<s>[INST] Let $a$, $b$, $c$ be positive integers such that $\frac{bc}{b+c}$, $\frac{ca}{c+a}$, $\frac{ab}{a+b}$ are integers. Show that $\gcd(a,b,c) >1$. Let $a$, $b$, $c$ be positive integers such that $\frac{bc}{b+c}$, $\frac{ca}{c+a}$, $\frac{ab}{a+b}$ are integers. Show that $\gcd(a,b,c) >1$.
I can see that $\gcd(b,c)$, $\gcd(c,a)$, and $\gcd(a,b)$ are all greater than 1, but how can we show $\gcd(a,b,c)>1$? Thanks!
 [/INST] Let's assume $(a,b,c)=1$. Since $\frac{ab}{a+b}$ is a positive integer, we must have $(a,b)>1$. The reason is clear. If $(a,b)=1$, then:
$$(a+b,a)=(a+b,b)=(a,b)=1 \implies (a+b, ab)=1.$$
Now, suppose $a=dm$ and $b=dn$, where $d>1$ and $(m,n)=1$. We will have:
$$\frac{ab}{a+b}=\frac{dmn}{m+n};$$
but $(m+n, mn)=1$ because $(m,n)=1$. Therefore:
$$m+n|d \implies d=s(m+n), $$
where $s$ is a positive integer. So $a=(m+n)sm$ and $b=(m+n)sn$.
Since we have assumed that $(a,b,c)=1$, we must have: $(c, (m+n)s)=1$.
Similarly, since $\frac{ac}{a+c}$ is a positive integer, we have:
$$(a,c)=((m+n)sm, c)=(m,c)>1.$$
Suppose $c=hc_1$ and $m=hm_1$, where $h$ is a positive integer and $(c_1,m_1)=1$. So,
$$\frac{ac}{a+c}=\frac{h((m+n)sm_1)c_1}{c_1+(m+n)sm_1},$$
but $(c_1, (m+n)sm_1)=1$; as a result, $((m+n)sm_1c_1, (m+n)sm_1+c_1)=1$. Thus, we must have:
$$c_1+(m+n)sm_1|h \\ \implies h \geq  c_1+(m+n)sm_1  > m,$$
which is a contradiction because:
$$m=hm_1 \geq h.$$
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