Prove that for infinitely many $n$, the numbers $(a+n)$,$(b+n)$,$(c+n)$ are mutually prime The problem is -

Let $a$, $b$ and $c$ be three distinct integers such that they are mutually prime. Show that for infinitely many $n$, the numbers $(a+n)$,$(b+n)$,$(c+n)$ are mutually prime.

I have been able to show that for infinitely many $n$ we can find $\gcd (a,b,c)$ $=$ $1$. The proof is very simple. Since by hypothesis the integers are distinct therefore there should be an ordering among them. Let $a$ $>$ $b$ $>$ $c$. Now choose $n$ to be such that $(a+n)$ is a prime. It is now easy to see that $\gcd (a,b,c)$ $=$ $1$. 

We prove that if $a$, $b$, and $c$ are distinct, then there are infinitely many $n$ such that $a+n$, $b+n$, and $c+n$ are pairwise relatively prime. We do not need to assume that $a$, $b$ and $c$ are pairwise relatively prime.
Without loss of generality we may assume that $a=0$ and  $b$ and $c$ are positive, with $b\lt c$. Call the positive integer $n$ good if (i) $n$ is relatively prime to $b$, $c$, and $c-b$ and (ii) $n$ is divisible by any prime $\lt c$ that does not divide any of $b$, $c$, or $c-b$. There are infinitely many good $n$. We show that for any good $n$, the numbers $a+n$ (that is, $n$), $b+n$, and $c+n$ are pairwise relatively prime.
Suppose that $n$ and $b+n$ are not relatively prime. Then there is a prime $p$ that divides $n$ and $b+n$, and therefore that divides $b$. But this contradicts our choice of $n$. The same argument shows that $n$ and $c+n$ are relatively prime.
Finally, we show that $b+n$ and $c+n$ are relatively prime. If not, there is a prime $p$ that divides $b+n$ and $c+n$. It follows that $p$ divides $c-b$. So $p\lt c$, and therefore $p$ divides $n$. But then $p$  divides $b$ and $c$, contradicting the choice of $n$.
