Prove the following: 𝐹𝑜𝑟 𝑛 ∈ ℤ, 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛,𝑡ℎ𝑒𝑛 𝑛 3 + 2𝑛 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 4.

Answer 1

If N is even, then N = 2k for some integer k.

N^3 + 2n = (2k)^3 + 2(2k) = 8k^3 + 4k = 4(2k^2+ k)

By closure property of multiplication and addition,

2k^2 + k is also an integer.

Therefore, N^3 + 2n = 4(2k^2+k) is divisible by 4.

Answer 2

Thomas

Well if n is even, then n = 2x for some integer x

n³ + 2n = (2x)³ + 2. 2x = 2³x³ + 4x = 8x³ + 4x = 4 ( 2x³+x).

Clearly if x is an integer then so is 2x³+ x. And clearly 4 divides n³ + 2n, regardless of the value of x.

That's the proof.

Mike