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.
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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