# Hypotenuse as required

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1. Feb 21, 2016

### Terry Coates

I wonder if I have found the most efficient way of finding any pythagoras triple with a given hypotenuse that has not been published?
My method uses a trial and error search, but with the limits set to minimize the number of trials as follows:
Try odd values X within the limits of int(C^0.5) +1 to int(2.C)^0.5 inclusive such that
2.C - X^2 = Y^2. C being the required hypotenuse. {This is always possible and in a number of ways depending on the number (at least one) of prime factors of form 4n+1 in C.}
Then the required value of one side of the triangle is given by X.Y
Example C= 65
lower limit = 9 upper limit = 11
X = 11, Y = 3 or X = 9 Y = 7

33^2 + 56^2 = 65^2 or 63^2 + 16^2 = 65^2

These are primitive solutions, but non primitive ones can be found by applying the above method to C = 5 or 17 and then multiplying the three sides by 17 or 5.

2. Feb 21, 2016

### Staff: Mentor

I moved this thread from the textbooks section, as it seemed more related to the mathematics and very little to textbooks.

3. Feb 21, 2016

### Staff: Mentor

Your examples can all be found by the equation $(u^2-v^2)^2 + (2uv)^2 = (u^2+v^2)^2$ which is well known.
I can't remember and haven't looked up whether all triples can be found this way. Nevertheless, your's can be found, e.g. $65^2=8^2+1^2=7^2+4^2$.

4. Feb 21, 2016

### mathman

This gives all triples.