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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?

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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 a2+b2 always = c2 where c is the hypotenuse.  Unless you define c = half the length of the hypotenuse or forget anything about right angled triangles.

 

 

 

 

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6 hours ago, DrP said:

Yes - but not presumably for a right angled triangle where a2+b2 always = c2 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.

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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?

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Posted (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 by mathematic
typo

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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)

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