Question

A polynomial is called a Fermat polynomial if it can be written as the sum of the squares of two polynomials with integer coefficients. Suppose that $f\left(x\right)$ is a Fermat polynomial such that $f\left(0\right)=1000.$ Prove that $f\left(x\right)+2x$ is not a Fermat polynomial.

Answer 1

Let $p\left(x\right)$ be a Fermat polynomial such that $p\left(0\right)$ is divisible by $4.$
Suppose that $p\left(x\right)=g(x{)}^2+h(x{)}^2$ where $g\left(x\right)$ and $h\left(x\right)$ are polynomials with integer coefficients.
$\therefore g(0{)}^2+h(0{)}^2$ is divisble by $4.$
Since $g\left(0\right)$ and $h\left(0\right)$ are integers, their squares are either $1$ (mod 4) or $0$ (mod $4$). It therefore follows that $g\left(0\right)$ and $h\left(0\right)$ are even.
$\therefore$ the coefficents of $x$ in $g(x{)}^2$ and in $h(x{)}^2$ are both divisible by $4.$
In particular, the coefficient of $x$ in a Fermat polynomial $p\left(x\right)$ , with $p\left(0\right)$ divisible by $4$, is divisible by $4.$
Thus if $f\left(x\right)$ is a Fermat polynomial with $f\left(0\right)=1000$ then $f\left(x\right)+2x$ cannot be a Fermat polynomial.