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Pythagorean Triple with 71

Date: 12/07/97 at 00:46:00
From: Katherine Lok
Subject: Pythagorean Triple

How to find a Pythagorean triple that contains 71 or prove that 
there is none?

Date: 12/07/97 at 03:08:24
From: Doctor Pete
Subject: Re: Pythagorean Triple

There is indeed at least one Pythagorean triple with 71 as one of its 
elements: {71, 2520, 2521} is such a triple:

     71^2 + 2520^2 = 2521^2.

Is there another? Actually, no, there are no others. To see why, I 
will state the following fact (which I won't prove):

Every Pythagorean triple has the form {a, b, c}, where

     a = k(m^2-n^2),
     b = 2kmn,
     c = k(m^2 + n^2)},

where k, m, n, are positive integers, and m and n are relatively prime 
and not both odd. The triple is primitive if k = 1.

It is quite easy to show that this formula satisfies the Pythagorean 
relationship a^2 + b^2 = c^2.  It is somewhat more difficult to show 
that every triple has this form.

In any case, we immediately see that 71 is a prime number. Therefore, 
k must either be 71 or 1. But if k = 71, then one of m^2 - n^2, 2mn, 
or m^2 + n^2 must be 1, which is impossible. So k = 1; i.e., the 
triple is primitive. Then also notice that b = 2mn cannot be 71, so 
either m^2 - n^2 = 71, or m^2 + n^2 = 71. Suppose we have the first 
case. Then

     m^2 - n^2 = (m+n)(m-n) = 71

and since 71 is prime, we must have m+n = 71, m-n = 1, or m = 36, 
n = 35, and {a,b,c} = {71,2520,2521}. On the other hand, suppose

     m^2 + n^2 = 71

Then 0 < m, n < 9. Testing out a few cases, we see that there is no 
solution; hence this possibility does not yield a triple.  Since we 
have exhausted all cases, {71,2520,2521} is the only solution.

-Doctor Pete,  The Math Forum
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Associated Topics:
High School Number Theory

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