Can someone characterize the integer solutions to: ad + bc + bd = ac These correspond to pairs of complex numbers (a+bi)(c+di) such that the real part (ac-bd) is equal to the imaginary part (ad+bc). In Pari; { for (a=1,50, for (b=1,50, for (c=1,50, for (d=1,50, if (a*d+b*c+b*d==a*c,print(a","b","c","d)))))) } gives (e.g.) 10,8,9,1 10,8,18,2 10,8,27,3 10,8,36,4 10,8,45,5 Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
At 05:17 AM 9/8/03, Jon Perry wrote:
Can someone characterize the integer solutions to:
ad + bc + bd = ac
These correspond to pairs of complex numbers (a+bi)(c+di) such that the real part (ac-bd) is equal to the imaginary part (ad+bc).
You've nearly answered your own question: if the product has its real and imaginary parts equal, it must be 1+i times a real. So, given any a+bi, the corresponding c+di must be a real multiple of 1+i times the conjugate of a+bi, (1 + i)(a - bi) = a+b + (a-b)i. To generate all integer solutions, given any integers a,b, let g = gcd(a + b, a - b). Then the corresponding values of c and d are c = k(a + b)/g, d = k(a - b)/g, for any integer k. (If you want to restrict to positive integers, you need to have a > b > 0, k > 0.) -- Fred W. Helenius <fredh@ix.netcom.com>
'>Can someone characterize the integer solutions to:
ad + bc + bd = ac
These correspond to pairs of complex numbers (a+bi)(c+di) such that the
real
part (ac-bd) is equal to the imaginary part (ad+bc).
You've nearly answered your own question: if the product has its real and imaginary parts equal, it must be 1+i times a real. So, given any a+bi, the corresponding c+di must be a real multiple of 1+i times the conjugate of a+bi, (1 + i)(a - bi) = a+b + (a-b)i. To generate all integer solutions, given any integers a,b, let g = gcd(a + b, a - b). Then the corresponding values of c and d are c = k(a + b)/g, d = k(a - b)/g, for any integer k. (If you want to restrict to positive integers, you need to have a > b > 0, k > 0.)' VG. Can you explain the integer solutions when the real part is equal to 2*the imaginary part +1. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
At 07:52 AM 9/8/03, Jon Perry wrote:
Can you explain the integer solutions when the real part is equal to 2*the imaginary part +1.
In other words, you want to find integers a,b,c,d such that (a + bi)(c + di) = (2k + 1) + ki for some integer k. This has integer solutions iff gcd(a,b) = 1 and 2a+b is not a multiple of 5. (The latter condition is equivalent to saying that a+bi is not a Gaussian integer multiple of 2+i.) The corresponding values of k are the solutions to (2a + b)k == -a (mod a^2 + b^2). Not very enlightening, really. -- Fred W. Helenius <fredh@ix.netcom.com>
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