Formula for Pythagorean TriplesDate: 10/23/97 at 06:31:51 From: Tom Sunde Subject: Formula for Pythagorean Triples We have come across the formula for Pythagorean triples: a = (m^2-n^2) b = 2mn c = (m^2 + n^2) Is this correct for absolutely all Pythagorean triples? Can you find a specific triple with this formula? Can m and n be any two numbers as long as they are positive? Can you show us some examples? We have found all the sequences for Pythagorean triples (odd numbers). The problem is that we have used our own method to find these. In our project we have to include the method. What is this method? We have never learned this at school. Thanks for your help! Date: 10/23/97 at 09:11:49 From: Doctor Chita Subject: Re: Formula for Pythagorean Triples Dear Tom: Yes, you've discovered a very useful way to generate Pythagorean triples. The conditions for it to work are that m and n are integers, and m > n. To generate some triples, make a table like this: m n 2mn m^2 - n^2 m^2 + n^2 --------------------------------------------------- 2 1 4 3 5 3 1 6 8 10 3 2 12 5 13 etc. You see from this table that you not only get primitive triples, like (3, 4, 5), but some multiples of them as well: (6, 8, 10). These two right triangles would be similar. Looking at the generators, you can also see why m has to be greater than n. For example, try m = 1 and n = 2 and see what happens. Although this set of expressions is very useful, it does not give you all the triples. For example, the (9, 12, 15) triple can not be generated using these expressions. Try it and see. You will also notice other patterns in this table. Can all three numbers of a triple be even? odd? Why or why not? Have fun playing with these triples. Since there's no limit to m and n, there is an infinite number of triples generated this way. How many can you generate up to m = 50? -Doctor Chita, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 10/23/97 at 09:22:35 From: Doctor Rob Subject: Re: Formula for Pythagorean Triples These formulas do give Pythagorean triples. They do not give all of them, however. If you want to find all Pythagorean triples, it suffices to find all "primitive" Pythagorean triples, that is, ones such that the set {a,b,c} has greatest common divisor 1. Then any multiple d*a, d*b, and d*c will be another Pythagorean triple. The formulas you give are the ones which give primitive Pythagorean triples, provided appropriate restrictions are placed on m and n. First of all, if m and n have a common factor bigger than 1, then a, b, and c have the square of that factor in common, so the triple would not be primitive. Thus we must make the restriction that m and n be relatively prime. Furthermore, if m and n are both odd, then 2 is a divisor of a, b, and c, so the triple would not be primitive. Thus one of m and n must be odd and the other even. Lastly, we must have |m| > |n|, lest a < 0, and m and n have the same sign, lest b < 0. Since (-m,-n) gives the same Pythagorean triple as (m,n), we can assume that both m and n are positive. It turns out that these three conditions on m and n will guarantee that the triple is primitive. Thus the formulas that give all Pythagorean triples are these: a = d*(m^2 - n^2), b = 2*d*m*n, c = d*(m^2 + n^2), where d is any positive integer, m > n > 0 are integers of opposite parity and relatively prime. Examples: The smallest allowable parameter values are m = 2, n = 1, d = 1. They give a = 3, b = 4, c = 5. Another allowable set of parameter values is m = 5, n = 2, d = 3. They give a = 63, b = 60, c = 87. If you are given a triple a, b, and c, you can find d, m, and n in the following way. First of all, d = GCD(a,c). Then m = Sqrt[(c+a)/(2*d)] and n = Sqrt[(c-a)/(2*d)]. These are found by solving the first and last equations simultaneously for m^2 and n^2. Example: Given the Pythagorean triple a = 28, b = 96, c = 100, we find that d = GCD(28,100) = 4, m = Sqrt[(100+28)/(2*4)] = Sqrt[16] = 4, and n = Sqrt[(100-28)/(2*4)] = Sqrt[9] = 3. Sure enough, m > n > 0, they are of opposite parity, and they are relatively prime. You can make a table of primitive triples (d = 1) by starting with the smallest example above. You increase m by one, and run through all possible values for n satisfying the conditions (0 < n < m, n has opposite parity to m, and GCD(n,m) = 1). Repeat this until tired. This will give you the complete list of primitive Pythagorean triples. To get all Pythagorean triples, add to the list {d*a,d*b,d*c} for all d > 1. m n a b c 2 1 3 4 5 3 2 5 12 13 4 1 15 8 17 4 3 7 24 25 5 2 21 20 29 5 4 9 40 41 6 1 35 12 37 6 5 11 60 61 7 2 45 28 53 etc. Notice that a is always odd, b is always even, and c is always odd. Notice that a may be either larger or smaller than b. You may ask how these equations are derived. This works as follows. Suppose {a,b,c} is a primitive Pythagorean triple. Then GCD(a,c) = 1, because otherwise that common factor would divide b, too, and the triple wouldn't be primitive. Similarly GCD(a,b) = 1 and GCD(b,c) = 1. Also a and b cannot both be odd, since if that happened, then a^2 and b^2 would leave remainder 1 when divided by 4, so c^2 = a^2 + b^2 would leave remainder 2 when divided by 4. Then 2 would be a divisor of c^2, but 2^2 = 4 wouldn't, which is impossible. Thus one of a and b is even and the other is odd. Let b be the even one. Then c is odd, too. Rewrite the equation in the form b^2 = c^2 - a^2, divide both sides by 4, and factor, to get (b/2)^2 = [(c+a)/2]*[(c-a)/2]. Both (c+a)/2 and (c-a)/2 are integers, because a and c are both odd, and b/2 is an integer because b is even. Furthermore GCD[(c+a)/2,(c-a)/2] = 1, since any common factor would divide their sum, c, and their difference, a, which have no common factor greater than 1. Since they are relatively prime, and their product is a square, it follows that each of them is a square. Let (c+a)/2 = m^2 and (c-a)/2 = n^2, where m > 0 and n > 0. Observe that m^2 > n^2, so m > n. Also GCD(m^2,n^2) = 1, so GCD(m,n) = 1. Finally, one of m^2 and n^2 is odd and the other even, since they are half the sum and difference of two odd numbers, so m and n have opposite parity. Then c = m^2 + n^2, a = m^2 - n^2, b = Sqrt[c^2 - a^2] = Sqrt[(c+a)*(c-a)] = Sqrt[(2*m^2)*(2*n^2)] = 2*m*n, which are the formulas above that you found. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/