Problem 1589

The factorisation $(2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$, where one of the numbers $k$ and $k+1$ is always even, shows that the square of any odd number gives a remainder of $1$ upon division by $8$. The square of an even number not divisible by $4$ gives a remainder of $4$ and the square of an even number divisible by $4$ gives a remainder of $0$. Thus the only possible remainders of squares modulo $8$ are $0$, $1$ and $4$.

Therefore the sum of squares of three odd numbers gives a remainder of $3$ modulo $8$. The sum of squares of one number divisible by $4$, one number divisible by $2$, but not by $4$, and one odd number gives a remainder of $5$. Thus there are infinitely many interesting numbers giving remainders $3$ and $5$ modulo $8$, and therefore also infinitely many products of distinct interesting numbers with a remainder of $3 \cdot 5 \equiv 7 \pmod{8}$. These numbers are indeed special, as three numbers with remainders of $0$, $1$ or $4$ cannot add up to a remainder of $7$ modulo $8$.

The smallest example is obtained by choosing the interesting numbers $1^2 + 3^2 + 5^2$ and $0^2 + 1^2 + 2^2$, whose product is $35 \cdot 5 = 175$.