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The polynomial $f(x)$ has degree $10$. We know also that $f(a)=f(-a)$ for $a\in \{1,2,3,4,5\}$. Prove that $f(r)=f(-r)$ for all $r\in \mathbb R$ (i.e. that the polynomial is even)

Note: this is a radical edit which I hope captures the sense of the original question, which was rather unclear.

Mark Bennet
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Eugene Korotkov
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2 Answers2

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$g(x)=f(x)-f(-x)$ is a polynomial of degree $\le 9$ (since even powers, including the term in $x^{10}$ cancel out during subtraction) and $g(\pm1)=g(\pm2)=g(\pm3)=g(\pm4)=g(\pm5)=0$. Therefore $g$ has $10$ distinct roots i.e. more than its degree, which implies $g \equiv 0 \iff f(x)=f(-x)$.

[ EDIT ]  Alternative proof, based on the same idea: $g(x)=f(x)-f(-x)$ is a polynomial of degree $\deg g\le \deg f = 10$. But $g(\pm1)=g(\pm2)=g(\pm3)=g(\pm4)=g(\pm5)=0$ and also $g(0)=f(0)-f(0)=0$ so $g$ has $11$ distinct roots, which implies $g \equiv 0 \iff f(x)=f(-x)$
dxiv
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If you plug one of the values in the polynomial, you get a linear relation between its $11$ coefficients. If you plug the opposite value to get the same linear relation, with alternating signs. Subtracting the two, every other term vanishes.

In the end, you get a non-singular system of five equations in five unknowns (the coefficients of the odd powers of $x$), which has a zero right-hand side. Hence all these coefficients are zero.