- #1
1+1=1
- 93
- 0
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 ?
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 ?