Thanks for that. I have been trying to work on question 6 (x^n - y^n / y - n) but the long line of xs and ys keep catching me off guard. Also with the one about being divisible by 6, Im a bit stuck because after factoring the equation of n = k+ 1, 6 was not a common factor.
Sorry for the late reply!
With the divisibility question, I'm going to put the answer up, but I'm going to give a few hints for Q6.
Skipping the first step, since I'm sure you've done that
<br />\\ \text{Now, prove that } (k+1)^3-(k+1)=6b \ \ (b \in \mathbb{Z}^+)<br />\\ \begin{align*} \text{LHS } &= (k+1)^3-(k+1)<br />\\&= k^3+3k^2+2k<br />\\&= k^3-k+3k^2+3k<br />\\&= 6a+3k(k+1) \ \ \text{(from assumption)}<br />\end{align*})
I think it's safe to assume you got up to here, then realised hang on, that's a 3, not a 6. But consider k(k+1) for a second. In any two consecutive integers (we gave the condition that k was an integer!) one must be even, the other must be odd. That means we can actually express two consecutive integers as 2c(2c+1) for some integer c, or if k was odd, (2c+1)(2c+2) for some integer c. Essentially, what this all means is that k(k+1) must be even, and thus 3k(k+1) is an integer divisible by 6. Does this make sense? I think it's relatively straightforward from here since we now have some integer multiplied by 6
For Q6, here's the hint (which basically gives away the answer, so look at your own discretion):
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