Math Fun Facts!
hosted by the Harvey Mudd College Math Department created, authored and ©1999-2010 by Francis Su
Subscribe to our RSS feed   or follow us on Twitter.
Get a random Fun Fact!
or
No subject limitations
Search only in selected subjects
    Algebra
    Calculus or Analysis
    Combinatorics
    Geometry
    Number Theory
    Probability
    Topology
    Other subjects
  Select Difficulty  
Enter keywords 

  The Math Fun Facts App!
 
  List All : List Recent : List Popular
  About Math Fun Facts / How to Use
  Contributors / Fun Facts Home
© 1999-2010 by Francis Edward Su
All rights reserved.

From the Fun Fact files, here is a Fun Fact at the Medium level:

Pythagorean Triples

Which triples of whole numbers {a, b, c} satisfy

a2 + b2 = c2 ?

Such triples are called Pythagorean triples because they are integer solutions to the Pythagorean theorem. You probably know {3, 4, 5} and {5, 12, 13}. But can you classify all possible Pythagorean triples?

Answer: it is possible to prove that all Pythagorean triples are of the form

{ M2-N2, 2MN, M2+N2 }
for some integers M and N, or they are multiples of this form.

Thus setting M=2,N=1 gives {3,4,5} and M=3,N=2 gives {5,12,13}.

Presentation Suggestions:
If you are really motivated and have time to practice this, you can try to following. Before telling students the rule for construction, tell them to give you any number and that in your head you will construct a Pythagorean triple using that number. If they give you an even number K=2M, let N=1; if they give you an odd number K=2N+1, let M=N+1. If you can do this quickly for several examples, you can say "Well, since I'm not that good with mental calculations, there's obviously a trick. It turns out that all Pythagorean triples are of this form..."

The Math Behind the Fact:
Simple number theory arguments using parity will give this conclusion. Assume a2 + b2 = c2 for an integer triple (a, b, c). By removing any common factors, if needed, we may assume a, b, and c have no common factor.

Since odd perfect squares must be congruent to 1 mod 4, and even squares are congruent to 0 mod 4, we can conclude that c must be odd, and at exactly one of a or b must be even. Suppose b is even. Then b=2k for some integer k, hence

4k2 = b2 = c2 - a2 = (c+a)(c-a).
Since (c+a) and (c-a) must have the same parity (evenness or oddness), they must both be even. Then c+a=2r, c-a=2s and rs=k2. It is easy to check that c=r+s, and a=r-s. But r and s can have no common factors because otherwise c and a would both share that common factor as well. So they must both be perfect squares, say a=M2 and b=N2. This gives the desired result.

How to Cite this Page:
Su, Francis E., et al. "Pythagorean Triples." Math Fun Facts. <http://www.math.hmc.edu/funfacts>.

Keywords:    Pythagoras
Subjects:    number theory
Level:    Medium
Fun Fact suggested by:   Francis Su
Suggestions? Use this form.
4.61
 
current
rating
Click to rate this Fun Fact...
    *   Awesome! I totally dig it!
    *   Fun enough to tell a friend!
    *   Mildly interesting
    *   Not really noteworthy
and see the most popular Facts!
New: get the MathFeed iPhone App!

Brings you news and views on math:
showcasing its power, beauty, and humanity

Want another Math Fun Fact?

For more fun, tour the Mathematics Department at Harvey Mudd College!