All primes greater than 3 can be expressed in the form 6n+1 or 6n-1.
For example, 5 = 6 - 1,
7 = 6 + 1,
11 = 2.6 - 1,
13 = 2.6 + 1,
17 = 3.6 - 1
19 = 3.6 + 1,
23 = 4.6 - 1,
and so on.
There are a whole load of numbers for which 6n plus or minus 1 is not prime, but the important point is that 6n plus or minus one covers all of the primes greater than 3.
Consider then
$$N = 7^{\, 6n+1}-6^{\,6n+1}-1=(7\times7^{\,6n})-(6\times6^{\,6n})-1.$$
Now
$$7^{\,6}=117649=2736\times 43 + 1\equiv1(\bmod 43),$$
so
$$(7^{\,6})^{2}, (7^{\,6})^{3},\:\text{and in general}\:(7^{\,6})^{n}=7^{\,6n}\equiv 1(\bmod 43),$$
in which case
$$7\times7^{\,6n}\equiv7\times 1\equiv7(\bmod 43).$$
Similarly, since
$$6^{\,6}=46656=1085\times 43 + 1\equiv1(\bmod 43),$$
$$6\times6^{\,6n}\equiv 6(\bmod43).$$
It follows that
$$N\equiv7-6-1\equiv 0(\bmod 43),$$
that is , there is a remainder of zero if N is divided by 43.
To cover the 6n - 1 case, consider
$$42N =7\times 6(7^{\,6n-1}-6^{\,6n-1}-1)=(6\times7^{\,6n})-(7\times6^{\,6n})-42.$$
Proceeding as above,
$$42N\equiv6-7+1\equiv0(\bmod43),$$
implying that 42N is divisible by 43, and since 42 isn't, it follows that N is.
