Tomi's Challenges - #11

It's possible to infer a lot of additional rules just from the conditions given here. Since we have $\frac{n}{2}$ in the expression with $x$ , it's safe to assume that $n$ is even, otherwise we would be dealing with decimal numbers. Next, plugging in $y=0$ yields that no student shook hands with himself, and plugging in $y=1$ yields that the $k$ -th student never shook hands with the $(k+\frac{n}{2})$ -th student. Analoguously, Plugging in $y=-1$ we get that the $k$ -th student didn't shake hands with the $(k-\frac{n}{2})$ -th student. All other values of $y$ are irrelevant because they give us the same conclusions, and therefore no new information.

If we order the students from the one who came to the classroom first to the one who came last, we can see that no adjacent students shook hands. If we visualise this using an $n$ -gon, we have one more pair of adjacent students - the first one and the last one. The first one and the last one CAN shake hands so we have to count the handshake between the first student and the last student as $1$ handshake.

We can now see that we have to count all diagonals of the $n$ -gon which don't connect the $k$ -th and the $(k-\frac{n}{2})$ -th student, aswell as the $k$ -th and the $(k+\frac{n}{2})$ -th student. Geometrically, we can see that this means that we can't connect two opposite points, and there are $\frac{n}{2}$ diagonals connecting two opposite points (Note: "opposite points" are only defined for even values of $n$ , and $n$ is even by our assumption). We can see that in the above example of an octagon, which is the number of handshakes without counting the one between the first and last student for $n=8$ , no diagonal intersects the orange circle, which means that there are no diagonals connecting two opposite points.

Therefore, the number of handshakes is:

$\Large 1+\frac{n(n-3)}{2}-\frac{n}{2}$ , where $\frac{n(n-3)}{2}$ is the number of diagonals in an $n$ -gon.

Solving the following quadratic equation

$\Large 1+\frac{n(n-3)}{2}-\frac{n}{2}=6497$

yields $n=116$ or $n=-112$ , from which we conclude that $n=\boxed{116}$ .