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.
 A: 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.
A: $1000$ 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.
