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Suppose that x,y, and z are three integers with x²+y²+z²=2xyz. Prove that x=y=z=0.

User Davor
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Final Answer:

The equation x² + y² + z² = 2xyz has a unique solution: x = y = z = 0. Assuming otherwise, squaring both sides leads to a contradiction, as the left side, a sum of non-negative squares, is always greater than or equal to zero, while the right side is positive for any non-zero x, y, or z. This contradiction proves that the only solution is x = y = z = 0.

Step-by-step explanation:

The given equation x² + y² + z² = 2xyz suggests a connection between the squares of the variables and their product. To prove x = y = z = 0, let's assume the contrary, that there exists a solution where at least one of x, y, or z is non-zero. Without loss of generality, let's consider the case where x is non-zero.

Now, square both sides of the equation to obtain x⁴ + y⁴ + z⁴ + 2x²y² + 2y²z² + 2z²x² = 4x²y²z². Observe that each term on the left side is a sum of non-negative squares. The equality x⁴ + y⁴ + z⁴ ≥ 0 holds true for all x, y, and z. Additionally, the terms 2x²y², 2y²z², and 2z²x² are also non-negative. Therefore, the left side of the equation is always greater than or equal to zero.

However, for any non-zero values of x, y, or z, the right side of the equation is positive (4x²y²z² > 0). This contradiction implies that our assumption of a non-zero solution is incorrect. Therefore, the only solution to x² + y² + z² = 2xyz is x = y = z = 0. This conclusion is reached by demonstrating that any other scenario leads to a contradiction, affirming the uniqueness of the solution.

User Mattyohe
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