 All you could ever want from pythagoreas!!!  

Author  Message 

Dojo Admin
Number of posts : 154 Location : Probably somewhere near a computer Registration date : 20080812
 Subject: All you could ever want from pythagoreas!!! Sun Aug 17, 2008 3:39 pm  
 Post pythagoreas proofs/integer triples you can find!!!! _________________ ~Dojo


 
mathblitz Developer
Number of posts : 29 Age : 20 Location : GET AWAY STALKER Registration date : 20080816
 Subject: Re: All you could ever want from pythagoreas!!! Sun Aug 17, 2008 4:42 pm  
 heres some pythagoream triples: 3,4,5 5 12 13 7 24 25 9 40 41 11 60 61 13 84 85 see if you spot the pattern... (its easy i spotted it in third grade or fourth) 

 
herefishyfishy1 Developer
Number of posts : 36 Age : 22 Location : An insignificant little bluegreen planet 93 million miles from the sun Sol. Registration date : 20080816
 Subject: Re: All you could ever want from pythagoreas!!! Sun Aug 17, 2008 4:57 pm  
 You can assign an ordered pair to each of the triples. Call the pair (x,y). The first number in the triple is x^{2}y^{2}. The second is 2xy. The third is x^{2}+y^{2}. 

 
AIME15 Hardcore TMF user
Number of posts : 163 Age : 20 Location : Pleasanton, CA Registration date : 20080813
 Subject: Re: All you could ever want from pythagoreas!!! Mon Aug 18, 2008 5:50 pm  
 This looks like special binomial products, like (ab)^2, (a+b)^2, etc.
Also, at least one number in each triple is divisible by 4 and at least one is divisible by 5. I've never been able to prove it, although i know there is one. 

 
pythag011 School Math Nerd
Number of posts : 15 Registration date : 20080817
 Subject: Re: All you could ever want from pythagoreas!!! Mon Aug 18, 2008 7:04 pm  
  AIME15 wrote:
 This looks like special binomial products, like (ab)^2, (a+b)^2, etc.
Also, at least one number in each triple is divisible by 4 and at least one is divisible by 5. I've never been able to prove it, although i know there is one.
The (x,y) thing actually generates all the pythagorean triples with all three numbers relatively prime. So then if x, y are both odd, then x^2y^2 and x^2+y^2 and 2xy are all even and it's not relatively prime. So , when it is relatively prime, one of x and y must be even, so the 2xy is a multiple of 4.
For the divisibility by 5 thing, take the equation $x^2 + y^2 = z^2$ modulo 5, and there are no solutions if none of them are divisible by 5. 

 
AIME15 Hardcore TMF user
Number of posts : 163 Age : 20 Location : Pleasanton, CA Registration date : 20080813
 Subject: Re: All you could ever want from pythagoreas!!! Tue Aug 19, 2008 1:14 pm  
 So you could do the same thing with 4? 

 
pythag011 School Math Nerd
Number of posts : 15 Registration date : 20080817
 Subject: Re: All you could ever want from pythagoreas!!! Tue Aug 19, 2008 2:57 pm  
  AIME15 wrote:
 So you could do the same thing with 4?
Read the first sentence. Thouh I wrote it in a really ba d way... 

 
AIME15 Hardcore TMF user
Number of posts : 163 Age : 20 Location : Pleasanton, CA Registration date : 20080813
 Subject: Re: All you could ever want from pythagoreas!!! Tue Aug 19, 2008 4:04 pm  
 Oh yeah, thanks.
Anyone find proofs of this yet? 

 
pythag011 School Math Nerd
Number of posts : 15 Registration date : 20080817
 Subject: Re: All you could ever want from pythagoreas!!! Tue Aug 19, 2008 4:28 pm  
 Here's a better proof of the fact that at least one element has to be a multiple of 4.
Take the equation x^2 + y^2 = z^2 in modulo 8.
A perfect square is either 1 ( if it is odd) (4 if it is even but not x is not a multiple of 4) or 0(if it is a multiple of 4) modulo 8.
If none of them are 0 modulo 8, Then x^2 + y^2 can either be 1+1 = 2, which is not a value a perfect square can have, 4+1=5(same thing) or 4+4 = 0, which implies z is a multiple of 4. Therfore, one of them has to be a multiple of 4. 

 
AIME15 Hardcore TMF user
Number of posts : 163 Age : 20 Location : Pleasanton, CA Registration date : 20080813
 Subject: Re: All you could ever want from pythagoreas!!! Tue Aug 19, 2008 5:44 pm  
 i meant proofs ot eht ehorem, not the proofs of my statement earlier :DDD 

 
pythag011 School Math Nerd
Number of posts : 15 Registration date : 20080817
 Subject: Re: All you could ever want from pythagoreas!!! Tue Aug 19, 2008 7:03 pm  
 The thereom that all pythagorean triples can be expressed as (c(x^2y^2), 2xcy, and c(x^2+y^2))?
Here is a proof.
First, divide gcd(x,y,z) out. This will be c.
After that, one of (x,y) will be odd, and z will be odd. Why? If any two of them are even, all of them are even, and that can't happen 'cause we divide the gcd out. If z is even, then we consider the equation modulo 4, and since squares can only be 0 or 1 modulo 4, then x and y are both eve, contradiction.
Let's say y is even.
Let m be (x+z)/2, and let n be (zx)/2. Then x= mn and z = m+n. We get y the square of y is 4mn. If m and n are not relatively prime, then their sum and difference are also not relatively prime. Furthermore, y cannot be relatively prime because y^2 = z^2x^2. So m and n are relatively prime. Since 4mn is a square and 4 is a square, mn must be a square. Because m and n are relatively prime, m must be a square and n must be a square. Now we get the thereom. QED 

 
ternary0210 School Math Nerd
Number of posts : 41 Age : 19 Registration date : 20080817
 Subject: Re: All you could ever want from pythagoreas!!! Mon Sep 08, 2008 4:58 pm  
 15 112 113 16 30 34 32 60 68 64 120 136 17 144 145 19 180 181 21 220 221 

 
Sponsored content
 Subject: Re: All you could ever want from pythagoreas!!!  
 

 
