Is there some trick to manipulating an equation? (adding 0s, multiplying by 1, etc..)

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I have such a hard time doing this sort of thing that it's annoying me. I'm not very mathematically inclined but it frustrates me that a solution with such a small answer takes me more than a page to figure out and at times I HAVE to look up the answer. For example for the following question:

$11^n - 4^n$ is always divisible by 7

Here is how I did it and I most likely made a mistake somewhere along the way too:

base case: 1

$11^1 -4^1 = 7$ so it holds true

let n $\in \mathbb{N}$, Assume $11^n - 4^n$ is divisible by 7. (Induction hypothesis)

$11^{n+1} -4^{n+1}$ = $11^n(11) - 4^n(4)$

=$11^n(10+1) - 4^n(3+1)$ #this here is just me basically hoping that it works out

= $11^n(10) +11^n - 3(4^n) - 4^n$

= $11^n(10) - 3(4^n) + (11^n - 4^n)$

since we know $(11^n - 4^n)$ by the I.H, figure if $11^n(10) - 3(4^n)$ is divisible by 7

$11^n(10) - 3(4^n)$ = $11^n(10) - (2+1)(4^n)$ #again this is me trying something random and hoping it works

= $11^n(10) - 2(4^n) - 4^n$

= $11^n(10) - (10-8)(4^n) - 4^n$

= $11^n(10) -10(4^n) + 8(4^n) - 4^n$

ok so looks like midway into looking at my answer to this question i found a mistake...every question like this i always end up going in circles or go nowhere.. is there some sort of trick to see just WHAT i'm suppose to change in the equation to get a result which satisfies the claim?

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you are trying to end up with a term that is divisible by 7 and a term that contains $11^n-4^n$.

Write $11^{n+1}-4^{n+1} = 11 \cdot 11^n - 4 \cdot 4^n = 7 \cdot 11^n + 4 \cdot 11^n - 4 \cdot 4^n = 7 \cdot 11^n + 4 \cdot (11^n - 4^n )$.

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Use modular arithmetic. Since $11\equiv4\mod7$, $$ 11^{n}-4^{n}\equiv4^{n}-4^{n}\mod7. $$