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.

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

Contents

Problem

Solutions

Solution

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