Prove there exist infinitely many $n\in\mathbb N$ for any $a,b,c\in\mathbb N$ which are distinct such that $a+n,b+n,c+n$ are pairwise coprime Homework problem: prove there exist infinitely many $n\in\mathbb N$ for any $a,b,c\in\mathbb N$ which are distinct such that $a+n,b+n,c+n$ are pairwise coprime.
First off, I am aware of a similar post (Prove that for infinitely many $n$, the numbers $(a+n)$,$(b+n)$,$(c+n)$ are mutually prime), but the problem is a bit different and the answers don't seem to help me. Please correct me if I'm wrong.
I really don't know where to start with this problem. I tried constructing such $n$, for example taking $n=p-max(a,b,c)$ where $p$ is any prime larger than $max(a,b,c)$, but it only proves that $GCD(a+n,b+n)=GCD(a+n,c+n)=1$, while leaving $GCD(b+n,c+n)=1$ unproven. I am really stuck here. Any help? Thank you.
 A: Claim:$\;$If $a,b,c$ are distinct integers, there exist infinitely positive integers $n$ such that $a+n,b+n,c+n$ are pairwise coprime.

Proof:

Without loss of generality, we can assume $a < b < c$, and by an appropriate shift of $a,b,c$, we can assume $a=0$.

Letting $g=\gcd(c-1,c-b)$, write $c-1=gx$ and $c-b=gy$, where $x,y$ are positive integers with $\gcd(x,y)=1$.

Let $m=Q^ky-x$ and let $n=gm$, where $k$ is an arbitrary positive integer, and $Q$ is the product of all positive integers which are both less than or equal to $c$ and relatively prime to $c-1$.

Since there are infinitely choices for $k$, to prove the claim, it will suffice to show that $n,b+n,c+n$ are pairwise coprime (recall that $a=0$).

First suppose $p$ be a prime factor of $n$.

If $p{\,\mid\,}c$, then.
\begin{align*}
&
p\le c\;\,\text{and}\;\,p{\,\not\mid\,}(c-1)
\\[4pt]
\implies\;&
p{\,\mid\,}Q\;\,\text{and}\;\,p{\,\not\mid\,}gx
\\[4pt]
\implies\;&
p{\,\not\mid\,}g\;\,\text{and}\;\,p{\,\not\mid\,}x
\\[4pt]
\implies\;&
p{\,\mid\,}m
&&\bigl(\text{since $n=gm$}\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}(Q^ky-x)
\\[4pt]
\implies\;&
p{\,\mid\,}x
&&\bigl(\text{since $p{\,\mid\,}Q$}\bigr)
\\[4pt]
\end{align*}
contradiction.

Hence $p{\,\not\mid\,}c$, and therefore $p{\,\not\mid\,}(c+n)$.

It follows that $n$ and $c+n$ are coprime.

If $p{\,\mid\,}b$, then
\begin{align*}
&
p{\,\not\mid\,}(c-b)
&&\bigl(\text{since $p{\,\not\mid\,}c$}\bigr)
\\[4pt]
\implies\;&
p{\,\not\mid\,}gy
\\[4pt]
\implies\;&
p{\,\not\mid\,}g\;\,\text{and}\;\,p{\,\not\mid\,}y
\\[4pt]
\implies\;&
p{\,\mid\,}m
&&\bigl(\text{since $n=gm$}\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}(Q^ky-x)
\\[4pt]
\end{align*}

From $p{\,\mid\,}b$, we get $p < c$, hence $p{\,\mid\,}Q$ or $p{\,\mid\,}(c-1)$, but not both.

But if $p{\,\mid\,}Q$, then
\begin{align*}
&
p{\,\mid\,}x
&&\bigl(\text{since $p{\,\mid\,}(Q^ky-x)$}\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}gx
\\[4pt]
\implies\;&
p{\,\mid\,}(c-1)
\\[4pt]
\end{align*}
contradiction.

