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The largest integer say $n$ for which $(n+5)$ divides $n^5+5$ is

$3115$

$3120$

$3125$

$3130$

I just want a hint not the solution.

3 Answers 3

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$n+5$ divides $n^5+5$ if $$ n^5 + 5 \equiv 0 \pmod{n+5} $$ and since $n\equiv -5 \pmod{n+5}$ you're looking for the largest $n$ such that $$ (-5)^5 + 5 \equiv 0 \pmod{n+5} $$ The left-hand side of this evaluates to $-3120$, so $n+5$ must be the largest divisor of $3120$ ...

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First, we have this conculsion:

$n+5$ divides $n^5+5^5$.

It is an application of polynomial remainder theorem.

Now the question becomes simple. Because $n+5$ divides $n^5+5$, $n+5$ must divides $5^5-5$. As a result, the largest value is $n=5^5-5-5=3115$

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HINT:

$\dfrac{n^5+5}{n+5}=n^4-5n^3+25n^2-125n+625-\dfrac{3120}{n+5}$

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    So the answer should be 3115 – 2017-01-25
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    @Resorcinol: Well... Do you have any other option for $n$ such that $\frac{3120}{n+5}$ is integer? – 2017-01-25
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    @Resorcinol: When you submit this, I guess that you might want to show the explicit calculation (using long-division of two polynomials). – 2017-01-25