(In reply to
Something cool I never knew about polynomials (hint) by Jer)
A good tip, permitting an easyish solution:
It was shown before somewhere that if the roots of ax^4+bx^3+cx^2+dx+e be p,q,r,s, then p+q+r+s = b/a (in this case, m/16)
Applying your rule: 1/u+1/v+v+u = m/16
Applying the given rule: 1/v=1/u*t, v=1/u*t^2,u=1/u*t^3, so 1/u+t/u+t^2/u+t^3/u = m/16
If t^3/u=u, then t=u^(2/3)
1/u+u^(2/3)/u+(u^(2/3))^2/u+u = m/16, equivalently u+u^(1/3)+1/u^(1/3)+1/u = m/16
Bearing in mind those cube roots, and that integer under m, u=8 seems a pretty obvious choice.
Then m=16*(8+2+1/2+1/8)=170 and the roots are : 1/8,1/2,2,8, with t=4
Checking:
16(x1/8)(x1/2)(x2)(x8) = (8x1)(2x1)(x2)(x8) = 16x^4(170)x^3+(2*170+17)x^2(170)x+16, the given equation.
Edited on April 5, 2015, 1:42 am

Posted by broll
on 20150405 01:36:32 