Unity+ Posted April 16, 2014 Share Posted April 16, 2014 Let's take a look at the following picture: Of course, a^2+b^2 = c^2 has infinite primitive(I think) Pythogorean triples where a,b, and c are whole numbers, proven by Euclid. However, are there infinite primitive Pythagorean triples where one of the elements, when square rooted, is also a whole number? I am asking this because, for example, let us take the primitive Pythagorean triple (3,4, 5). Notice how the square root of 4 is 2, obviously. Let us apply this to a geometric visual. This means that it takes a real whole number sided square to get two squares that have irrational sides. In interesting phenomena, I say. Link to comment Share on other sites More sharing options...
imatfaal Posted April 16, 2014 Share Posted April 16, 2014 To get two sides being perfect squares you are basically looking for solutions of the diophantine equations[latex]x^4+y^4=z^2[/latex]or[latex]x^4+y^2=z^4[/latex]Which I don't think have solutions. This problem is connect to the question of whether the area of a pythagorean triangle made from integer sides ever has a perfect square area - it doesn't.When I was checking up my answer I came across this on mathworld - BTW seriously cool site In 1643, Fermat challenged Mersenne to find a Pythagorean triplet whose hypotenuse and sum of the legs were squares. Fermat found the smallest such solution: ,, He worked that out by hand!!!!http://mathworld.wolfram.com/PythagoreanTriple.htmlYou have also gotta love the idea of Fermat and Mersenne setting each other challenges - bit too superhuman-difficulty for ordinary mortals. Link to comment Share on other sites More sharing options...
mathematic Posted April 16, 2014 Share Posted April 16, 2014 (edited) There are an infinite number of integer Pythagorean triples. Let m > n > 0 be any integers. [Latex]a = 2mn, b = m^2 - n^2, c = m^2 + n^2[/Latex]. Edited April 16, 2014 by mathematic Link to comment Share on other sites More sharing options...
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