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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?

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What do you mean by "the integer part"? Do you mean, the real part? – Gerry Myerson Jan 27 '13 at 9:32
@gerrymyerson yes. – Siona Jan 27 '13 at 9:46
up vote 0 down vote accepted

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 $n\ge2$.

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