Q: Show that there is no polynomial $p_5$ such that $p_5(-1)=1$, $p_5(0)=0$, $p_5(1)=1$, $p_5''(-1)=0$, $p_5''(0)=0$, $p_5''(1)=0$, but that if the first condition is replaced by $p_5(-1)=-1$, then there are infinitely many polynomials.
My attempt was to form a Hermite interpolating polynomial. Although I don't know how to deal with the second derivative data points. The polynomials $H_i$'s that are multiplied with the non-derivative data points in Hermite polynomials came to be
\begin{align*} H_0=& \frac{1}{4}x^2(x-1)^2(-\frac{9}{2}x-\frac{7}{2})\\ H_1=&(1-x^2)^2\\ H_2=&\frac{1}{4}x^2(x+1)^2(4-3x)& \end{align*}
I also determined the $K_i$'s that are multiplied with the first-derivative data points. But since those points aren't given I got a feeling that the $K_i$'s aren't necessary to answer this particular question. So
$$p_5(x)=\sum_{i=0}^2H_i(x)p_5(x_i)+K_i(x)p_5'(x_i)$$
How do proceed form here?
Since $p_5$ is a fifth degree polynomial, $p''_5$ is of third degree and we know its three roots. Then $$p''_5(x)=ax(x-1)(x+1)=ax^3-ax$$ We can now integrate twice to find $p_5$ \begin{align}p'_5(x)&=\frac{ax^4}{4}-\frac{ax^2}{2}+b\\p_5(x)&=\frac{ax^5}{20}-\frac{ax^3}{6}+bx+c\end{align} Since $p_5(0)=0$, then $c=0$. $$p_5(x)=\frac{ax^5}{20}-\frac{ax^3}{6}+bx$$ Now $p_5$ is an odd function where $$p_5(-x)=-p_5(x)$$ If $p_5(-1)=p_5(1)\neq0$, there is no solution. If $p_5(-1)=-p_5(1)$, the number of solutions is infinite.