Pythagorean Triple with 71Date: 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 Check out our web site! http://mathforum.org/dr.math/ |
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