Fp1 proof by mathematical induction that a expression is divisible by an interger...

First you show that f(1) is divisible by the interger, then assume f(k) is divisible by the interger.

In my book, for proving f(k+1) is divisible by the interger, they show this in all the examples:

f(k+1) - f(k) = whatever those values are substituted in

does it have to be f(k+1) - f(k) since it doesn't seem to be divisible by the interger for some questions I'm doing. Can it instead be e.g. f(k+1) + f(k), or would that no longer be proof?

Showing f(k+1)+f(k) is divisible by what ever (lets call it n) would work. Since f(k) is divisible by n, and you've show that f(k+1)+f(k) is divisble by n, then f(k+1) = "f(k+1)+f(k)" - f(k) would also be divisible by n; you would need to add that step to your working.

HOWEVER, f(k+1)-f(k) will also be divisible by n, so can you give an example where you think it isn't.

f(n)=7^n+4^n+1
divisible by 6

I get f(k+1) - f(k) = 6x7^k + 3x4^k

but 3 isn't divisible by 6, so I'm stuck.

But $4^k$ is divisible by 2, so $3 \times 4^k$ is divisible by...