The remainder of $x^{p-1}+\ldots+x+1$ when divided by $x-1$ is $p$, so $a=p$ works. It's also the smallest one, since plugging $1$ into $(x-1)f(x)+(x^{p-1}+\ldots+x+1)g(x)=a$ gives $p|a$. Why do we need $p$ to be prime? The remainder when we divide $\frac{x^n-1}{x-1}$ to $x-1$ is always $n$ (since $x^{n-1}+\ldots+x+1=(x-1)(x^{n-2}+2x^{n-3}+\ldots+(n-2)x+(n-1))+n$), and the exact same argument as above shows that $n|a$ in the general case.. Sorry if I'm missing something obvious.

Are we to assume the degree of $f(x),g(x)$ are $\geq 1$?

Fermat's Little Turtle wrote: Are we to assume the degree of $f(x),g(x)$ are $\geq 1$? It doesn't matter.