And if $p{\,\mid\,}(c-1)$, then
\begin{align*}
&
p{\,\mid\,}gx
\\[4pt]
\implies\;&
p{\,\mid\,}x
&&\bigl(\text{since $p{\,\not\mid\,}g$}\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}Q^ky
&&\bigl(\text{since $p{\,\mid\,}(Q^ky-x)$}\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}Q
&&\bigl(\text{since $p{\,\not\mid\,}y$}\bigr)
\\[4pt]
\end{align*}
contradiction.

In both cases, we have a contradiction, hence $p{\,\not\mid\,}b$, and therefore $p{\,\not\mid\,}(b+n)$.

It follows that $n$ and $b+n$ are coprime.

Next suppose $\gcd(b+n,c+n) > 1$.

Let $p$ be a common prime factor of $b+n$ and $c+n$.
\begin{align*}
\text{Then}\;\;&
p{\,\mid\,}(c+n)
\\[4pt]
\implies\;&
p{\,\mid\,}(c+gm)
\\[4pt]
\implies\;&
p{\,\mid\,}\bigl(c+g(Q^ky-x)\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}\bigl((c-gx)+Q^kgy\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}\Bigl(\bigl(c-(c-1)\bigr)+Q^kgy\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}(1+Q^kgy)
\\[4pt]
\implies\;&
p{\,\not\mid\,}gy
\\[4pt]
\implies\;&
p{\,\not\mid\,}(c-b)
\\[4pt]
\end{align*}
contradiction, since any common factor of $b+n$ and $c+n$ must divide $c-b$.

Hence $\gcd(b+n,c+n)=1$.

Thus we've shown that $n,b+n,c+n$ are pairwise coprime, which completes the proof of the claim.
A: First note that any common factor of $x+n$ and $y+n$ is also a factor of $|x-y|$. Therefore, all we have to do is avoid small prime factors $\le\max\{a,b,c\}-\min\{a,b,c\}$.
Consider a single prime $p$.

*

*If at most one of $a$, $b$, $c$ is a multiple of $p$, then for every $n\equiv 0\pmod p$, we are sure that no two of $a+n$, $b+n$, $c+n$ have $p$ as a common factor.

*If on the other hand two or all three of $a$, $b$, $c$ are multiples of $p$, then for every $n\equiv 1\pmod p$, we are sure that no two of $a+n$, $b+n$, $c+n$ have $p$ as a common factor.

By repeating the above for the finitely many primes $p_1,\ldots, p_k\le \max\{a,b,c\}-\min\{a,b,c\}$ and using the Chinese Remainder Theorem, we  conclude that for suitable $r$, all $n$ with $n\equiv r\pmod{p_1\cdots p_k}$ prevent any two of $a+n$ ,$b+n$, $c+n$ from having any of the $p_i$ as common factor. As noted above, this is sufficient to make $a+n$, $b+n$, $c+n$ pairwise co-prime.
A: Easy way: $ $ wlog assume $\,c\!=\!0\,$ via variable change $(c+n\to \color{#0a0}n),\,$ so our coprime conditions are $\,\color{#0a0}{(n,a) \!=\! (n,b)} \!=\! \color{#c00}{(a\!+\!n,a\!-\!b)\!=\!1}.\,$ By Stieltjes $\rm\color{#c00}{this}$ holds by choosing $\,n\,$ so every prime factor of $\,a\!-\!b\,$ divides exactly one of $\,a\,$ or $\,n,\,$ e.g. $\,n = de^k$ where $\,(e,ab)\!=\!1\,$ and $\,d=$ product of all prime factors of $\,a\!-\!b\,$ that don't divide $\,a.\,$ $\color{#0a0}{(n,a)\!=\!1\!=\!(n,b)}\,$ by $\,p\mid a\!\iff\! p\mid b,\,$ by $\,p\mid a\!-\!b$.
