| 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!
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Problem Moderated by: Douglas |
| Problem 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 |
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