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Complex Conclusion (Posted on 2012-08-29) Difficulty: 3 of 5
Each of a and c is a positive integer and each of b and d is an integer with b ≥ d.

Determine all possible pairs (a + bi, c + di) of complex numbers whose sum equals their product.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution re: Playing around on the complex plane. - Proof and Solution Comment 7 of 7 |
(In reply to Playing around on the complex plane. by Jer)

We can prove it.  The first sentence of the problem statement establishes that each of a+bi and c+di are Gaussian integers with positive real parts.

Lets say a+bi=x and c+di=y.  Then we have xy = x+y.  Rearrange that into (x-1)*(y-1) = 1.

Now we must have abs[x-1] * abs[y-1] = 1.  Also x-1 and y-1 are Gaussian integers: x, y, 1 are all Gaussian integers and Gaussian integers are closed over addition, subtraction, and multiplication thus x-1 and y-1 are Gaussian integers.

For any nonzero Gaussian integer z we have abs[z]>=1.  So then abs[x-1]>=1 and abs[y-1]>=1.  But for abs[x-1] * abs[y-1] = 1 we then must have abs[x-1]=1 and abs[y-1] = 1.

But on a complex plane abs[x-1]=1 describes a circle centered at 1+0i with radius 1, therefore x is complex number a distance 1 from 1+0i (also true of y).

At this point we have proven Jer's the assertion that solutions to the original problem will have x=a+bi and y=c+di be a unit distance from 1+0i.

Now since x and y are Gaussian integers and the only Gaussian integers that are a unit distance away from 1+0i are 0, 2, 1+i, and 1-i.  We are to discard zero per the problem statement asking for Gaussian integers with positive real part.  Then x and y must be limited to just 2, 1+i or 1-i. 
 
At this point it is trivial to construct the possible identities 2*2 = 2+2 and (1+i)*(1-i) = (1+i)+(1-i).

  Posted by Brian Smith on 2022-12-12 13:15:17
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