# Thread: Mod 5

1. ## Mod 5

I am having trouble with this homework question and was hoping someone could help. Cheers

Prove that 5|n(n2-1)(n2=1)

2. Hello,

Look for every possibility of n if 5 divides the product.

Try $n \equiv 0 mod 5$, then $n \equiv 1 mod 5$ and so on

3. Originally Posted by asw-88
I am having trouble with this homework question and was hoping someone could help. Cheers

Prove that $5|n(n^2-1)(n^2+1)$ (at least, I think that's what is meant)
If you multiply out the brackets then you get $5|n^5-n$, which is true by Fermat's little theorem.

4. Hello, asw-88!

Prove that: . $5\,|\,n(n^2-1)(n^2+1)$
Let $N \:=\:n(n^2-1)(n^2+1)$

$n$ must one of five possible forms: . $5k-2,\:5k-1,\:5k,\:5k+1,\:5k+2$

$[1]\;n \,= \,5k-2\!:\;\;N \:=\5k-2)([5k-2]^2-1)([5k-2]^2 + 1)" alt="[1]\;n \,= \,5k-2\!:\;\;N \:=\5k-2)([5k-2]^2-1)([5k-2]^2 + 1)" />
. _ . . . . . . . . . . $=\;(5k+2)(25k^2-20x+3)(25k^2-20k + 5)$
. _ . . . . . . . . . . $= \;(5k+2))(25k^2+20k + 3){\color{red}5}(5k^2 - 4k + 1)$ . . . a multiple of 5

$[2]\;n \,=\,5k-1\!:\;\;N \:=\5k-1)([5k-1]^2-1)([5k-1]^2+1)" alt="[2]\;n \,=\,5k-1\!:\;\;N \:=\5k-1)([5k-1]^2-1)([5k-1]^2+1)" />
. . . . . . . . . . . . $= \;(5k-1)(25k^2 - 10k)(25k^2-10k + 2)$
. . . . . . . . . . . . $= \;(5k-1){\color{red}5}k(5k- 2)(25k^2 - 10k+2)$ . . . a multiple of 5

$[3]\;n \,=\,5k\!:\qquad N \:=\5k)([5k]^2 - 1)([5k]^2+1)" alt="[3]\;n \,=\,5k\!:\qquad N \:=\5k)([5k]^2 - 1)([5k]^2+1)" />
. . . . . . . . . . . . $= \;{\color{red}5}k(25k^2 - 1)(25k^2+1)$ . . . a multiple of 5

$[4]\;n\,=\,5k+1\!:\;\;N \;=\;(5k+1)([5k+1]^2-1)([5k+1]^2+1)$
. . . . . . . . . . . . $= \;(5k+1)(25k^2 + 10k)(25k^2+10k + 2)$
. . . . . . . . . . . . $= \;(5k+1){\color{red}5}k(5k + 2)(25k^2 + 10k + 2)$ . . . a multiple of 5

$[5]\;n\,=\,5k+2\!:\;\;N \;=\;(5k+2)([5k+2]^2-1)([5k+2]^2+1)$
. . . . . . . . . . . . $= \;(5k+2)(25k^2 + 20k + 3)(25k^2 + 20k + 5)$
. . . . . . . . . . . . $= \;(5k+2)(25k^2 + 20k+3){\color{red}5}(5k^2 + 4k + 1)$ . . . a multiple of 5

. . . . . . . . . . . . . . . . . Q.E.D.

5. I'm not going to do it, but I suspect it would be relatively simple to do an induction proof of this as well.

-Dan

6. Originally Posted by topsquark
I'm not going to do it, but I suspect it would be relatively simple to do an induction proof of this as well.

-Dan
I'll do the initial condition:
$
5|1^5-1$

$5|0
$

Because I care, I'll even show the next iteration:
$
5|2^5-2$

$5|30
$

The rest of it is almost as easy as this.