Pythagorean TriplesDate: 07/10/97 at 19:04:43 From: Jason Osborne Subject: Pythagorean triples Can you tell me why (3,4,5), (20,21,29), (119,120,169), and (696,697,985) are Pythagorean triples? I noticed the following pattern: (5^2+4)/1 = 29, (29^2+4)/5 = 169, (169^2+4)/29 = 985 and so on. Why does this occur? Date: 07/11/97 at 15:33:26 From: Doctor Rob Subject: Re: Pythagorean triples Good question! Obviously you have made an interesting observation here. And yes, I can explain why this happens, but my explanation is somewhat complicated. For some relatively prime integers u and v of opposite parity (i.e., one odd, one even), primitive (that is, with no common GCD) Pythagorean triples (a,b,c) satisfy the relations: a = 2*u*v b = u^2 - v^2 c = u^2 + v^2 The triples you are looking at have |a - b| = 1. When we substitute in this last equation, we find that for some integer n: u^2 - 2*u*v - v^2 = (-1)^n If we complete the square on u, we get: (u - v)^2 - 2*v^2 = (-1)^n Letting w = u - v, this becomes: w^2 - 2*v^2 = (-1)^n This equation is called Pell's Equation, for no very good reason (but that's a different story). Its solutions are known, and here are the first few: (w,v) = (1,0), (1,1), (3,2), (7,5), (17,12), (41,29), (99,70), ... (Of course if (w,v) works, so will (w,-v), (-w,v), and (-w,-v).) Where n = 0, 1, 2, 3, 4, 5, 6, ..., respectively, the general formula is: w(n) = [(1 + Sqrt[2])^n + (1 - Sqrt[2])^n]/2 v(n) = [(1 + Sqrt[2])^n - (1 - Sqrt[2])^n]/(2*Sqrt[2]) Now it may appear miraculous that all the square roots of 2 simply drop out when we expand these expressions, and that the denominators do divide evenly into the numerators, and when all is said and done, the values of w(n) and v(n) are integers. Miracle or not, it happens! We let r = 1 + Sqrt[2] and s = 1 - Sqrt[2], and note that r + s = 2, r - s = 2*Sqrt[2], and r*s = -1. Then: w(n) = (r^n + s^n)/(r + s) v(n) = (r^n - s^n)/(r - s) Given this solution, it is easy to see that both w(n) and v(n) satisfy the same recursion: w(n) = 2*w(n-1) + w(n-2) v(n) = 2*v(n-1) + v(n-2) Recall that u(n) = v(n) + w(n), so it, too, satisfies the same recursion: u(n) = 2*u(n-1) + u(n-2) (You can verify these by substituting the expressions in terms of r and s and simplifying using the relations between r and s noted in the last paragraph). Furthermore, they also satisfy the mixed recursions: w(n) = w(n-1) + 2*v(n-1) v(n) = w(n-1) + v(n-1) And so: u(n) = 2*u(n-1) + v(n-1) v(n) = u(n-1) (Again, you can verify this using the r-and-s-form above). Don't forget that w(n)^2 - 2*v(n)^2 = (-1)^n. Now for each n >= 0, we have: a(n) = 2*u(n)*v(n) b(n) = u(n)^2 - v(n)^2 c(n) = u(n)^2 + v(n)^2 This gives the following sets of triples (which are the ones you started with): (a,b,c) = (0,1,1), (4,3,5), (20,21,29), (120,119,169), (696,697,985) That's a good sign! So far, everything we have done should be correct. Now you observed that c(n) = [c(n-1)^2 + 4]/c(n-2). To prove this, we will express everything in sight in terms of v(n-1) and u(n-1), then in terms of u(n) and v(n), then in terms of v(n) and w(n), and finally replace v(n)^2 with [w(n)^2 - (-1)^n]/2 wherever we can. We start with: [c(n-1)^2 + 4]/c(n-2) = [(u(n-1)^2 + v(n-1)^2)^2 + 4]/[u(n-2)^2 + v(n-2)^2] = [u(n-1)^4 + 2*u(n-1)^2*v(n-1)^2 + v(n-1)^4 + 4]/ [u(n-1)^2 - 4*u(n-1)*v(n-1) + 5*v(n-1)^2] = [u(n)^4 - 8*u(n)^3*v(n) + 26*u(n)^2*v(n)^2 - 40*u(n)*v(n)^3 + 25*v(n)^4 + 4]/[5*u(n)^2 - 24*u(n)*v(n) + 29*v(n)^2] = [5 + 4*(-1)^n*v(n)*w(n) - 6*(-1)^n*w(n)^2 - 8*v(n)*w(n)^3 + 6*w(n)^4]/[-5*(-1)^n - 14*v(n)*w(n) + 10*w(n)^2] = -(-1)^n + 2*v(n)*w(n) + 2*w(n)^2 = u(n)^2 + v(n)^2 = c(n) You can check the next-to-next-to-last equality by cross-multiplying, then setting v(n)^2 = [w(n)^2 - (-1)^n]/2. So your conjecture is true: it is a theorem! There may be a shorter, easier proof, but I don't know it. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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