In: Advanced Math
Prove that for every n ∈ N:
a) (10^n + 3 * 4^(n+2)) ≡ 4 mod 19, [note that 4^3 ≡ 1 mod 9]
b) 24 | (2*7^(n) + 3*5^(n) - 5),
c) 14 | (3^(4n+2) + 5^(2n+1) [Note that 3^(4n+2) + 5^(2n+1) = 9^(2n)*9 + 5^(2n)*5 ≡ (-5)^(2n) * 9 + 5^(2n) *5 ≡ 0 mod 14]