## Wednesday, July 15, 2015

### Determine if n²-21n+111 is or is not a perfect square.

Determine if $n^2-21n+111$ is or is not a perfect square.

I have forgotten about how I mentioned in the slideshow # 9 (creative teaching methodology) that I would post back to discuss the question being raised at the end of that presentation...I guess it is better late than never, here it goes the solution:

The heuristic approach that we're going to use to attacking this Olympiad level mathematics question is the technique of "divide into cases".

First case:

When $n\lt 10$:

Note that $n^2-21n+111$ can be rewritten in two ways:

\begin{align*}n^2-21n+111&=(n-10)^2-n+11\\&=(n-10)^2+(11-n)\\&=(10-n)^2+(11-n)\\&=(10-n)^2+\text{always positive}\gt (10-n)^2 \end{align*}

\begin{align*}n^2-21n+111&=(n-11)^2+n-10\\&=(11-n)^2+(n-10)\\&=(11-n)^2+\text{always negative}\lt (11-n)^2 \end{align*}

Combining the results we have:

$(10-n)^2 \lt n^2-21n+111 \lt (11-n)^2$

Since $(11-n)^2$ is the next square after $(10-n)^2$, then it's impossible that $n^2-21n+111$ is a square for $n\lt 10$.

Second case:

When $n\lt 11$:

We tackle this case using pretty much the same way as we just did above, note that $n^2-21n+111$ can be rewritten in two ways:

\begin{align*}n^2-21n+111&=(n-10)^2-n+11\\&=(n-10)^2+(11-n)\\&=(n-10)^2+\text{always negative}\lt (n-10)^2 \end{align*}

\begin{align*}n^2-21n+111&=(n-11)^2+n-10\\&=(n-11)^2+(n-10)\\&=(n-11)^2+\text{always positive}\gt (n-11)^2 \end{align*}

Combining the results we have:

$(n-11)^2 \lt n^2-21n+111 \lt (n-10)^2$

Since $(n-10)^2$ is the next square after $(n-11)^2$, then it's impossible that $n^2-21n+111$ is a square for $n\lt 11$.

Now, we just have to consider the cases when $n=10$ and $n=11$.

When $n=10$, we have:

$n^2-21n+111=(10)^2-21(10)+111=1=1^2$

When $n=11$, we have:

$n^2-21n+111=(11)^2-21(11)+111=1=1^2$

We can conclude by now that $n^2-21n+111$ is a square iff $n=10$ and $n=11$.