A sequence of pseudorandom numbers is generated as follows Xo = 4 x₁ = ( 6x₁-1 + 5 ) mod 13 if i > 0 Find x6

If n 2 1, then the Fermat number Fn = 22" + 1 is prime if and only if 3(Fn-1)/2 = -1 mod (Fn) .

O Find 5675 (mod 19)

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?