If x^2 + y^2 = z^2, prove xyz is a multiple of 60 Watch

billydisco
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Ok i dont even know what the methods of proof are, but if i had to literally show this i would simply draw 3-4-5, 5-12-13 triangles

anyhow- can someone show me how to do this algebraically? I was going to say z=(x^2+y^2)^1/2 and use binomial
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Gaz031
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Note that 60=(2^2)(3)(5)=(3)(4)(5) and use modular arithmetic.
As an example let's show xyz is divisible by 4 by case analysis.
1) Suppose z=0 (mod 4). Then xyz=0 (4) trivially.
2) Suppose z=1 (4). Then z^2=1 (4) so x^2+y^2=1 (4). Hence one of x or y must be 0 (4) while the other must be 1 (4). It follows xyz=0 (4).
3) Suppose z=2 (4). Then z^2=4 (4) so x^2+y^2=0 (4). Hence x=y=0 (4) and so xyz=0 (4).
4) Suppose z=3 (4). Then z^2=9=1 (4). See 2).
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jb_sweden
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I.e. one side of every such triplet is divisible by 3, another by 4, and another by 5.
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Chewwy
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(Original post by jb_sweden)
I.e. one side of every such triplet is divisible by 3, another by 4, and another by 5.
what's a side? if you're relating it to triangle, then 5,12,13 is a counter-example to what you're saying...
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generalebriety
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(Original post by chewwy)
what's a side? if you're relating it to triangle, then 5,12,13 is a counter-example to what you're saying...
Not really. 5 * 12 * 13 = 3 * 4 * 5 * 13. I think by side he meant "number"... and the use of the word "another" is certainly wrong... but he has the right idea.
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billydisco
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(Original post by Gaz031)
Note that 60=(2^2)(3)(5)=(3)(4)(5) and use modular arithmetic.
As an example let's show xyz is divisible by 4 by case analysis.
1) Suppose z=0 (mod 4). Then xyz=0 (4) trivially.
2) Suppose z=1 (4). Then z^2=1 (4) so x^2+y^2=1 (4). Hence one of x or y must be 0 (4) while the other must be 1 (4). It follows xyz=0 (4).
3) Suppose z=2 (4). Then z^2=4 (4) so x^2+y^2=0 (4). Hence x=y=0 (4) and so xyz=0 (4).
4) Suppose z=3 (4). Then z^2=9=1 (4). See 2).
mod 4???
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quinny132
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yup i agree fully.:cool:
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generalebriety
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(Original post by love2learn7)
mod 4???
Mod = modulo. It's a system of counting where the numbers repeat themselves every certain interval. For example, clocks count mod 12 - they don't count the exact number of hours that have gone past since the first time it was ever 12 o'clock, they count it since the last time it was 12 o'clock. So any number n = n ± km (mod m). For example, 0 = 4 = 8 = 12 = 444 = -16 (mod 4). (It often helps to visualise it in this 'clock' way, but starting at 0, not 12. So counting up from 0, mod 12, would go: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, ...)

This type of arithmetic often helps to test for divisibility. Since the natural (counting numbers), mod 4, can easily be written:
0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2...
(0 1 2 3 4 5 6 7...)
then it's obvious that any number n = 0 (mod 4) is divisible by 4. For example, 4 = 0 (mod 4), and 4 is divisible by 4. 27 = 1 (mod 4), so is not divisible by 4 (and will, of course, leave a remainder of 1). What Gaz said is that anything divisible by 60 will be divisible by 3, 4 and 5, and hence xyz = 0 (mod 3), xyz = 0 (mod 4), and xyz = 0 (mod 5). He's proven xyz = 0 (mod 4) [for all x^2 + y^2 = z^2] for you - now try it for mod 3 and 5.

http://en.wikipedia.org/wiki/Modular_arithmetic probably gives a better explanation than me.
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DDCA
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(Original post by Gaz031)
Note that 60=(2^2)(3)(5)=(3)(4)(5) and use modular arithmetic.
As an example let's show xyz is divisible by 4 by case analysis.
1) Suppose z=0 (mod 4). Then xyz=0 (4) trivially.
2) Suppose z=1 (4). Then z^2=1 (4) so x^2+y^2=1 (4). Hence one of x or y must be 0 (4) while the other must be 1 (4). It follows xyz=0 (4).
3) Suppose z=2 (4). Then z^2=4 (4) so x^2+y^2=0 (4). Hence x=y=0 (4) and so xyz=0 (4).
4) Suppose z=3 (4). Then z^2=9=1 (4). See 2).
Yupe. That's right. The complete proof can be found here:
http://www.cut-the-knot.org/pythagor...ripleDiv.shtml

by making use of the fact of the Pythagorean triples. More information on the Pythagorean triples can be found here:

http://mathforum.org/dr.math/faq/faq...g.triples.html
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Christophicus
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(Original post by generalebriety)
Mod = modulo. It's a system of counting where the numbers repeat themselves every certain interval. For example, clocks count mod 12 - they don't count the exact number of hours that have gone past since the first time it was ever 12 o'clock, they count it since the last time it was 12 o'clock. So any number n = n ± km (mod m). For example, 0 = 4 = 8 = 12 = 444 = -16 (mod 4). (It often helps to visualise it in this 'clock' way, but starting at 0, not 12. So counting up from 0, mod 12, would go: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, ...)

This type of arithmetic often helps to test for divisibility. Since the natural (counting numbers), mod 4, can easily be written:
0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2...
(0 1 2 3 4 5 6 7...)
then it's obvious that any number n = 0 (mod 4) is divisible by 4. For example, 4 = 0 (mod 4), and 4 is divisible by 4. 27 = 1 (mod 4), so is not divisible by 4 (and will, of course, leave a remainder of 1). What Gaz said is that anything divisible by 60 will be divisible by 3, 4 and 5, and hence xyz = 0 (mod 3), xyz = 0 (mod 4), and xyz = 0 (mod 5). He's proven xyz = 0 (mod 4) [for all x^2 + y^2 = z^2] for you - now try it for mod 3 and 5.

http://en.wikipedia.org/wiki/Modular_arithmetic probably gives a better explanation than me.
I've never really understood modulo. Can someone explain why it's -16?
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JeremyC
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(Original post by Widowmaker)
I've never really understood modulo. Can someone explain why it's -16?
Rather than (what I would consider to be) the usual, for instance, \large 5 \, \equiv \, 1 \, (\mbox{mod } 4) (where the (often positive) number after the congruent sign is typically chosen to be as close to zero as possible), I believe generalebriety was simply showing that any multiple of four is congruent to any other multiple of four (mod 4).

To write what he wrote in what I would consider to be the usual way, "4 = 8 = 12 = 444 = -16 = 0 (mod 4)" (but with congruent signs).

Hopefully that helps. (although I fear I haven't explained it particularly well)
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Christophicus
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(Original post by fatuous_philomath)
Rather than (what I would consider to be) the usual, for instance, \large 5 \, \equiv \, 1 \, (\mbox{mod } 4) (where the (often positive) number after the congruent sign is typically chosen to be as close to zero as possible), I believe generalebriety was simply showing that any multiple of four is congruent to any other multiple of four (mod 4).

To write what he wrote in what I would consider to be the usual way, "4 = 8 = 12 = 444 = -16 = 0 (mod 4)" (but with congruent signs).

Hopefully that helps. (although I fear I haven't explained it particularly well)
Thanks. Very useful.
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JeremyC
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You're welcome, and thanks for the rep.
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