Pythagorean Triples (x,c,y) with Fixed cDate: 05/09/2003 at 11:42:14 From: Shanon Subject: Polynomials Is there a shortcut to finding the integer solutions to equations of the type x^2 + c = y^2, where c is a constant of known value? Obviously, (1/2)(c-1) = x and (1/2)(c+1) = y are solutions for odd values of c. But what about other whole number solutions? It seems that there should be some relation between the solutions and if we know some of them it should be possible to deduce or at least zero in on the others. I am most interested in finding the smallest integer solution. Date: 05/09/2003 at 14:46:33 From: Doctor Luis Subject: Re: Polynomials Hi Shanon, Yes, there is a way. Numbers a,b,c that satisfy the equation a^2 + b^2 = c^2 are called Pythagorean Triples. To read more about them, I suggest the following links from our site: Formula for Pythagorean Triples - Dr. Math archives http://mathforum.org/library/drmath/view/55811.html Pythagorean Triples - Dr. Math FAQ http://mathforum.org/dr.math/faq/faq.pythag.triples.html Now, the way you generate triples (a,b,c) is by specifying any two integers m,n and then using the following formulas a = m^2 - n^2 b = 2 m n c = m^2 + n^2 If you do quick check on these formulas, you can verify that a^2 + b^2 = c^2 holds true for all values of m and n. So, what you're asking, then, is to generate Pythagorean Triples (x,c,y) such that c is fixed. Let's try plugging these in the formulas above: x = m^2 - n^2 c = 2 m n y = m^2 + n^2 Therefore, x^2 + c^2 = y^2 definitely have integer solutions if c is even. You pick m and n from the factors in c/2. If c is odd, you might want to switch the formulas above c = m^2 - n^2 x = 2 m n y = m^2 + n^2 In which case, you'll have to express the odd number c as a difference of squares. A quick way to do this is by noticing that c = (m + n)(m - n) Therefore, if you write c as the product of two numbers, say p and q, you can solve the two simultaneous equations m+n=p, m-n=q for m,n. Once you have m,n, then you can find x and y. Does this make sense? I hope this helped answer your question. Let us know if you have any others. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
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