Divisibility Of Power Differences

The prime factorization of $7!$ is $5040 = 2^4\cdot 3^2 \cdot 5 \cdot 7.$

Suppose $7!$ divides $x^a-x^b = x^b(x^{a-b}-1)$ for all $x.$ If $x=2,$ the highest power of $2$ dividing $7!$ must be $\le$ the highest power of $2$ dividing $2^b(2^{a-b}-1).$ Since $2^{a-b}-1$ is odd, and the highest power of $2$ dividing $7!$ is $2^4,$ we must have $b\ge 4.$

Now, if $\text{gcd}(x,7!)=1,$ then $7!|x^b(x^{a-b}-1)$ implies $7!|(x^{a-b}-1).$ Since this is true for all such $x,$ then $a-b$ must be divisible by the Carmichael number $\lambda(7!).$ By properties of $\lambda$ (which follow from the Chinese Remainder Theorem and some facts about the arithmetic modulo prime powers--details are in the wiki ), $\lambda(7!)=\text{lcm}(\lambda(2^4),\lambda(3^2),\lambda(5),\lambda(7)) = \text{lcm}(4,6,4,6) = 12.$ So $12|(a-b).$

The smallest $a$ and $b$ that satisfy these requirements is $a=16,b=4.$ All that remains is to prove that this choice of $a$ and $b$ satisfy the conditions of the problem.

For any integer $x,$ consider the prime powers $p^k$ in the factorization of $7!.$ If $p|x,$ then $p^k|x^4$ (because $k\le 4).$ If $p \nmid x,$ then $x^{12} \equiv 1 \pmod{p^k}$ because $\lambda(p^k)$ divides $12,$ as seen above. So in all cases, either $x^4$ or $x^{12}-1$ is divisible by $p^k.$ So $p^k|x^4(x^{12}-1).$ Since this is true for all the $p^k,$ their product divides $x^4(x^{12}-1)$ as well, so $7!|x^4(x^{12}-1).$

So the answer is $16+4 = \fbox{20}.$

In general, if we replace $7!$ by any positive integer $n,$ then the minimum $a+b$ is obtained by letting $b$ equal the maximum exponent of all the primes in the factorization of $n,$ and $a = b+\lambda(n).$ The proof is exactly the same.