For which integers $n$ is the expression $n^{2}-6 n-2$ divisible by a) 8 ; b) 9; c) 11 ; d) 121 ?

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a) $n^{2}-6 n-2 \equiv(n+1)^{2}-3(\bmod 8)$, but a square cannot give a remainder of 3 when divided by 8 (see problem $\left.\underline{60685}\right)$.

b) $n^{2}-6 n-2=(n-3)^{2}-2(\bmod 9)$, but a square cannot give a remainder of 2 when divided by 9 (see problem 60685).

c) $n^{2}-6 n-2=(n-3)^{2}(\bmod 11)$.

d) $n^{2}-6 n-2=(n-3)^{2}-11$. If this number is divisible by 121, then $n-3$ is divisible by 11. But then $(n-3)^{2}$ is divisible by 121, while 11 is not.

## Answer

a), b), d) never; c) for $n=3+11 k(k \in \mathbf{Z})$.

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