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.