1971 Canadian MO Problems/Problem 6

Problem

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

Solution

$n^2 + 2n + 12 = (n+1)^2 + 11$ . 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 $121$ .

1971 Canadian MO (Problems)
Preceded by Problem 5	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •	Followed by Problem 7

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1971 Canadian MO Problems/Problem 6

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