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.

$1000$ can be expressed as a sum of two squares in two different ways: either $1000 = 10^2 + 30^2$ or $1000 = 18^2 + 26^2$. Thus if  $f(x)$ is a Fermat polynomial with $f(0) = 1000$ and $f(x) = [a(x)]^2 + [b(x)]^2$, then $a(0)^2 + b(0)^2 = 1000$ so that (wlog) either $a(0) = 10$ and $b(0) = 30$, or $a(0) = 18$ and $b(0) = 26$.
Writing $a(x) = xp(x) + 10$ and $b(x) = xq(x) + 30$ for $p,q \in \mathbb Z[x]$ yields $$f(x) = a(x)^2 + b(x)^2 = x^2 (p(x)^2 + q(x)^2) + 80x(p(x) + q(x)) + 1000,$$ and writing $a(x) = xp(x) + 18$ and $b(x) = xq(x) + 26$ for $p,q \in \mathbf Z[x]$ yields $$f(x) = a(x)^2 + b(x)^2 = x^2 (p(x)^2 + q(x)^2) + 88x (p(x) + q(x)) + 1000.$$ 
In both cases we have that $\displaystyle \left.\frac{f(x) - 1000}{x}\right|_{x = 0}$ is a multiple of $8$. 
$f(x)$ and $f(x) + 2x$ can't both have this property.
