Question

n^4-6n^3-5n^2+2n=0 有几个正整数解?

Answer 1

我来试试吧...希望LZ能给点分

n^4-6n^3-5n^2+2n=n^4+n^3-7n^3-7n^2+2n^2+2n=n^3(n+1)-7n^2(n+1)+2n(n+1)

=(n^3-7n^2+2n)(n+1)=n(n+1)(n^2-7n+2)=0

故正整数解只有一个 n=1