## Would this be a proof?

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### Would this be a proof?

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.

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.

### Re: Would this be a proof?

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

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.

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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### Re: Would this be a proof?

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].

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].

### Re: Would this be a proof?

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.

- heyitsguay
**Posts:**118**Joined:**Thu Oct 16, 2008 6:21 am UTC

### Re: Would this be a proof?

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.

3

^{2}+4

^{2}=5

^{2}

5

^{2}+12

^{2}=13

^{2}

Perhaps pay more attention.

### Re: Would this be a proof?

Ah, but you had one 5

Nope...

15

15

21

21

^{2}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...

15

^{2}+8^{2}=17^{2}15

^{2}+112^{2}=113^{2}21

^{2}+20^{2}=29^{2}21

^{2}+220^{2}=221^{2}### Re: Would this be a proof?

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]

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.

### Re: Would this be a proof?

Unique is the wrong word. 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.

### Re: Would this be a proof?

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.

### Re: Would this be a proof?

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

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:**

Edit: No nested spoilers?

### Re: Would this be a proof?

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.

### Re: Would this be a proof?

mdyrud wrote:Unique is the wrong word. 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:**

### Re: Would this be a proof?

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.

### Re: Would this be a proof?

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 2

^{k-1}distinct PPTs.

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