Is it a Fermat polynomial? A Fermat polynomial is a polynomial which can be written as the sum of squares of two polynomials with integer coefficients. Let $f(x)$ be a Fermat polynomial such that $f(0)=1000$. Prove that $f(x)+2x$ is not a Fermat polynomial.

Let $f(x) =  [p(x)]^2 + [q(x)]^2$.  We have $p(0)^2 + q(0)^2 = f(0) = 1000 \equiv 0\pmod{4}$, so $p(0)$ and $q(0)$ must both be even.
Consider the coefficient $c_1$ of $x$ in $f(x)$.  It must be $c_1 = 2ap(0) + 2bq(0)$, where $a$ and $b$ are integers (the coefficients of $x$ in $p(x)$ and $q(x)$, respectively).  Since $p(0)$ and $q(0)$ are even, it follows that $c_1$ is divisible by $4$.
Then the coefficient of $x$ in $g(x) = f(x)+2x$ is $c_1+2$ and is not divisible by $4$.  But $g(0) = f(0)$, and so if $g(x)$ were a Fermat polynomial, then the same argument would show that the coefficient of $x$ in $g(x)$ would be divisible by $4$.  Hence $g(x)$ is not a Fermat polynomial.
