# Prove to me that the pythagorean triples

1. Jun 23, 2004

### 1+1=1

prove to me that the pythagorean triples are from the form 3k,4k,or 5k. for k >=1.

proof: cases for 3K:
3k ^2 9k^2 factor out 3k 3k(3k)
3k+1 ^2 9k^2+1 3k(3k)+1
3k+2 ^2 9k^2+1+3 3k(3k)+4

cases for 4k:
4k ^2 16k^2 factor out 4k 4k(4k)
4k+1 ^2 16k^2+1 4k(4k)+1
4k+2 ^2 16k^2+4 4k(4k)+4
4k+3 ^2 16k^2+9 4k(4k)+9

cases for 5k:
5k ^2 you get the idea...
5k+1
5k+2
5k+3
5k+4

would this be the easiest way of doing this type proof? anyone w/ info/suggestions please reply. i know that there are remainders involved, but is there any way of finding out w/o looking at it? i mean, should i ^2 or ^3 them ?

2. Jun 23, 2004

### mathman

I am not sure what your question is. There are other triples besides (3,4,5) - the factor of k is irrelevant. In general you can get all triples using the following:

x=m2-n2
y=2mn
Then z=m2+n2

Where m>n (both integers).

3. Jun 23, 2004

### 1+1=1

well i am trying to prove that the only pythagorean triples are of the form 3k,4k, or 5k. i think those cases may work, but that seems as though that is a lot of work. any suggestions?

4. Jun 23, 2004

### Hurkyl

Staff Emeritus
A pythagorean triple is three numbers. What can you possibly mean when you say it's "of the form 3k"?

5. Jun 23, 2004

### jcsd

I think he means that all solutions to

s^2 = x^2 + y^2

are of the form

s = 5k, x = 4k, y = 3k

where k is an arbiatry (inerger) constant, of course it can easily be verified that this false.

6. Jun 23, 2004

### 1+1=1

jcsd-you are correct. how would this proof look? by just picking one counterexample? i am unclear as to how to prove this please suggestions...

7. Jun 23, 2004

### Hurkyl

Staff Emeritus
Any counterexample would disprove it.

8. Jun 23, 2004

### 1+1=1

i thought that any number multiplied by 3,4,5 would work for the right triangle, but it seems as though when k=13, 39^2+52^2 does =70^2. i don't understand how this example works, yet any counterexample would disprove that the only triples must be of the form 3k,4k or 5k. i think i am reading too much into this and i am just going to do the cases for each and look for a pattern.

here is another perplexing problem...

if (s,t)=1 and one of s and t is even,one odd, prove (x,y)=(x,z)=1 and x=2st, y=t^2-s^2, z=t^2+s^2. could my proof say this...

assume (s,t)=1. so there exists a p prime that divides both s and t, p divides s and p divides t. case 1. s is even. then s=2k and t=2j+1. after this i get side tracked. does this sound like it works so far?

p.s. i must say as a side note, i appreciate all the wonderful assistance from yall on here!

9. Jun 23, 2004

### Hurkyl

Staff Emeritus
Ah, I see; you've stated your conjecture backwards.

"Every triple is of the form (3k, 4k, 5k)"

But what you mean is:
"Everything of the form (3k, 4k, 5k) is a triple"

Or so I think. Have you tried simply plugging 3k, 4k, and 5k into the equation x^2 + y^2 = z^2 and see if it holds?

10. Jun 24, 2004

### matt grime

If s and t are coprime you may not assume there is a prime p with that property, in fact we know the exact opposite is true, that there is no prime with that property.

x=2st
y=t**2-s**2

so if p is a prime dividing x it divides 2,s or t. as one of s and t is even it divides s or t. let's assume it divides s, then it divides s**2. If it divides y too, that is if it divides hcf(x,y) then it divides y+s**2, which is t**2, so it divides t, but that can't happen as it must divide hcf(s,t)=1 #

so no such prime exists and (x,y)=1.

Last edited: Jun 24, 2004
11. Jun 24, 2004

### 1+1=1

yes i think i get it now, since (s,t)=1, it is relatively prime, by definition. as far as a proof, could i do a contradiction? would that be the least painful way of doing this? also, if (x,y)=(x,z)=1, that would mean that x and y and z are all = to 1, which would disprove this,correct?

12. Jun 24, 2004

### matt grime

I offered you a proof of the statement. it involved contradiction.

do you understand that if z**2=x**2+y**2 then if r is a prime factor of any two of the three numbers x,y,z then it is a factor of all three, which follows from simply rearranging the terms (if r divides, say, x and y it divides the RHS so it divides the RHS so it divides z)

Suppose (x,y)=(x,z)=1, show by finding a simple example that this does NOT imply x=y=z=1.

I'm also really rather confused as to what it is you are attempting to disprove. it appears you are wanting to disprove something that is well known to be true, that al primitive pythagorean triples are generated by coprime pairs s,t with

x=2st y=s**2-t**2 and z=s**2+t**2

Last edited: Jun 24, 2004