1973 USAMO Problems/Problem 4
Let , , and be the roots of the cubic polynomial . Let , , and . From this, , , and . Solving each of these, , , and . Thus , , and are the roots of the polynomial . Thus , and there are no other solutions.
Let have roots x, y, and z. Then using our system of equations, so . Thus, at least one of x, y, and z is equal to 1; without loss of generality, let . Then we can use the system of equations to find that as well, and so is the only solution to the system of equations.
Let and Then We have Then one of and has to be 0, and easy to prove the other two are also 0. So is the only solution to the system of equations.
We are going to use Intermediate Algebra Techniques to solve this equation.
Let's start with the first one: . This will be referred as the FIRST equation.
We are going to use the first equation to relate to the SECOND one () and the THIRD one (.
Squaring this equation:
Subtracting this equation from the 2nd equation in the problem, we have , so .
Now we try the same idea with the cubed terms. Cube the first equation:
. Plug in and factor partially:
Now here is the key step. Note that . So we are going to substitute for each of the expressions and we get: (I rearranged it a bit).
Resubstituting in the second and third equation: . So .
So now we have three equations for the elementary symmetric sums of :
Equation 4: (this is also equation 1)
Equation 6: .
If we call the solutions of (Equation 7) , then are the three roots but in some order. Notice that Equation 7 can be factored as , which means that . Therefore are permutations of in some order, which can be only . (This step uses Vieta's formulas)
Therefore, the only solution is .
From the question
Equation 1 :
Equation 2 :
Equation 3 :
From Eq(1) and Eq(2),
Equation 4 :
From Eq(1), Eq(3), and the fact that ,
Equation 5 :
As from Eq(1), ,,
Equation 6 :
Now let there be a cubic function with the roots so from the relation between roots and coefficients,
Equation 7 : (As we don't know the product let the constant be )
As then at ,
Equation 8 :
And from Eq(6),
Equation 9 :
Equation 10 :
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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