# Difference between revisions of "1971 Canadian MO Problems/Problem 6"

## Problem

Show that, for all integers $n$, $n^2+2n+12$ is not a multiple of $121$.

## Solutions

### Solution 1

$n^2 + 2n + 12 = (n+1)^2 + 11$. Consider this equation mod 11. $(n+1)^2 + 11 \equiv (n+1)^2 \mod 11$. The quadratic residues $mod 11$ are $1, 3, 4, 5, 9$, and $0$ (as shown below).

If $n \equiv 0 \mod 11$, $(n+1)^2 \equiv (0+1)^2 \equiv 1\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 1 \mod 11$, $(n+1)^2 \equiv (1+1)^2 \equiv 4\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 2 \mod 11$, $(n+1)^2 \equiv (2+1)^2 \equiv 9\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 3 \mod 11$, $(n+1)^2 \equiv (3+1)^2 \equiv 5\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 4 \mod 11$, $(n+1)^2 \equiv (4+1)^2 \equiv 3\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 5 \mod 11$, $(n+1)^2 \equiv (5+1)^2 \equiv 3\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 6 \mod 11$, $(n+1)^2 \equiv (6+1)^2 \equiv 5\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 7 \mod 11$, $(n+1)^2 \equiv (7+1)^2 \equiv 9\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 8 \mod 11$, $(n+1)^2 \equiv (8+1)^2 \equiv 4\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 9 \mod 11$, $(n+1)^2 \equiv (9+1)^2 \equiv 1\mod 11$, thus not a multiple of 11, nor 121.

If $n \equiv 10 \mod 11$, $(n+1)^2 \equiv (10+1)^2 \equiv 0\mod 11$. However, considering the equation $\mod 121$ for $n \equiv 10 \mod 11$, testing $n = 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120$, we see that $(n+1)^2 + 11$ always leave a remainder of greater than $1\mod 121$.

Thus, for any integer $n$, $n^2+2n+12$ is not a multiple of $121$.

### Solution 2

$n^2+2n+12=(n+1)^2+11$, so if this is a multiple of 11 then $(n+1)^2$ is too. Since 11 is prime, $n+1$ must be divisible by 11, and hence $(n+1)^2$ is divisible by 121. This shows that $(n+1)^2+11$ can never be divisible by 121.