Ventus 0 Posted June 12 So everyone knows that a^2+b^2=c^2, but is there a set of numbers where a^2+b^2=2c^2? How would one go about finding these different set of numbers? 0 Share this post Link to post Share on other sites

DrP 588 Posted June 12 2 hours ago, Ventus said: everyone knows that a^2+b^2=c^2 This is for the specific case of a right angled triangle with hypotenuse c. 2 hours ago, Ventus said: but is there a set of numbers where a^2+b^2=2c^2? Yes - but not presumably for a right angled triangle where a^{2}+b^{2} always = c^{2} where c is the hypotenuse. Unless you define c = half the length of the hypotenuse or forget anything about right angled triangles. 0 Share this post Link to post Share on other sites

Ventus 0 Posted June 12 6 hours ago, DrP said: Yes - but not presumably for a right angled triangle where a^{2}+b^{2} always = c^{2} where c is the hypotenuse. Unless you define c = half the length of the hypotenuse or forget anything about right angled triangles. Yes, if we completely disregard right angled triangles and just look at a, b, and c as isolated variables. 0 Share this post Link to post Share on other sites

uncool 218 Posted June 12 There are countably many triplets of integers (a, b, c) such that a^2 + b^2 = 2c^2. There's even a method to find them. Do you know how to find a triplet (a, b, c) such that a^2 + b^2 = c^2? 0 Share this post Link to post Share on other sites

mathematic 82 Posted June 12 (edited) 14 minutes ago, uncool said: There are countably many triplets of integers (a, b, c) such that a^2 + b^2 = 2c^2. There's even a method to find them. Do you know how to find a triplet (a, b, c) such that a^2 + b^2 = c^2? a=2mn, b=m^2-n^2,c=m^2+n^2 Edited June 12 by mathematic typo 0 Share this post Link to post Share on other sites

uncool 218 Posted June 12 Do you know why that formula works, in a geometric sense? (I don't plan to simply give an answer at the moment; simply giving an answer is uninformative, and someone recently asked the same question in the Homework Help section) 0 Share this post Link to post Share on other sites