Prove that ifa≡b(modn)andb≡c(modn), thena≡c(modn). Note that you must use thedivides definition of congruence in your proof, (that is,a≡b(modn)⇐⇒n|(a−b))

amb (mod n) and d (mod n), then a+cmb+d (mod n) (d) bd (mod n). (e) If a = b (mod n), then a + c = b +c (mod n) and ac be (mod n). (f) If a = b (mod n), then a* = bk (mod n) for any positive integer k.

In the following question: " Let a, b ∈ Z. Show that if a ≡ 5 (mod 6) and b ≡ 3 (mod 4), then 4a + 6b ≡ 6 (mod 8)." why do we ignore the -6 when substituting the variables of a and b, if we expand "4a + 6b ≡ 6 (mod 8)" we get (4a+6b)-6=8y, when we substitute we have to end up with the 8y right? Im just confused why the -6 is ignored?

Let x, y ∈Z. Prove that if x≡1(mod 5) and y ≡2(mod5 ), then x^2-y^2≡0(mod 5).  Give 2 examples to show that this scenario either does or does not work. Give the proof with an explanation to each step.