Prove this there exist a lattice triangle $A_{1}B_{1}C_{1}$,such$\Delta A_{1}B_{1}C_{1}\sim\Delta A_{0}B_{0}C_{0}$ Question: 

Give the  $\Delta A_{0}B_{0}C_{0}$ is lattice triangle,and $\gcd(|A_{0}B_{0}|^2,|B_{0}C_{0}|^2,|C_{0}A_{0}|^2)=d>1$,where $B_{0}(0,0)$,

Prove that:

there exist a lattice triangle $A_{1}B_{1}C_{1}$,such$\Delta A_{1}B_{1}C_{1}\sim\Delta A_{0}B_{0}C_{0}$,and $|A_{0}B_{0}|^2=d|A_{1}B_{1}|^2,|A_{0}C_{0}|^2=d|A_{1}C_{1}|^2,|B_{0}C_{0}|^2=d|B_{1}C_{1}|^2$,where $B_{1}(0,0)$


 A: Although posed as a geometry question, this is actually a number theory question.  Let us consider lattice points as Gaussian integers and let $x$ and $y$ be the Gaussian integers corresponding to your points $A_0$ and $C_0$.  Let $z$ be the GCD of $x$ and $y$ in $\mathbb{Z}[i]$ and let $A_1$ and $C_1$ be the points corresponding to $x/z$ and $y/z$.   It suffices to show that the squared lengths $|x/z|^2$, $|y/z|^2$, and $|x/z-y/z|^2$ have no common prime factors, since then we must have $|z|^2=d$.
So we are reduced to the proving the following statement (with $a=x/z$ and $b=y/z$).

Let $a,b\in\mathbb{Z}[i]$ be relatively prime.  Then $|a|^2$, $|b|^2$, and $|a-b|^2$ are relatively prime integers.

To prove this, suppose $p\in\mathbb{Z}$ is a common prime factor of $|a|^2$, $|b|^2$, and $|a-b|^2$.  If $p$ is prime in $\mathbb{Z}[i]$, then $p$ itself must divide $a$ and $b$.  If $p$ is not prime in $\mathbb{Z}[i]$, then $p=q_1q_2$ where $q_1,q_2\in\mathbb{Z}[i]$ are prime.  Each of $a$, $b$, and $a-b$ must be divisible by either $q_1$ or $q_2$.  In particular, two of $a$, $b$, and $a-b$ must be divisible by the same one of $q_1$ and $q_2$.  It then follows that the third is as well.  So one of the primes $q_1$ and $q_2$ must divide both $a$ and $b$, as desired.
