Linear Diophantine Equation in the Complex number system?

Question

The specific question that I have is (1+2i)x+(-2+3i)y=1, although I would like to have more insight in how to solve a generalized linear diophantine in the complex number system, such that:

(a+bi)x+(c+di)y=e where a,b,c,d,e are intergers, and e is most likely 1.

Help would be greatly appreciated.

I should have been more specific, since it is for number theory the answers must be gaussian intergers or intergers.

Also I really would like a generalization to the formula to solve other similar problems.

Answer 1

In fact, if you choose any number field, and choose two integers in the field, say x1 and x2, which generate coprime principal ideals, then you can find a linear combination of x1 and x2 with algebraic integer coefficients in your field that equals 1, e.g., there exist r and s that are integers in your field such that r*x1+s*x2=1. This follows since the set of linear cominations is the ideal I generated by x1 and x2, and this ideal contains both principal ideals (x1) and (x2). Since the ring of integers is a Dedekind domain, I is a common ideal factor of (x1) and (x2), and since we assumed these principal ideals were coprime, I is actually the whole ring of integers. In particular, 1 is in I. By the definition of I, that shows that there exist r and s as above.

Now for your particular case, you are asking to solve ax+by=1 with a and b integers in the field Q(i). From above, this will be possible if a and b generate coprime principal ideals. Conversely, if this equation is solvable, then if the ideals (a) and (b) had a some common prime ideal factor P, then P contains a and P contains b, so P contains ax+by for any integers x and y. But then since we assumed such x and y could be found so that ax+by = 1, this would imply that P contains 1, a contradiction. Thus, in general, a necessary and sufficient condition that ax+by=1 is solvable in a general number field is that (a) and (b) are coprime ideals.

It is not hard to adapt this to show that ax+by=d is solvable if and only if the ideal (d) is divisible by the greatest common ideal divisor of the ideals (a) and (b).

When the number field is Q, the method for finding this linear comination is the Euclidean algorithm. Unfortunately, the integers in most number fields do not form Euclidean domains, but luckily for us, the Gaussian integers do. Thus, for your particular problem, we do the Euclidean algorithm, where at each step, the remainder needs to have modulus (i.e. absolute value) less than the modulus of the number we are dividing by. Notice that -2+3i has modulus 13 and 1+2i has modulus 5. It follows that they generate relatively prime principal ideals (both of which are prime ideals, by the way), so your equation is solvable.

We can write:
-2+3i = i*(1+2i) + 2i, and 2i has modulus 4 which is less than 5.
Then 1+2i=1*2i+1, which is the greatest common divisor of -2+3i and 1+2i. Working backwards by solving for 1, we find:

1=(1+2i) - 2i = (1+2i) - ((-2+3i) - i (1+2i))
= (1+2i)*(1+i) + (-2+3i)*(-1), so that
(1+2i)x + (-2+3i)y = 1 has the solution x=1+i, y=-1.

Furthermore, the totally of solutions is given by:
x=(1+i) + (-2+3i)*t, y = -1 - (1+2i)*t where t is any Gaussian integer.

Answer 2

excuse my answer if it seems like i'm bumbling through this at points. it's been 3 or 4 years since i've done number theory.

in regard to the specific question:
(1+2i)x + (-2+3i)y = 1..................solve for x:
(1+2i)x= 1 - (-2+3i)y
x = [1 - (-2+3i)y]/(1+2i)................ multiply through by (1-2i):
(messy steps)
x = [1 - 2i + (-4 -7i)y]/5

Let x = a + bi, and let y = c + di. Substitute.
(messy steps)
a + bi = (1 - 4c + 7d)/5 + (-2 - 7c - 4d)i/5

So, from this we can say:

a = (1 - 4c + 7d)/5
and
b = (-2 - 7c - 4d)/5

Since these are Diophantine equations, i.e. the solutions must be integers or Gaussian integers, a and b are both integers. Therefore, we conclude from the equation for a:

1 - 4c +7d = 0 (mod 5)
-4c + 7d = -1 = 4 (mod 5)
c + 2d = 4 (mod 5)

Doing this with the equation for b gives us the same result, so this tells us nothing specifically about c or d, but only about their relation to one another. However, it does signify that for any values of c and d that satisfy the above congruence in modulo 5, the corresponding value for x in the original equation will be an integer or Gaussian integer, i.e. all values of y that fulfil the above congruence are partial solutions to the diophantine equation. Therefore, there are as many solutions to this diophantine equation as there are solutions to this congruence. Hence, there are infinite solutions to this diophantine equation.

A handful of solutions:
a b c d
3 -2 0 2
5 -5 1 4
0 -4 2 1
2 -7 3 3
-3 -6 4 0

As for the general solution, I'm going to work on it a little now. I'll add something if I make any headway.

I hope this was somewhat helpful.

Answer 3

1x-2y=1 2ix+3iy=0, so 2x+3y=0


Mult 1st eqn by -2:

-2x+4y= -2
2x +3y = 0

7y = -2 y = -2/7

x-2y = 1
x + 4/7 = 1
x = 3/7


So I get x = 3/7, y = -2/7