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Calling all pythagoreans (Posted on 2006-07-03) Difficulty: 3 of 5
The triangle with sides 3, 4, and 5, is the smallest integer sided pythagorean triangle. Can you prove that in every such triangle:
  • at least one of its sides must be multiple of 3?
  • at least one of its sides must be multiple of 4?
  • at least one of its sides must be multiple of 5?

  Submitted by e.g.    
Rating: 5.0000 (1 votes)
Solution: (Hide)
The sides of a pythagorean triangle can be written as K(p^2-q^2), K(2pq), and K(p^2+q^2) for integer K; see this reference for this.

The "multiple of 3" part: if p or q is multiple of 3, 2pq is a multiple of 3; if both are multiples of 3 plus 1 or 2, then their squares are multiples of 3 plus 1, so p^2-q^2 is a multiple of 3.

The "multiple of 4" part: if p or q is even, 2pq is multiple of 4; if both are odd, their squares are multiples of 4 plus 1, so p^2-q^2 is a multiple of 4.

The "multiple of 5" part: if p or q is multiple of 5, 2pq is a multiple of 5; if they are multiples of 5 plus 1, 2, 3, or 4, their squares are multiples of 5 plus/minus 1, so either p^2+q^2 or p^2-q^2 will be a multiple of 5.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
nozachary scott2006-07-09 13:45:12
re: Interesting dialogRichard2006-07-03 23:14:48
Interesting dialogbrianjn2006-07-03 20:02:57
Some Thoughtsre: Somekind of solutionFederico Kereki2006-07-03 14:26:46
Somekind of solutionatheron2006-07-03 13:56:09
SolutionPuzzle ResolutionK Sengupta2006-07-03 13:36:04
Hints/TipsA more specific resultOld Original Oskar!2006-07-03 10:49:48
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