the largest integer $n$ for which $n+5$ divivides $n^5+5$?

Question

When we divide $n^5+5$ by $n+5$, we get a remainder $-620$, i.e., $n^5+5=K(n+5)-620$ , now how to proceed further?

Answer 1

It means that $$\frac{n^5+5}{n+5} = K -\frac{620}{n+5}$$Assuming $K$ is an integer when $n$ is, how large can $n+5$ be if we want this right-hand side to be an integer?

Answer 2

Hint $\,{\rm mod}\ n\!+\!5\!:\ \color{#c00}{n\equiv -5}\,\Rightarrow\, P(\color{#c00}n)\equiv P(\color{#c00}{-5})\ $ so $\,n\!+\!5\mid P(n)\iff n\!+\!5\mid P(-5),\ $ by the Polynomial Congruence Rule, for $\,P(x)\,$ any polynomial with integer coefficients.