About Gaussian Integers .

Question

If $a+bi$ is a Gaussian integer with $(a,b)=1$ call it 'viewable' ( as a line from $a+bi$ to $0$ can be drawn intersecting no other gaussian integers). Are there any gaussian composites that are viewable?

Answer 1

$a+bi$ is viewable as long as for every prime $p$ with $p\mid a^2+b^2$ we have $p\not\mid a$ or (in fact: and) $p\not\mid b$. Since for every prime $p\equiv 1\pmod 4$, there exists a solution to $x^2+y^2=p$, so correspondingly two numbers $x\pm iy$, we can form a product of arbitrarily many such factors (for same or different $p$), as long as we never use both a factor and its complex conjugate (which would make the product divisible by $p$).

There is no solution of $x^2+y^2=p$ for $p\equiv 3\pmod 4$. Therefore, we may not use any such primes.

The prime $p=2$ plays a special role as $1^2+1^2=2$ and the conjugates $1\pm i$ differ by a unit factor - so the factor $1+i$ may occur only once.

So one example of a severely composite viewable number is $$(1+i)\cdot (2+i)^{17}\cdot (3-2i)^5\cdot (4+i)\cdot (5-2i)=12875424699 - 10655124593i$$ arising from solutions of $x^2+y^2=2,5,13,17,29$.

Answer 2

There are. Experimentation will show that the Gaussian non-prime $(1-i)(1+2i)=3+i$ is viewable. There are many others.