by 121, and exactly 90 numbers of the form $x^{2}+x^{2}$ divisible by 121. Consequently there can be at most 4095 numbers of the stated form. Their number is, however, smaller, since there are many numbers that admit different representations as a sum of two squares. Unfortunately the problem setter had not taken this into account when formulating the problem. All the more gratifying was the fact that some participants supplied very interesting approaches to a solution.