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.