1
$\begingroup$

I want to show that Chinese Remainder Theorem(CRT) has a unique solution over gaussian integers. But when I choose $x=4+5i$ and determine coprime modulo numbers $1+2i$ and $1+4i$, the simultaneous congruences

$4 + 5i \equiv - 1\,\bmod (1 + 2i)$

$4 + 5i \equiv - 1 + 2i\,\bmod (1 + 4i)$ is obtained. If I use CRT for modulo $1+2i$,$1+4i$ and remainders $-1$,$-1+2i$, I obtain $ - 2 + 2i \ne 4 + 5i$. What is the problem? When does a system have unique solution?

$\endgroup$
  • 1
    $\begingroup$ I think not many are going to be able to understand what you wrote. Somewhere in this place there's a guide how to write mathematics correctly. $\endgroup$ – DonAntonio Feb 23 '16 at 8:44
  • $\begingroup$ I don't know what you mean by "the CRT is not provided". Anyway, you know the solution of a system of congruences is not a single number, but a congruence class full of numbers, right? $\endgroup$ – Gerry Myerson Feb 23 '16 at 9:08
  • $\begingroup$ I have edited my question. The right solution of question is $4+5i$, But , when we use Chinese remainder theorem for remainder $-1,-1+2i $ and modulo $1+2i, 1+4i$, the solution is obtained $-2-2i$. $\endgroup$ – ibrhm özbk Feb 23 '16 at 9:35
  • 2
    $\begingroup$ $-2-2i$ and $4+5i$ are both right, since $4+5i-(-2-2i)=6+7i$ is divisible by both $1+2i$ and $1+4i$, Indeed it is $-i(1+2i)(1+4i) $\endgroup$ – André Nicolas Feb 23 '16 at 10:07
  • $\begingroup$ Why is the solution not unique? Nicolas $\endgroup$ – egrtomath Feb 23 '16 at 10:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.