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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?

M47145
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user128932
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2 Answers2

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$a+bi$ 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 $x^2+y^2=2,5,13,17,29$.

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

André Nicolas
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  • Is there a way to tell the viewable gaussian composites from the non-viewable ones? – user128932 Apr 28 '14 at 06:08
  • Yes, I think these are the composites that are of the shape a unit times a product of $\beta_i^{k_i}$ where the $\beta_i$ are Gaussian primes that divide $2$ or an ordinary prime of the shape $4m+1$, with the proviso that the $\beta_i$ are "really distinct", meaning that is $\beta_i$ is in, then a unit times the conjugate if $\beta_i$ is not in the list of $\beta_j$. – André Nicolas Apr 28 '14 at 06:29
  • Can a viewable gaussian integer + another viewable gaussian integer = a third viewable gaussian integer? Or is this trivial..? – user128932 May 01 '14 at 03:45
  • It can be, but need not be: (i) $2+3i$ and $2-3i$ are viewable, but their sum $4$ is not. (ii) $1$ and $i$ are viewable, and so is their sum. – André Nicolas May 01 '14 at 03:56
  • Can a viewable gaussian composite have properties like a gaussian prime in certain 'situations'? – user128932 May 01 '14 at 04:00
  • There is too much freedom of construction. For any ordinary primes $p_i$ of the form $4k+1$, we can take one of its Gaussian prime factors $\pi_i$, and let $n=\prod\pi_i^{a_i}$ for arbitrary $a_i$. Doesn't feel very prime-like. – André Nicolas May 01 '14 at 04:30
  • Can any gaussian integer be a product of only viewable gaussian bi-primes? Does every gaussian integer have a unique factorization of viewable gaussian bi-primes? Can a viewable gaussinan composite have primitive root? – user128932 May 01 '14 at 04:36
  • The questions could use some clarification. We get into trouble with Gaussian integers with prime factors of the form $4k+3$. – André Nicolas May 01 '14 at 04:43
  • Is every gaussian integer the average of two gaussian primes? This is related to Goldbach's conjecture. – user128932 May 10 '14 at 06:50
  • Does every 'viewable' gaussian integer ( with no prime factors of the form 4k+3) have a unique factorization as 'viewable' gaussian primes or gaussian integers that are the product of two gaussian primes? – user128932 Nov 25 '14 at 19:03
  • Gaussian integers have the unique factorization property, with the caveat that there are $4$ units instead of the usual $2$. There are viewable Gaussian integers, such as $(1+2i)^3$, that are not the product of distinct Gaussian (viewable) primes. – André Nicolas Nov 25 '14 at 21:47
  • Is every viewable gaussian integer a product of gaussian primes no two of which are conjugates? – 201044 Sep 14 '15 at 04:41
  • The definition of viewable includes the condition $(a,b)=1$. If a product of Gaussian primes includes a pair of conjugates, then this condition is violated. – André Nicolas Sep 14 '15 at 04:48