1968 IMO Problems/Problem 3
Consider the system of equations with unknowns where are real and . Let . Prove that for this system
(a) if , there is no solution,
(b) if , there is exactly one solution,
(c) if , there is more than one solution.
Adding the equations together yields
(a) If , then there is no solution to the quadratic equation , as the determinant is negative. This implies that either for all , or for all . In either case the above summation cannot be 0, which implies that there are no solutions to the given system of equations.
(b) If , then there is exactly one solution to the quadratic equation (let it be ), and either for all , or for all . The only way that the above summation is 0 is if for all . As there is exactly one that makes (namely ), then the only possible solution to the system of equations is . It's not hard to show that this works, so when the system of equations has exactly one solution.
(c) If , then there are exactly two solutions to the quadratic equation . Let the roots be and . If for all , then . This shows that is a solution. We can show that is another solution using the same reasoning, which shows that the equation has more than one solution.
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