Question

Prove that for any natural n the value of the expression (4n+5)^2–9 is divisible by 4.

Answer 1

squaring (4 n + 5) ... 16 n^2 + 40 n + 25

subtracting 9 ... 16 n^2 + 40 n + 16

all the terms are divisible by 4

Answer 2

(4n+5)^2–9 = (4n+5)^2-3^2
= (4n+5+3)(4n+5-3)
= (4n+8)(4n-2)
= 4(n+2)(4n-2)
which is divisible by 4