# 1971 Canadian MO Problems/Problem 6

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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

n^2+2n+12 =(n+1)^2+11 = (n+1)(n+1)+11 Now 11 itself is a multiple of 11, Therefore there are 2 cases for the value of n Case 1: n+1 isn't a multiple of 11 if n isn't a multiple of 11 then (n+1)^2 isn't a multiple of 121 i.e n^2+2n+12 Isn't divisible by 11 Case 2: if n+1 is a multiple of 11 then (n+1)^2 is a multiple of 121, let (n+1)^2=121k But 121k+11 can't be equal to a multiple of 11 hence proved

### Solution 3

In order for $121$ to divide $n^{2} + 2n + 12$, $11$ must also divide $n^{2} + 2n + 12$.

Plugging in all numbers modulo $11$:

$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,$ or $0, 1, 2, 3, 4, 5, (-5), (-4), (-3), (-2), (-1)$ to make computations easier,

reveals that only $10$ satisfy the condition ${n^{2} + 2n + 12} \equiv 0 \pmod{11}$.

Plugging $10$ into ${n^{2} + 2n + 12}$ shows that it is not divisible by $121$.

Thus, there are no integers $n$ such that $n^{2} + 2n + 12$ is divisible by $121$.

~iamselfemployed