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Reciprocal Diophantine Equation
By Anthony Cardell Tony on Monday, October 08, 2001 - 07:11 pm:

The problem is this: Find solution(s) in integers of (1/a)2+(1/b)2=(1/c)2

The solution listed a=15, b=20, c=12 as the smallest solution (with regards to a,b,c), and the general formula as being a=m^4-n^4 , b=2mn(m2+n2) , c=2mn(m2-n2) where m and n are integers. It is not too difficult to prove that these solutions work, but how would you get to them???

By Kerwin Hui on Monday, October 08, 2001 - 10:36 pm:

I think the general formula should be

a=k(m4-n4), b=2kmn(m2+n2), c=2kmn(m2-n2).

Anyway, first note that if (a,b,c) is a solution, then so is (ka, kb, kc). Hence we can assume WLOG that hcf(a,b,c)=1. Multiply both sides by a2b2c2 and you get (bc, ca, ab) as a Pythagorean triple. Do you know the general solution of a pythagorean triple?

Kerwin

By Anthony Cardell Tony on Wednesday, October 10, 2001 - 12:47 pm:

Actually, the original problem I had was (rs)2+(st)2=(rt)2 and reduced it to the reciprocal pythagorean equation (since I had seen the general solution to the reduced expression in a book).
I'm not sure what the general solution of the pythagorean triple is. Could you provide a website or some background info?

thanx

By Emma McCaughan on Wednesday, October 10, 2001 - 07:19 pm:

There are several articles on Pythagorean triples on the NRICH site. Look on the menu bar at the left, under Archive, and click on Articles. Alternatively, they are indexed in Asked NRICH under Frequently Discussed Topics.