How $u | a,b,c,d$ and thus $u=1$?

Question

If $a=a_1t, b=b_1t, c=c_1t, d=d_1t$ and -

$a_1+b_1i=\frac{u}{v}m\bar p$,

$a_1-b_1i=\frac{u}{v}p\bar m$,

$c_1=\frac{u}{v}p\bar p, d_1=\frac{u}{v}m\bar m $

then how can we imply that $u | a,b,c,d$ and thus $u=1$?

Here, $ \bar a $ is complex conjugate of $a$, $|a|$ is the modulus/absolute value of complex number of $a$, $\mathbb{Z}[i]$ is an Integral Domain and commutative ring.

possibility of misprint: It might be that $u|a_1,b_1,c_1,d_1$ instead of $u | a,b,c,d$, similarly for $v$. In that case my question remain same.

Source of the problem:

Reference:

1. Page 160 of An Introduction to Diophantine Equations by Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

Answer 1

If we add $(a_1+b_1i)$ and $(a_1-b_1i)$,

we get $2a_1=\frac{u}{v}(m\bar p +p\bar m).$

Let, $m=(e_1+f_1i), p=(e_2+f_2i)$, then-

$2a_1=\frac{u}{v}(m\bar p +p\bar m) \implies 2a_1=\frac{u}{v}\{(e_1+f_1i)(e_2-f_2i)+(e_2+f_2i)(e_1-f_1i)\}$

$ \implies 2a_1=\frac{u}{v} \times 2 \times (e_1e_2+f_1f_2 )$

$ \implies a_1=\frac{u}{v} (e_1e_2+f_1f_2 )$

Thus, $u|a_1$, so, we can write $a_1=ua_2.$

From, $a_1+b_1i=\frac{u}{v}m \bar p \implies b_1i=\frac{u}{v}m \bar p-a_1 $

$\implies b_1i=\frac{u}{v}m \bar p-ua_2 \implies b_1i=u(\frac{m \bar p}{v}-a_2)$

Thus, $u|b_1$.

Again, by inspection, from $c_1= \frac{u}{v} p \bar p, c_1= \frac{u}{v} m \bar m $, it is clear, $u | c_1,d_1.$

So, $u | a_1,b_1,c_1,d_1.$

But note, $t$ is chosen in a way such that $\text{gcd}(a,b,c,d)=t$ implies $\text{gcd}(a_1,b_1,c_1,d_1)=1$.

From $\text{gcd}(a_1,b_1,c_1,d_1)=1$, $u$ has to be $1.$ Similarly, $v=1$ .

Answer 2

Jim has already provided a nice answer, but it seems that you want an answer with more details.

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We have $$\begin{align}a_1+b_1i&=\frac{u}{v}m\overline p\tag1 \\\\a_1-b_1i&=\frac uvp\overline m\tag2 \\\\c_1&=\frac uvp\overline p\tag3 \\\\d_1&=\frac uvm\overline m\tag4\end{align}$$

where $u$ and $v$ are relatively prime positive integers.

• Proof for $u\mid a_1$ :
  
  $(1)+(2)$ gives $$2a_1=\frac uv(m\overline p+p\overline m)$$ Multiplying the both sides by $v$ gives $$2a_1v=u(m\overline p+p\overline m)\tag5$$ Here, let us show that $m\overline p+p\overline m$ is an even integer.
  
  Since $m,p\in\mathbb Z[i]$, there exist integers $m_1,m_2,p_1,p_2$ such that $m=m_1+m_2i,p=p_1+p_2i$. Then,$$\begin{align}m\overline p+p\overline m&=(m_1+m_2i)(p_1-p_2i)+(p_1+p_2i)(m_1-m_2i) \\\\&=2(m_1p_1+m_2p_2)\end{align}$$ which is an even integer.
  
  Since there exists an integer $M$ such that $m\overline p+p\overline m=2M$, we get, from $(5)$, $$2a_1v=u\times 2M,$$ i.e. $$a_1v=uM$$Since $a_1v$ has to be divisible by $u$, and $u,v$ are relatively prime positive integers, it follows that $u\mid a_1$.
  
  

• Proof for $u\mid b_1$ :
  
  $(1)-(2)$ gives $$2b_1i=\frac uv(m\overline p-p\overline m)$$ Multiplying the both sides by $-iv$ gives $$2b_1v=ui(p\overline m-m\overline p)\tag6$$Here, let us show that $i(p\overline m-m\overline p)$ is an even integer.
  
  Since $m,p\in\mathbb Z[i]$, there exist integers $m_1,m_2,p_1,p_2$ such that $m=m_1+m_2i,p=p_1+p_2i$. Then, $$\begin{align}i(p\overline m-m\overline p)&=i\{(p_1+p_2i)(m_1-m_2i)-(m_1+m_2i)(p_1-p_2i)\} \\\\&=2(p_1m_2-p_2m_1)\end{align}$$ which is an even integer.
  
  So, we see that there exists an integer $N$ such that $i(p\overline m-m\overline p)=2N$, we get, from $(6)$, $$2b_1v=u\times 2N,$$ i.e. $$b_1v=uN$$Since $b_1v$ has to be divisible by $u$, and $u,v$ are relatively prime positive integers, it follows that $u\mid b_1$.
  
  

• Proof for $u\mid c_1$ :
  
  Multiplying the both sides of $(3)$ by $v$ gives$$c_1v=up\bar p$$Since $c_1v$ has to be divisible by $u$, and $u,v$ are relatively prime positive integers, it follows that $u\mid c_1$.
  
  

• Proof for $u\mid d_1$ :
  
  Multiplying the both sides of $(4)$ by $v$ gives$$d_1v=um\bar m$$Since $d_1v$ has to be divisible by $u$, and $u,v$ are relatively prime positive integers, it follows that $u\mid d_1$.
  
  

Therefore, we get $$u\mid a_1,b_1,c_1,d_1$$

Now, note that the first sentence of the Solution says

$\qquad$"Let $t=\gcd(a,b,c,d),\ a=ta_1,\ b=tb_1,\ c=tc_1$, and $d=td_1$"

This means that $\gcd(a_1,b_1,c_1,d_1)=1$.

It follows from this and $u\mid a_1,b_1,c_1,d_1$ that $$u=1$$

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We also have $$a_1+b_1i=\frac{v}{u}n\overline q,\quad a_1-b_1i=\frac vuq\overline n,\quad c_1=\frac vun\overline n,\quad d_1=\frac vuq\overline q$$ Similarly as above, we get $v\mid a_1,b_1,c_1,d_1$ and $v=1$.