2122.14 – Divisibility by 16

Show that x4−1{x^4}-1 is divisible by 1616 whenever xx is an odd integer.

Solution

Take a deep breath, and proceed.

Since xx is odd, let x=2k+1,k≥0x = 2k +1, k \geq 0.

x4−1=(2k+1)4−1=(2k+1)2(2k+1)2−1=(4k2+4k+1)(4k2+4k+1)−1=16k4+32k3+24k2+8k=16(k4+2k3)+24k2+8k\begin{align*}{x^4}-1 &= {(2k+1)^4}-1 \\&= {(2k+1)^2}{(2k+1)^2}-1 \\&= (4{k^2}+4k+1)(4{k^2}+4k+1)-1 \\&= 16{k^4}+32{k^3}+24{k^2}+8k \\&= 16({k^4}+2{k^3})+24{k^2}+8k\end{align*}

All we need to worry about is the 24k2+8k=8k(3k+1)24{k^2}+8k = 8k(3k+1) part.

Now, if k is even, we have 8×even×odd=8×(2×something)8 \times \text{even} \times \text{odd} = 8 \times (2 \times \text{something}), so we're OK.

And if k is odd, then 3k+13k + 1 is even, and we have 8×odd×even8 \times \text{odd} \times \text{even}, and we're OK again. Whew.

Or, you could go x4−1=(x2+1)(x2−1)=(x2+1)(x+1)(x−1){x^4}-1 = ({x^2}+1)({x^2}-1)=({x^2}+1)(x+1)(x-1) and proceed in the same way.