Functional equation with the property $P(x+1)=P(x)+2x+1$ 
Find all polynomials with real-valued coefficients for which the following property $$P(x+1)=P(x)+2x+1$$ holds for all $x \in \mathbb{R}.$

This seems to be a functional equation, so the initial approach would be to try some values. For $x=0$ we would have that $$P(1)=P(0)+1.$$
For $x=1$ $$P(2)=P(1)+3.$$
For $x=2$ $$P(3) = P(2)+5.$$
Now if I would know something about $P(0)$ that would seem to be helpful. If I assume that $P(0)=0$ I would get that
$$P(1)=1$$ $$P(2)=4$$ $$P(3)=9.$$
From here it would seem that $P(n)=n^2,$  however I’m not sure if I can make the assumption that $P(0)=0$ or can I?

Big hint: suppose $P(0)=c$ and work towards proving $P(x)=x^2+c$
Solution:
$$P(x+1)=P(x)+2x+1$$
Well our initial assumption from the looks of it that $P(x)=x^2+c$, but we'll work towards proving it's the only solution. First of all:
$$P(x+1)-P(x)=2x+1$$
because $P$ is a polynomial, this directly means that $P(x)=ax^2+bx+c$, because of the well known method of differences (check it on brilliant wiki). It means that the second difference for the values of the polynomial is constant. (one can also know that by differentiating) So
$$a((x+1)^2-x^2)+b((x+1)-x)=2x+1$$
$$\iff 2ax+a+b=2x+1$$
Thus, by equating coefficients, $a=1$ and $b=0$
so
$$P(x)=x^2+c$$
for all $c \in \mathbb{R}$
