$P$ is a polynomial of degree 4. $P(0)=P(1)=1, P(2)=4, P(3)=9, P(4)=16$. Calculate $P(-1)$
$P$ looks a lot like $X^{2}$, so we will study $P(X)-X^{2}$. This polynomial has 4 roots $(1,2,3,4)$. Thus $P(X)-X^{2}=(X-1)(X-2)(X-3)(X-4) Q(X)$, where $Q$ is a polynomial. Let's study the degrees:

$\operatorname{deg}\left(P(X)-X^{2}\right)=4=\operatorname{deg}((X-1)(X-2)(X-3)(X-4) Q(X))=4+\operatorname{deg}(Q) \Longleftrightarrow \operatorname{deg}(Q)=0$ Therefore $Q$ is a constant $c$, which we will determine with $P(0) . P(0)-0^{2}=c(0-$ $1)(0-2)(0-3)(0-4) \Longleftrightarrow c=\frac{1}{4!}=\frac{1}{24}$. So $P(-1)=(-1)^{2}+\frac{(-2) \cdot(-3) \cdot(-4) \cdot(-5)}{4!} \Longleftrightarrow P(-1)=$ $1+5=6$