Thanks, Soroban

After last week's problem, Soroban emailed me a more 'generalized' statement, from which this week's problem is derived.

Let x, y be integers.

Prove: If a + bi = (x + yi)³ then

a² + b² is a perfect cube.

For those of you who solved last week's problem, this one should be easy; if you had trouble with last week's, check the solution, and that should help you solve this one!

View the solution

Thanks, Soroban Solution

Long Solution
(x + yi)³ = x³ + 3x²yi - 3xy² - y³i

a = x³ - 3xy²
b = 3x²y - y³

a² = x⁶ - 6x⁴y² + 9x²y⁴
b² = 9x⁴y² - 6x²y⁴ + y⁶

a² + b² = x⁶ + 3x⁴y² + 3x²y⁴ + y⁶

But this simplifies to: (x² + y²)³

And this completes the proof, since x² + y² is an integer.

Short Solution (supplied by Sasha)
The short solution is truly beautiful in that it proves far more than was required.

(x+yi)n=a+bi

|(x+yi)ⁿ|=|a+bi|
|x+yi|ⁿ=|a+bi|

(√(x²+y²))ⁿ=√(a²+b²)

a²+b²=(x²+y²)ⁿ

This final statement is the generalized statement offered by Soroban, and the problem given is a specific case of this statement where n=3

Complex Proof

Euler's Number

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