2122.14 – Divisibility by 16

Take a deep breath, and proceed. {k^4}

Let $x = 2k +1, k\geq0$.

${x^4}-1 = {(2k+1)^4}-1 = {(2k+1)^2}{(2k+1)^2}-1=(4{k^2}+4k+1)(4{k^2}+4k+1)-1$

$=16{k^4}+16{k^3}+4{k^2}+16{k^3}+16{k^2}+4k +4{k^2}+4k +1 - 1$

$=16{k^4}+32{k^3}+24{k^2}+8k = 16({k^4}+2{k^3})+24{k^2}+8k$.

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

Now, if k is even, we have $8$ x even x odd = $8$ x ($2$ x something), so we're OK.

And if k is odd, then $3k + 1$ is even, and we have $8$ x odd x even, and we're OK again. Whew.

OR, you could go ${x^4}-1 = ({x^2}+1)({x^2}-1)=({x^2}+1)(x+1)(x-1)$ and proceed in the same way.