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

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 $(n+1)^{2} = 11(11x - 1)$ . 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.

Solution 2

Assume that $n^2+2n+12=121k$ for some integer $k$ then $n^2+2n+(12-121k)=0$ $\begin{align*} x&=\frac{-2\pm\sqrt{4-4(12-121k)}}{2} \\ &=\frac{-2\pm2\sqrt{484k-44}}{2} \\ &=\sqrt{11(11k-1)} \\ \end{align*}$ By the assumption that $n$ is an integer, $11k-1$ must has a factor of $11$ , which is impossible, contradiction.