Prove that 2(6^2n)-5(6^n)-2 is divisible by 10 for all values of n.

At n=0 that looks like 2(6^0)-5(6^0)-2 = -5, a counterexample.

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It works for n=1 and n=2 so let's try to prove it for natural n>0. 

I've been warned off using "mod" in answers so we'll kind of handwave around it.

We substitute n=m+1 so m is a non-negative integer, m >= 0.

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