The problem of this contest was again a set of 2 equations:
Apart from the obvious solution (0, 0, 0) there is also an infinite number of other solutions:
1: x² + y² = z²
2: z = x² + y → x² = z - y
x² in 1: y² - y + z = z² → y² - y + z -z² = 0
→ 3:y = ½ ± √(¼ + z² - z)
(z-½)² = z² -z + ¼ in 3: y = ½ ± (z - ½)
→ Either y = z(this leads to x = 0) or y = 1-z:
y in 2: z = x² + 1 - z → x² + 1 = 2z
For z to be an integer x² and therefor x need to be odd:
x = 2n + 1, n is any integer.
→ z = 2n² + 2n + 1
→ y = -2n² - 2n
So the solutions are:
(0, n, n), (2n + 1, -2n²-2n, 2n²+2n+1), n is any integer.
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p(You Win) = 1/n, n = number of entries
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| Name | solutions found | comment |
|---|---|---|
 @tonimontana | Only one of the general solutions | You found the trivial case of y=z, but not the general case. |
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not needed
Congratulations
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