Find d when d|n2+n-2, d|n3+2n-1 & d=1 (mod 2), d > 1

[kuahji]

Let d=GCD(n²+n-2,n³+2n-1). Find d if d=1(mod 2) & d > 1.

So we know d|n²+n-2 & d|n³+2n-1.

My question is simply this, the professor wrote down hence d|n³+n²-2n, right after what is written above. But I'm just not seeing how you get that combination. I understand how to work the problem, just not that one step & I'm probably just over-looking something really simple.

 

[jeffasinger]

$n^3 + n^2 - 2n = n(n^2 + n -2)$
and d|(n^2 + n - 2)

But this is probably the wrong subforum for that question.

 

[kuahji]

Yes, I meant to post in under homework. I must have been surfing too many forums at once. Thanks though! I knew it was something silly.