# Finding the largest $n \in \mathbb{N}$ for which $n-7$ divides $n^3-7$

I want to find the largest $n \in \mathbb{N}$ for which $n-7$ divides $n^3-7$. In other words, I am looking for the largest $n$ such that $\frac{n^3-7}{n-7}$ is an integer. Can anyone provide me with a hint? Please do not post a full solution.

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Hint: Again, the same idea, but avoiding polynomial division. Write $$n\equiv 7\pmod {n-7}$$ so $$n^3-7 \equiv 7^3-7\pmod {n-7}$$

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Hint: $n^3-7=(n-7)\cdot(n^2+\ldots)+\ldots$

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This answer is equivalent to the one by Hagen von Eitzen. Let $w=n-7$.

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More or less the same hint as Hagen's:

$$n^3-7=\left( n^3-7^3\right)+336 \,.$$

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Here are $\,2\,$ proofs of the key idea. If $\,a\,$ is an integer, $\,f(x)\,$ is an integer coef polynomial, then

$\ n\!-\!a\mid f(n)\iff n\!-\!a\mid f(a),\,$ by $\,{\rm mod}\ n\!-\!a\!:\ f(n)\equiv f(a),\:$ so $\ f(n)\equiv 0\!\iff\! f(a)\equiv 0$

Or, proved more simply, without using congruences, but using the Factor Theorem,

$\ n\!-\!a\mid \color{#c00}{f(n)\!-\!f(a)},\$ so $\ n\!-\!a\mid f(n) = \color{#c00}{f(n)\!-\!f(a)}\!+\!f(a)\iff n\!-\!a\mid f(a)$

Your problem is the special case $\ a = 7,\ \ f(x) = x^3-7.$

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