Suppose that there are $n$ numbers that are in the following format: $a+bi$. Each number has different combination of $a$ and $b$. $a,b$ must be non-zero integers.

Suppose that we impose the following rule: $a^2-b^2 = k_1x$ and $2ab = k_2y$.

$k_1$ and $k_2$ are free non-zero integers - by free, I mean that they can be different for different numbers. However, $x$ and $y$ are fixed (set).

We want to set the number so that when all of these $n$ numbers are multiplied, the integer part of the multiplication result cannot equal to the form of $sx+ty$ where $s$ and $t$ can be any non-zero integers.

The question is,

is this possible? If so, would this be possible for any cardinality of the set of numbers less than infinite?

  • $\begingroup$ What do you mean by "the integer part"? Do you mean, the real part? $\endgroup$ – Gerry Myerson Jan 27 '13 at 9:32
  • $\begingroup$ @gerrymyerson yes. $\endgroup$ – Siona Jan 27 '13 at 9:46

If $x$ and $y$ are relatively prime, then every integer can be put in the form $sx+ty$. So, let $p$ be a prime dividing both $x$ and $y$. Indeed, suppose the power of $p$ in $\gcd(x,y)$ is $p^e$.

First take the case $p\ne2$.

From $2ab=k_2y$ we get $p^r$ divides $a$, $p^{e-r}$ divides $b$. Then from $a^2-b^2=k_1x$ we get $p^s$ divides both $a$ and $b$ for some $s\ge e/2$. It follows that for $n\ge2$, $p^e$ divides the real part of the product of the $n$ numbers.

The same argument actually works for $p=2$ despite the factor of $2$ in $2ab=k_2y$. So for $n\ge2$, $\gcd(x,y)$ divides the real part of the product of the $n$ numbers, so that real part can always be written as $sx+ty$.

So, what you are asking for is impossible for $n\ge2$.

| cite | improve this answer | |

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.