Divisors and polynomials

Could someone please help me understand why the following is true and how you work it out.
"There are 4 numbers between 0 and 91 such that 91 divides (x^2)-1".
Likewise can anyone give an example of 2 irreducible polynomials who's sum is reducible.
Thanks (for reference this is for algebra revision)

For the first one, notice x^2-1 is the difference of two squares (x+1)(x-1)
Also 91 has two (prime) factors 7, 13.
Can you take it from there?

Sorry I'm still not sure where to go from here. I just can't work out what part of my course this links to

It may surprise you that I've got no idea how it relates to your course, probably more so than you.

What equation must exist if x^2-1 is divisible by 91?

I don't know if this is what you mean but there exists a constant say m such that 91m=x^2-1

Yes. This is "hard" because its a single quadratic in two integer variables. The obvious thing to try is using factors, so sub in the two relationships given earlier ...

I realise that I'm probably overthinking this and missing something obvious but i cant see where this is going, sorry🙃

Pls substitute the two given relationships in post 2 into your equation.

How to spot where its going? It should be easy to see x^2-1 is the difference of two squares, it's shouting at you to use this fact. Therefore the answer has something to do with what are the two factors that give (x+1)(x-1). 91 has two prime factors 7 & 13, again the two factors thing pops up again. At least write down the next line.

Am I supposed to be replacing (x+1)(x-1) with 7 and 13?

I could only find 8 and 45 as values of m that work. This only gives me two x values opposed to the 4 I need

How did you come up with those values? I'm off bed shortly so ...

I just spotted them from the difference between the two numbers for example m of 8 would give x of 27 and 28=7*4 and 26=13*2. 2*4=8. Are these values incorrect then?