Would this be a proof?

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Timefly
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Would this be a proof?

Postby Timefly » Sat Dec 04, 2010 10:35 pm UTC

I'm still getting used to the idea of proofs and I'm just wondering would this be a proof that there are infinite Pythagorean triples?

Consider that the numbers are a Pythagorean triple 3, 4 and 5.
When a = 3, b = 4, c = 5
a^2 + b^2 = c^2
as 9 + 16 = 25.

x is a positive integer.

When x = 1 the formula can be rewritten as so.
(3x)^2 + (4x)^2 = (5x)^2
as (3*1)^2 + (4*1)^2 = 3^2 + 4^2 = 9 + 16 = (5*1)^2 = 5^2 = 25.

Now let x be any integer. Consider, the product of integers is an integer and the cardinality of the set of all integers is infinite.
So when x is a positive integer a set of Pythagorean triples can be found by
a = 3x, b = 4x and c = 5x.
As x can be any integer there are at least the same amount of pythagorean triples as there are integers.

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Eastwinn
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Re: Would this be a proof?

Postby Eastwinn » Sat Dec 04, 2010 11:10 pm UTC

Next step: prove that there is an infinite amount of coprime Pythagorean triples...

:D

Edit: By the way, I agree with the post below me. The way you wrote it was neither concise nor particularly clear. But the idea was definitely there.
Last edited by Eastwinn on Sat Dec 04, 2010 11:35 pm UTC, edited 1 time in total.
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gorcee
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Re: Would this be a proof?

Postby gorcee » Sat Dec 04, 2010 11:22 pm UTC

It is, but it is needlessly messy, and invokes terms that don't need to be invoked.

It's simpler to say:

A Pythagorean triple is a set of positive integers (a,b,c) such that [imath]a^2 + b^2 = c^2[/imath]. The most simple Pythagorean triple, (3,4,5), satisfies [imath]3^2 + 4^2 = 5^2[/imath]. Multiply both sides of this equation by [imath]x^2[/imath], where [imath]x[/imath] is some positive integer. Then, a new Pythagorean triple (3x, 4x, 5x) can be generated: [imath](3x)^2 + (4x)^2 = (5x)^2[/imath].

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mdyrud
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Re: Would this be a proof?

Postby mdyrud » Sun Dec 05, 2010 12:42 am UTC

One of my favorite parts of learning proofs in high school was proving that all odd numbers are part of a unique Pythagorean triple during one of my classes when I should have been paying attention. So once you get the hang of some of the basics, see what other stuff you can show with the Pythagorean theorem.

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heyitsguay
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Re: Would this be a proof?

Postby heyitsguay » Sun Dec 05, 2010 12:58 am UTC

mdyrud wrote:One of my favorite parts of learning proofs in high school was proving that all odd numbers are part of a unique Pythagorean triple during one of my classes when I should have been paying attention. So once you get the hang of some of the basics, see what other stuff you can show with the Pythagorean theorem.


32+42=52

52+122=132

Perhaps pay more attention.

kbltd
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Re: Would this be a proof?

Postby kbltd » Sun Dec 05, 2010 3:44 am UTC

Ah, but you had one 52 on the left and one on the right of the equals sign. Maybe mdyrud meant that all odd numbers appear in the summands of a unique, coprime, Pythagorean triple?

Nope...
152+82=172
152+1122=1132

212+202=292
212+2202=2212
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gorcee
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Re: Would this be a proof?

Postby gorcee » Sun Dec 05, 2010 3:53 am UTC

It's pretty easy to show that elements of a Pythagorean triple are non-unique without having to find a counter example.

Any number n that is evenly divisible by both 3 and 4 will appear once in [imath](3x)^2+n^2=(5x)^2[/imath] and again in [imath]n^2 + (4y^2) = (5y^2)[/imath]

Example: n = 12

3*y = 12 -> y = 4.
[imath]12^2+(4\cdot4)^2 = (5\cdot4)^2[/imath]

4*x = 12 -> x = 3.
[imath](3\cdot3)^2+12^2 = (5\cdot3)^2[/imath]
Last edited by gorcee on Sun Dec 05, 2010 4:20 am UTC, edited 1 time in total.

