1971 Canadian MO Problems/Problem 6

Revision as of 22:06, 14 December 2011 by Airplanes1 (talk | contribs) (Solution)

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 $n \equiv 10 \mod 11$, $(n+1)^2 \equiv (10+1)^2 \equiv 121 \mod 11$, thus a multiple of 11. However, considering the equation $\mod 121$, $(n+1)^2 + 11 \equiv (10+1)^2 + 11 \equiv 121+ 11 \equiv 132 \equiv 11 \mod 121$, thus not a multiple of 121, even though it is a multiple of 11.

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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