We shall show that the statement is true for all $f(0)$ divisible by $4$. Let $f(x)=p(x)^2+q(x)^2$. Then $f(0)=p(0)^2+q(0)^2$ implies $4\mid p(0)^2+q(0)^2$, so taking quadratic residues $\pmod{4}$ we have $2\mid p(0)$ and $2\mid q(0)$. Let the coefficients of $p(x)$ and $q(x)$ be $(p_n,p_{n-1},\ldots,p_0)$ and $(q_m,q_{m-1},\ldots,q_0)$ respectively. Then the coefficient of $x$ in $p(x)^2+q(x)^2$ is $2p_0p_1+2q_0q_1$. Since $p_0$ and $q_0$ are both even, this quantity is divisible by $4$. Therefore, it is necessary for a Fermat polynomial to have the coefficient of $x$ divisible by $4$. But, $f(x)+2x$ has the coefficient of $x$ to be $\equiv 2 \pmod{4}$. Therefore it can't be a Fermat Polynomial.

Just for completion : dibyo_99 wrote: Therefore, it is necessary for a Fermat polynomial with $4|f(0)$ dibyo_99 wrote: to have the coefficient of $x$ divisible by $4$. $f(x)+2x$ has $f(0)+2*0=f(0)$ divisible by $4$ but dibyo_99 wrote: the coefficient of $x$ to be $\equiv 2 \pmod{4}$. Contradiction.

Let $p(x)$ be a Fermat polynomial such that $p(0)$ is divisible by 4. Suppose that $p(x) = g(x)^2$$+h(x)^2$ where $g(x)$ and$h(x)$ are polynomials with integer coefficients. Therefore $g(0)^2$ + $h(0)^2$is divisible by 4. Since $g(0)$ and $h(0)$ are integers, their squares are either 1 (mod 4) or 0 (mod 4). It therefore follows that $g(0)$ and $h(0)$ are even. Therefore the coefficients 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(x)$, with $p(0)$ divisible by 4,. Thus if $f(x)$ is a Fermat polynomial with $f(0) = 1000$ then $f(x) + 2x$ cannot be a Fermat polynomial

we can strengthen the problem to mod 11.

@above If we can do that, we can strengthen it to \(\pmod{44}\).

vjdjmathaddict wrote: we can strengthen the problem to mod 11. In fact, we can generalise to any prime which of form 4k+3