Which triples of whole numbers {a, b, c} satisfy <\/p>\n\n\n\n
a2<\/sup>\u00a0+ b2<\/sup>\u00a0= c2<\/sup>\u00a0?<\/p>\n\n\n\n Such triples are called Pythagorean triples<\/em> 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?<\/p>\n\n\n\n Answer: it is possible to prove that\u00a0all<\/em>\u00a0Pythagorean triples are of the form<\/p>\n\n\n\n { M2<\/sup>-N2<\/sup>, 2MN, M2<\/sup>+N2<\/sup>\u00a0}<\/p>\n\n\n\n for some integers M and N, or they are multiples of this form.<\/p>\n\n\n\n Thus setting M=2, N=1 gives {3,4,5} and M=3, N=2 gives {5,12,13}.<\/p>\n\n\n\n Presentation Suggestions:<\/strong> The Math Behind the Fact:<\/strong> 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 <\/p>\n\n\n\n 4k2<\/sup>\u00a0= b2<\/sup>\u00a0= c2<\/sup>\u00a0– a2<\/sup>\u00a0= (c+a)(c-a). <\/p>\n\n\n\n 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<\/sup>. It is easy to check that c=r+s, and a=r-s. But r and s can have\u00a0no common factors\u00a0because otherwise c and a would both share that common factor as well. So they must both be perfect squares, say a=M2<\/sup>\u00a0and b=N2<\/sup>. This gives the desired result.<\/p>\n\n\n\n How to Cite this Page:<\/strong> Fun Fact suggested by<\/strong>: Which triples of whole numbers {a, b, c} satisfy a2\u00a0+ b2\u00a0= c2\u00a0? Such triples are called Pythagorean triples because they are integer...<\/p>\n","protected":false},"author":7,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[3,4,10,154],"class_list":["post-461","page","type-page","status-publish","hentry","tag-easy","tag-medium","tag-numtheory","tag-pythagoras"],"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/461","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/users\/7"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/comments?post=461"}],"version-history":[{"count":6,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/461\/revisions"}],"predecessor-version":[{"id":1591,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/461\/revisions\/1591"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/media?parent=461"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/tags?post=461"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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…”<\/p>\n\n\n\n
Simple number theory arguments using parity will give this conclusion. Assume a2<\/sup> + b2<\/sup> = c2<\/sup> 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.<\/p>\n\n\n\n
Su, Francis E., et al. “Pythagorean Triples.” Math Fun Facts<\/em>. <http:\/\/www.math.hmc.edu\/funfacts>.<\/p>\n\n\n\n
Francis Su<\/p>\n","protected":false},"excerpt":{"rendered":"