# Thread: problem from some USA contest in 1977.

1. ## problem from some USA contest in 1977.

I just can't solve the problem or find the solution online:
1.(1977.)Let a and b be 2 solutions of .Prove that is the solution of

I tried Vietes formulas and I tried to divide the second equation with $x-a*b$

Thanks

2. Originally Posted by myro111
I just can't solve the problem or find the solution online:
1.(1977.)Let a and b be 2 solutions of .Prove that is the solution of

I tried Vietes formulas and I tried to divide the second equation with $x-a*b$

Thanks
You can certainly do it using Vieta's formulas, but it will be laborious. I don't see any simpler way, though.

Let $a,\, b,\, c,\, d$ be the roots of $x^4+x^3-1=0$. The Vieta formulas tell you that $\sum a = -1$, $\sum ab = \sum abc = 0$ and $abcd=-1$. You want to find the equation whose roots $x_1,x_2,...,x_6$ are $ab,\, ac,\, ad,\, bc,\, bd$ and $cd$.

The coefficients of that equation are again given by the Vieta equations. For example, $\sum x_i = \sum ab = 0$ and $x_1x_2x_3x_4x_5x_6 = (abcd)^3 = -1$. Those two are easy, but the others are harder. For example, $\sum x_ix_j = \sum a^2bc + 3abcd$. To evaluate that, notice that $0 = \Bigl(\sum a\Bigr)\Bigl(\sum abc\Bigr) = \sum a^2bc + 4abcd$, and it follows that $\sum x_ix_j = -abcd = 1$. I haven't even tried to do $\sum x_ix_jx_k$ or $\sum x_ix_jx_kx_l$, but it should be possible, using the standard relations among symmetric polynomials.

3. Can you please do the last part for me because I can't get it.Thanks,and how do I know that the solutios are ab,bc,...?

Thank you.

4. Originally Posted by myro111
Can you please do the last part for me because I can't get it.Thanks,and how do I know that the solutions are ab, bc,...?
You are told that a and b are two of the solutions of the fourth-degree equation $x^4+x^3-1=0$, but you don't know which two. So your solution must be an equation whose roots contain all possible products of pairs of roots. That means that if the four roots of the original equation are a, b, c, d, then the solution equation must have all six products ab, ac, ad, bc, bd, cd as its roots.

I can't do the last part of the problem for you, because I haven't done it myself (and I don't have the energy to try). Maybe someone else can step in here? The link that I gave to the Wikipedia page on symmetric polynomials should have all the background information needed.

There is one additional coefficient of the sixth-degree polynomial that I verified: the coefficient of x should be (the negative of) the sum of the product of the roots taken five at a time. You can check that this is equal to $\sum(abcd)^2ab = (abcd)^2\sum ab = 0$. As I said before, it's the remaining two coefficients (those of $x^3$ and $x^2$) that are going to be more difficult.