Really a fantastic polynomial problem

Since plugging $x=2$ gives the values $P(2)=4$ & $P(3)=16$ and we have the relation $\displaystyle \frac{P(x+1)}{P(x)} =\frac{x+2}{x-1}$

A set of relations show $\displaystyle \frac{P(x)}{P(x-1)} = \frac{x+1}{x-2},\frac{P(x-1)}{P(x-2)}=\frac{x}{x-3},\cdots,\frac{P(3)}{P(2)}=4$ and multiplying all of them we get $\displaystyle P(x) = P(3) \frac{5.6.7.8\cdots (x+1)}{2.3.4\cdots(x-2)} = 4\binom{x+1}{3}$

Thus $P(2016)=5462358720$ follows.

From the equation we see that

$\left( x+2 \right) |P(x+1) \\ \left( x-1 \right) |P(x)$

and from here we can see that

$\left( x+1 \right) |P(x)$ . This means that $1$ and $-1$ are roots of $P(x)$ . Plug in $x = 0$ and we see that $P(0) = 0$ so $0$ is also a root.

Suppose we had another root that is not one of those $3$ . Notice that the equation above indicates that if $r$ is a root then $r+1$ and $r-1$ is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three.

This means that $P(x) = cx(x-1)(x+1)$ . We can use $P(2)^2 = P(3)$ to get $\large\ c = \frac{2}{3}$ . Plugging in $2016$ is now trivial and we see that it is

$\large\ P(2016) = \frac { 2\times 2016\times 2015\times 2017 }{ 3 } = \boxed{5462358720}.$

We have $x-1$ dividing $P(x)$ and $x+2$ dividing $P(x+1)$ , so that $x+1$ divides $P(x)$ as well. If we write $P(x) = (x-1)(x+1)Q(x)$ , then $(x-1)x(x+2)Q(x+1) \; = \; (x+2)(x-1)(x+1)Q(x)$ and hence $xQ(x+1) = (x+1)Q(x)$ , so that $\frac{Q(x+1)}{x+1} \; =\; \frac{Q(x)}{x}$ and hence $Q(x) = Ax$ . Since $P(2)^2 = P(3)$ it follows that $A = \tfrac23$ , so that $P(x) = \tfrac23x(x-1)(x+1)$ , which makes $P(2016) = \boxed{5462358720}$ .

function it();

p=80;

for k=6 to 2016 do

p=p*(k+1)/(k-2);

end

return(p);

Function call: it()

ans22 = 5462358720.00000