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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!

Bill Dubuque
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dddxdddy
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    See for example art of problem solving for $\gcd(a,b)=\gcd(a,c)=\gcd(b,c)=d>1$, as you said , and use this duplicate to conclude that $\gcd(a,b,c)=d$. – Dietrich Burde Feb 04 '23 at 11:46
  • @DietrichBurde You wrote "... for $\gcd(a,b)=\gcd(a,c)=\gcd(b,c)=d\gt 1$, as you said, ...". However, note the OP just wrote "... $\gcd(b,c)$, $\gcd(c,a)$, and $\gcd(a,b)$ are all greater than 1, ...", i.e., they didn't state they were all necessarily equal. Nonetheless, without making that assumption, the AoPS thread solution technique is related to at least one way, that I've determined, to solve the problem. – John Omielan Feb 04 '23 at 12:11

1 Answers1

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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:

$$m=hm_1 \geq h.$$

Reza Rajaei
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  • Thanks! In this way, it seems that we only need $\frac{ab}{a+b}$ and $\frac{ac}{a+c}$ to be an integer – dddxdddy Feb 04 '23 at 19:35