How much can you grow?

When $f(x)$ is divided by $(x-a)$ , $f(a)$ is the remainder . Here $f(n) = n^7+1$ and it is divided by $(n+7) =n-(-7)$ hence $f(-7)$ is the remainder and it's $(-7)^7+1 = -823543 +1 = -823542$

As we want $f(n)$ divisible by $(n+7)$ , the remainder must be divisible by $(n+7)$ , i.e.

$(n+7) | -823542 \implies (n+7) | 823542$

As greatest divisor of $823542$ is $823542$ itself , $n = \boxed{823535}$

It's given that $n+7 \mid n^7+1$ and we know that $n+7 \mid n^7+7^7$

So let's eliminate the $n^7$ :

$n+7 \mid (n^7+7^7)-(n^7+1) \implies n+7 \mid 7^7-1$

Since we want the largest possible value of $n$ , we have to set $n+7 = 7^7-1$

$\implies n = 823543-8 = \boxed{823535}$

Whenever I see a mod in something other than a variable, $n+7$ instead of $n$ in this case, I like to use substitution. So like, set $m=n+7$ and doing this gives me:

$n^7+1 \equiv 0 \mod(n+7) \iff$

$(m-7)^+1 \equiv 0 \mod(m) \iff$

$-7^7+1 \equiv 0 \mod(m) \iff$

$7^7-1 \equiv 0 \mod(m)$

And if we want the largest value for $m$ that makes this true, all we have to do is set $m=7^7-1 \iff n=7^7-8=\boxed{823535}$

This is essentially the same solution as Aditya's except in a different form.

(n^7+1)/(n+7)=( (n+1)(n^6-n^5+n^4-n^3+n^2-n+1)/(n+7)) = n^6- 7n^5 +49n^4 -343n^3+2401n^2-16897n - 833542/(n+7)+117649. So we see that n+7 must be equal to 833542 according to fact that n^7+1 is divisible by n+7. Therefore n = 833535