# 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

Notice $n^{2} + 2n + 12 = (n+1)^{2} + 11$. For this expression to be equal to a multiple of 121, $(n+1)^{2} + 11$ would have to equal a number in the form $121x$. Now we have the equation $(n+1)^{2} + 11 = 121x$. Subtracting $11$ from both sides and then factoring out $11$ on the right hand side results in $(n+1)^{2} = 11(11x - 11)$. Now we can say $(n-1) = 11$ and $(n-1) = 11x - 1$. Solving the first equation results in $n=10$. Plugging in $n=10$ in the second equation and solving for $x$, $x = 12/11$. Since $12/11$ *$121$ is clearly not a multiple of 121, \$n^{2} + 2n + 12 can never be a multiple of 121.