Find all polynomials $p$ such that $p(x+1)=p(x)+2x+1.$ 
Find all polynomials $p$ such that $p(x+1)=p(x)+2x+1.$

I obtained $p(1)=p(-1).$ I claimed that $p(x)=x^2$ but was unable to prove it. 
Any help will be appreciated. 
 A: Let $p_n = p(n)$, we then have the recurrence
$$p_{n+1} = p_n + (2n+1) \implies p_{n+1} - (n+1)^2 = p_n - n^2$$
Hence, we have
$$p_n = n^2 + k$$
where $k$ is some constant. Since $p(n) = n^2+k$ at all integer points and the fact that $p(n)$ is a polynomial, we have $p(x) = x^2+k$. This is because we then have $p(x)-x^2-k$ has infinitely many roots and the only polynomial having infinitely many roots is the zero polynomial. Hence, we obtain that
$$p(x)=x^2+k$$
A: Let me try. Put $h(x) = p(x) - x^2$. Then you have
$$h(x+1) = h(x)$$, for all $x$. Then, you have $h(x)$ is constant.
Let $h(x) = c$, then $p(x) = x^2 + c$.
A: Hint If $P(x)=a_nx^n+...+a_1x+a_0$ with $a_n \neq 0$ show that 
$$P(x+1)-P(x)=na_{n-1}x^{n-1}+...$$
has degree $n-1$. 
Since in your case $P(X+1)-P(X)=2X+1$ it follows that $P(X)$ is a monic quadratic polynomial. 
Writing $P(X)=X^2+ax+b$ and plugging in the equation, you get the answer.
A: $$
\begin{gathered}
  p(x+1) = p(x) + 2x + 1\quad  \Rightarrow \quad \Delta _{\,x} p(x) = 2x + 1\quad  \Rightarrow  \hfill \\
   \Rightarrow \quad \left\{ \begin{gathered}
  p(x) = \sum\nolimits_x {2x + 1}  = x\left( {x - 1} \right) + x + c(x) = x^{\,2}  + c(x) \hfill \\
  c(x) = \text{whatever}\;\text{periodic}\,\text{function}\,\text{with}\,\text{period}\,1 \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} 
$$
A: $$p(x+1)=p(x)+2x+1$$
derive the equation two times
$$p''(x+1)=p''(x)$$
that means the two sides should be constant 
$$p''(x)=k_1$$
integrate two times to get
$$p(x)=\frac{k_1}{2}x^2+k_2x+k_3\tag1$$
so
$$p(x+1)=\frac{k_1}{2}(x+1)^2+k_2(x+1)+k_3 \tag2$$
subtract $1$ from $2$ 
$$p(x+1)-p(x)=k_1x+\frac{k_1}{2}+k2=2x+1$$
hence
$$k_1=2$$
$$k_2=0$$
so the solution is
$$p(x)=x^2+k_3$$
A: $$p(n+1)-p(n)=2n+1$$
Hence for $x \in \mathbb{N}^{+}$,
$$\sum_{n=0}^{x-1} \left(p(n+1)-p(n) \right)=\sum_{n=0}^{x-1} (2n+1)$$
Yet as we have a telescoping sum,
$$\sum_{n=0}^{x-1} \left(p(n+1)-p(n)\right)=p(x)-p(0)=\sum_{n=0}^{x-1} (2n+1)$$
But  $\sum_{n=0}^{x-1} (2n+1)=x^2$ for $x \in \mathbb{N}^{+}$.
So,
$$p(x)=x^2+p(0)$$
For $x \in \mathbb{N}^{+}$.
But $p(x)$ is a polynomial (finite degree) so by the Rene's comment we must have:
$$p(x)=x^2+p(0)$$
