A Fun Problem With Polynomials

[@XJWill1]

This is straightforward using the Newton-Girard identities, which effectively convert three simultaneous high-order equations in three unknowns to three lower-order equations in three unknowns that are separable and easier to solve. e1 = x + y + z e2 = x*y + x*z + y*z e3 = x*y*z p2 = 5 = x^2 + y^2 + z^2 p3 = 9 = x^3 + y^3 + z^3 p5 = 33 = x^5 + y^5 + z^5 Now using the NG identities: p4 = p3*e1 - p2*e2 + p1*e3 = 9*e1 - 5*e2 + e1*e3 p5 = p4*e1 - p3*e2 + p2*e3 {1} 33 = e3 * e1^2 + 5*e3 - 5*e1*e2 - 9*e2 + 9*e1^2 2*e2 = e1*p1 - e0*p2 = e1^2 - 5 {2} e2 = 1/2 *e1^2 - 5/2 3*e3 = e2*p1 - e1*p2 + e0*p3 = e1*e2 - 5*e1 + 9 {3} e3 = 1/3 *e1*e2 - 5/3 *e1 + 3 Solve {1}, {2}, {3} for real values of e1, e2, e3 e1 = 3 e2 = 2 e3 = 0 Vieta's formula t^3 - e1*t^2 + e2*t - e3 = 0 t^3 - 3*t^2 + 2*t = 0 t*(t - 1)*(t - 2) = 0 So {x, y, z} = {0, 1, 2}

[@georgesdermesropian1565]

You made a mistake in factoring at 4: 36

[@bobkurland186]

does it count if you guess the solution?

[@lagomoof]

Knowing the powers of 2 (or rather one more than them) definitely helped with this. All permutations of {0,1,2} immediately work by inspection.

[@Blabla0124]

At 12:22 you set s = 3. How do you know this is the only solution? If you key in the equation for s in desmos, you see there is a root for s = -5.743...(etc) (s=3 seems to be a double root btw)

[@lagomoof]

WolframAlpha reports one solution triplet that is a real negative number and a conjugate complex pair, for six more solutions and then things get really weird. There are 12 further solutions that revolve around three purely complex numbers and their conjugates, but not all combinations are present or else there would be 48, not 12, solutions. All the above complex solutions are apparently roots of a nonic (ninth power) equation, which I think might be bit beyond this channel. Definitely beyond me!

[@Blabla0124]

There is indeed an issue at 4:36 (more people commented about this) and it changes the equations for s and p significantly. The thing I talk about is later on, so of no consequence for the solution. I think this clip needs to be redone (sorry)

[@XJWill1]

@@lagomoof It is easier to use the Newton-Girard identities than to solve the equations directly. Then it comes down to a factorable polynomial of order 5, with a double root. (e1 - 3)^2 * (e1^3 + 6*e1^2 + 2*e1 + 3) = 0 e2 = 1/2 *e1^2 - 5/2 e3 = 1/6 *e1^3 - 5/2 *e1 + 3 There is one real solution set for e1 = 3 , e2 = 2 , e3 = 0 which corresponds to the solution set x, y, z = 0, 1, 2 which actually represents 6 possible solutions if all permutations are considered. The cubic e1 factor results in 3 more sets of solutions (with 6 permutations each). With 3 simultaneous equations for x, y, z of order 2, 3, and 5, we should expect at most 2*3*5 = 30 solutions. That corresponds to a 5th order equation for e1 with 6 permutations each. However, e1 = 3 is a double root so that eliminates 6 solutions. So there are 4 sets of solutions with 6 permutations each. Of the other 3 sets of solutions, one has a real-value and a pair of complex conjugates. The other two are sets of 3 complex-valued numbers.

[@XJWill1]

@@lagomoof It is easier to use the Newton-Girard identities than to solve the equations directly. Then it comes down to a factorable polynomial of order 5, with a double root. (e1 - 3)^2 * (e1^3 + 6*e1^2 + 2*e1 + 3) = 0 e2 = 1/2 *e1^2 - 5/2 e3 = 1/6 *e1^3 - 5/2 *e1 + 3

[@XJWill1]

@@lagomoof There is one real solution set for e1 = 3 , e2 = 2 , e3 = 0 which corresponds to the solution set x, y, z = 0, 1, 2 which actually represents 6 possible solutions if all permutations are considered. The cubic e1 factor results in 3 more sets of solutions (with 6 permutations each).

[@FisicTrapella]

I'm really lost 😅 But I think there's something wrong in 4:36

[@nasrullahhusnan2289]

As the constant in RHS is small, by scrutinizing (x,y,z)=(0,1,2) or their permutations

[@beaumatthews6411]

Brain says 0,1,2

[@phill3986]

👏👏👏 👍👍👍

[@yoav613]

Why?