Complex Proof

Let x and y be real integers such that x > y > 0. If a + bi is the square of x + yi, prove that a and b are the legs in a pythagorean triple -- that is, if a and b are whole number lengths of the legs in a right triangle, the hypotenuse will also be a whole number. (i is the imaginary unit.)

For 2 bonus points, specify less restrictive conditions on the integers x and y for which the above statement is still true.

View the solution

Complex Proof Solution

(x + yi)²=x²+2xyi+(yi)²=
(x²-y²) + 2xyi = a+bi

Since the imaginary and real coefficients must be equal, we know that a=x²-y² and b=2xy.

Therefore, a²=x⁴-2x²y²+y⁴, and
b²=4x²y²

So a² + b² = x⁴+2x²y²+y⁴, or (x²+y²)². Therefore, a² + b² is always the square of an integer.

However, in order for a and b to be the legs in a pythagorean triple, they must both be positive. Since a=x²-y², we know that x²-y²>0 (or, in other words, |x|>|y|) and xy>0. While x is positive, y must also be positive in order for the product of x and y to be greater than zero, and x must also be greater than y so that its square exceeds that of y. Therefore, while x is positive, x > y > 0, which is the condition defined in the problem.

However, x can be negative - but y must be negative as well, and must be closer to zero than x. It is also possible that x

Toothpick Polygon

Thanks, Soroban

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