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

Question

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.

Answer 1

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.

Answer 2

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 $$p(x)=x^2+k$$

Answer 3

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 $p(x) = x^2 + c$.

Answer 4

$$ \begin{gathered} p(x+1) = p(x) + 2x + 1\quad \Rightarrow \quad \Delta _{\,x} p(x) = 2x + 1\quad \Rightarrow \hfill \\ \Rightarrow \quad \left\{ \begin{gathered} p(x) = \sum\nolimits_x {2x + 1} = x\left( {x - 1} \right) + x + c(x) = x^{\,2} + c(x) \hfill \\ c(x) = \text{whatever}\;\text{periodic}\,\text{function}\,\text{with}\,\text{period}\,1 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $$

Answer 5

$$p(x+1)=p(x)+2x+1$$ 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 $$p(x)=x^2+k_3$$

Answer 6

$$p(n+1)-p(n)=2n+1$$

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:

$$p(x)=x^2+p(0)$$