Given Pythagorean triples that satisfy the Diophantine equation:<\/p>\n
a² + b² = c²<\/p>\n
where a, b and c share no common factors, one of a and b must be odd, the other must be even, and c is always odd.
\nFirst, note that squares of even numbers are always divisible by 4. [Lemma 1]<\/p>\n
(2n)² = 4n²<\/p>\n
And squares of odd numbers are of the form (4n + 1). [Lemma 2]<\/p>\n
(2n + 1)² = 4n² + 4n + 1
\n = 4n(n+1) + 1<\/p>\n
If a and b are both even, it is trivial to see that a, b and c share the factor 2.<\/p>\n
c² = (2n)² + (2k)²
\n = 4n² + 4k²
\n = 4(n² + k²)<\/p>\n
By Lemma 1, c is even. Therefore the solution reduces to the more primitive form:<\/p>\n
n² + k² = (c\/2)²<\/p>\n
Consider a and b both odd:<\/p>\n
c² = (2n + 1)² + (2k + 1)²
\n = 4n² + 4n + 1 + 4k² + 4k + 1
\n = 4n² + 4n + 4k² + 4k + 2
\n = 2(2(n² + n + k² + k) + 1)<\/p>\n
This (by Lemmas 1 & 2) is not a perfect square, so this cannot be a solution. In fact, in this case c must be irrational:<\/p>\n
Let p = n² + n + k² + k
\nc = √2√(2p + 1)<\/p>\n
2p + 1 cannot have 2 as a factor. And since √2 is irrational, so is c. Therefore one of a and b must be odd and the other even.<\/p>\n
c² = (2n + 1)² + (2k)²
\n = 4n² + 4n + 1 + 4k²
\n = 4n² + 4n + 4k² + 1
\n = 4(n² + n + k²) + 1<\/p>\n
And c is odd (Lemma 2). QED.<\/p>\n","protected":false},"excerpt":{"rendered":"
Given Pythagorean triples that satisfy the Diophantine equation: a² + b² = c² where a, b and c share no common factors, one of a and b must be odd, the other must be even, and c is always odd. First, note that squares of even numbers are always divisible…<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11],"tags":[],"class_list":["post-466","post","type-post","status-publish","format-standard","hentry","category-maths"],"_links":{"self":[{"href":"https:\/\/www.elbeno.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/466","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.elbeno.com\/blog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.elbeno.com\/blog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.elbeno.com\/blog\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.elbeno.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=466"}],"version-history":[{"count":0,"href":"https:\/\/www.elbeno.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/466\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.elbeno.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=466"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.elbeno.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=466"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.elbeno.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=466"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}