Math Contest #14 Results and Solution
Solution
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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What is your chance of winning:
p(You Win) = 1/n
, n = number of entries
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List of participants with their entries:
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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Winner draw:
not needed
Congratulations @tonimontana, you won 1 SBI!
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Hi @quantumdeveloper, a small upvote and a tip.
$trendotoken
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Congratulations @addax, you are successfuly trended the post that shared by @quantumdeveloper!
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