Twin Differences of Squares?

Algebra Level 2

$0^2=0$

$1^2=1, 1-0=1$

$2^2=4, 4-1=3, 3-1=\color{#3D99F6}2$

$3^2=9, 9-4=5, 5-3=\color{#3D99F6}2$

$4^2=16, 16-9=7, 7-5=\color{#3D99F6}2$

Is the difference of $n^2-(n-1)^2$ and $(n+1)^2-n^2$ always $2$ for any positive integer $n$ ?

No Yes

1 solution

Lâm Lê
Sep 12, 2020

If you choose an example, say $15$ , $15^2+15+16=16^2$ because $15^2=15\times15, 15\times15+15=15\times16, 15\times16+16=16\times16=16^2$

Now if you replace $15+1(16)$ with $15-1(14)$ it now differs by $2$ from the original.

You can also choose any other positive natural $n$ for this; it always works because $(n+1)-(n-1)$ always equals $2$ !