WakeUp wrote: Let $P$ be a polynomial of degree $6$ and let $a,b$ be real numbers such that $0<a<b$. Suppose that $P(a)=P(-a),P(b)=P(-b),P'(0)=0$. Prove that $P(x)=P(-x)$ for all real $x$. $P(a)=P(-a)$ $\implies$ $P(x)=(x^2-a^2)(u_4x^4+u_3x^3+u_2x^2+u_1x+u_0)+c$ $P'(0)=0$ $\implies$ $u_1=0$ and $P(x)=(x^2-a^2)(u_4x^4+u_3x^3+u_2x^2+u_0)+c$ $P(b)=P(-b)$ and $b\ne 0$ and $b\ne a$ $\implies$ $u_4b^4+u_3b^3+u_2b^2+u_0=u_4b^4-u_3b^3+u_2b^2+u_0$ and so $u_3=0$ And so $P(x)=(x^2-a^2)(u_4x^4+u_2x^2+u_0)+c$ is even. Q.E.D.

Let $Q(x)=P(x)-P(-x)$. Since $P$ has even degree, $Q(x)$ has degree at most 5. Note that $$Q(a)=Q(-a)=Q(b)=Q(-b)=Q(0)$$Since $Q'(x)=P'(x)+P'(-x)$, $Q'(0)=0$ and $0$ is a doubble root in $Q$. But this means that $Q$ has at least six roots counted with multiplicity and therefore $Q(x)=0$ and $P(x)=P(-x)$ for all $x$.

MF163 wrote: Let $Q(x)=P(x)-P(-x)$. Since $P$ has even degree, $Q(x)$ has degree at most 5. Note that $$Q(a)=Q(-a)=Q(b)=Q(-b)=Q(0)$$Since $Q'(x)=P(x)+P(-x)$, $Q'(0)=0$ and $0$ is a doubble root in $Q$. But this means that $Q$ has at least six roots counted with multiplicity and therefore $Q(x)=0$ and $P(x)=P(-x)$ for all $x$. Sorry if this is trivial, but I'm not good on derivatives, How did you find that $Q'(x)=P(x)+P(-x)$?

Sorry, it should be $Q'(x)=P'(x)+P'(-x)$.

MF163 wrote: Sorry, it should be $Q'(x)=P'(x)+P'(-x)$. Why? Shouldn't it be $Q'(x)=P'(x)-P'(-x)$ ?

Because $(P(-x))'=(-x)'P'(-x)=-P'(-x)$.

MF163 wrote: Let $Q(x)=P(x)-P(-x)$. Since $P$ has even degree, $Q(x)$ has degree at most 5. Note that $$Q(a)=Q(-a)=Q(b)=Q(-b)=Q(0)$$Since $Q'(x)=P'(x)+P'(-x)$, $Q'(0)=0$ and $0$ is a doubble root in $Q$. But this means that $Q$ has at least six roots counted with multiplicity and therefore $Q(x)=0$ and $P(x)=P(-x)$ for all $x$. how can Q has at least six roots?