I need to show that (n^2) - 1 is a multiple of 24. n is an odd number but not divisible by 3.
I have a general understanding of the problem, but i am getting tripped up on the not divisible by 3 part.
Any help would be much appreciated!
I need to show that (n^2) - 1 is a multiple of 24. n is an odd number but not divisible by 3.
I have a general understanding of the problem, but i am getting tripped up on the not divisible by 3 part.
Any help would be much appreciated!
Hello, dlee426!
Prove that $\displaystyle N \:=\:n^2 - 1$ is a multiple of 24,
where $\displaystyle n$ is an odd number but not divisible by 3.
We have: .$\displaystyle n$ is of the form $\displaystyle 3a \pm 1$, where $\displaystyle a$ is even.
. . That is, $\displaystyle a = 2b$ for some integer $\displaystyle b.$
Then: .$\displaystyle n \:=\:3(2b) \pm 1 \:=\:6b\pm 1$
And: .$\displaystyle N \:=\:n^2-1 \:=\:(6b\pm1)^2 - 1 \:=\:36b^2 \pm12b \:=\:12b(b\pm1)$
We see that $\displaystyle N$ is divisible 12.
Note that $\displaystyle b(b\pm1)$ is the product of two consecutive integers.
. . That is, one of them is even, the other odd.
Hence, the product $\displaystyle b(b\pm1)$ is divisible by 2.
Therefore, $\displaystyle N$ is divisible by 24.