Polynomial Modulus

Let $f(n) = n^3 - 7$ and $g(n) = n^2 + 3$ .

When we divide these polynomials with $n-4$ , the remainder of such polynomial division will equal to $f(4)$ and $g(4)$ respectively.

Then $f(4) = 4^3 - 7 = 64 - 7 = 57$ and $g(4) = 4^2 + 3 = 19$ .

And because of the given congruence, $n-4 | f(4) = 57$ and $n-4 | g(4) = 19$ .

The only factors that can divide both $57$ and $19$ are $1$ or $19$ .

However, since $n > 5$ , $n - 4 = 19$ .

Therefore, $n = \boxed{23}$ .

Checking the answers for modulus $n - 4 = 23-4 =19$ :

$23^3 - 7 = 12160 = 19\times 640 \equiv 0 \pmod{19}$

$23^2 + 3 = 532 = 19\times 32 \equiv 0 \pmod{19}$

Put $x = n - 4$

Then, $x > 1$ and the second congruence becomes $(x+4)^2 + 3 \equiv 0 \pmod x \\ \left(\text{expanding} \right) \\ \implies 19 \equiv 0 \pmod x \quad(\because x \equiv 0 \pmod x)$

It follows that $x = 19 \implies \boxed{n = 23}$

Note: The first congruence is satisfied for this value of $n$ (this can be seen by substitution).