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mdyrud
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Re: Would this be a proof?

Postby mdyrud » Sun Dec 05, 2010 4:06 am UTC

Unique is the wrong word. :oops: Primitive Pythagorean triple is the term I was looking for. So a triple where all of the sides are coprime. Not the most amazing proof ever, but as a freshman, I was pretty proud of it.

kbltd
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Re: Would this be a proof?

Postby kbltd » Sun Dec 05, 2010 5:10 am UTC

Uhh, gorcee, mdyrud specifically mentioned odd numbers, so your line of argument doesn't address the specific claim being made - it can only exhibit non-unique even numbers.
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mr-mitch
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Re: Would this be a proof?

Postby mr-mitch » Sun Dec 05, 2010 6:44 am UTC

Eastwinn wrote:Next step: prove that there is an infinite amount of coprime Pythagorean triples...

:D

Edit: By the way, I agree with the post below me. The way you wrote it was neither concise nor particularly clear. But the idea was definitely there.


This is fairly easy:

Spoiler:
The set of primes greater than 2 is infinite.
We have a coprime set of primitive pythagorean triples (p, 1/2 (p^2-1), 1/2(p^2+1))
Therefore there is a one-to-one function between primes and coprime pythagorean triples, and therefore |P| <= |CPT| and hence CPT is infinite.

For the proof that it's coprime, Only one of 1/2 (p^2 -1)and 1/2(p^2 + 1) is odd or even, and therefore have no common divisors as p is prime and p>2. If one does divide the other then there is an integer q such that (1-q) p^2 = 2, which is impossible. Hmm. I'm not too sure about this, as we're dealing with integers not rational/irrartional numbers.
Although if they do share a common divisor, it must divide their difference, and their difference is 2. But one of them is odd, and so any divisor of the even one can't divide the odd one and they are coprime?


Edit: No nested spoilers?

gorcee
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Re: Would this be a proof?

Postby gorcee » Sun Dec 05, 2010 2:19 pm UTC

kbltd wrote:Uhh, gorcee, mdyrud specifically mentioned odd numbers, so your line of argument doesn't address the specific claim being made - it can only exhibit non-unique even numbers.


Whoops. Simple enough.

Change the 4 to a 5. Put a restriction on n on odd numbers. Done.

Esquilax
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Re: Would this be a proof?

Postby Esquilax » Sun Dec 05, 2010 6:33 pm UTC

mdyrud wrote:Unique is the wrong word. :oops: Primitive Pythagorean triple is the term I was looking for. So a triple where all of the sides are coprime. Not the most amazing proof ever, but as a freshman, I was pretty proud of it.


What about

[math]3^2 + 4^2 = {\bf 5}^2[/math]
and
[math]{\bf 5}^2 + 12^2 = 13^2[/math]
Spoiler:
Image

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Kurushimi
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Re: Would this be a proof?

Postby Kurushimi » Sun Dec 05, 2010 6:49 pm UTC

He didn't say "unique primitive" he just said "primitive"

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mdyrud
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Re: Would this be a proof?

Postby mdyrud » Sun Dec 05, 2010 7:55 pm UTC

Basically, I showed that given an odd number, you can find two consecutive integers larger than the odd number that create a Pythagorean triple. Not that difficult, but I was pumped. I was just throwing that out as an example of things that the original poster could look at proving. Things that aren't too tough, or useful, but give you experience with proving things.

kbltd
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Re: Would this be a proof?

Postby kbltd » Sun Dec 05, 2010 9:33 pm UTC

Basically, I showed that given an odd number, you can find two consecutive integers larger than the odd number that create a Pythagorean triple.


I agree that could be a nice example for the OP to have a look at and yes, phrased like that, there's a PT that's both primitive and unique.

The question of how many distinct PPTs a given odd number appears in is interesting too and (I think) not too difficult ... I think it works out as:

An odd number which is the product of powers of k distinct primes appears as a summand, i.e. as one of the two smaller integers in the triple, in 2k-1 distinct PPTs.
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