| 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. |
Problem Moderated by: Sasha |
| Problem 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 < y < 0. |
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