If $n$ is odd and $abc=(n-a)(n-b)(n-c),$ then $LCM((n,a),(n,b),(n,c))=n.$ 
Prove that if $a,b,c,n\in \mathbb Z^{+},2\not\mid n$ and $$abc=(n-a)(n-b)(n-c),$$
  $x=(n,a),y=(n,b),z=(n,c),$ then $LCM(x,y,z)=n.$

If $n=35,a,b,c=5, 21, 28,$ then $x=(35,5)=5,y=(35,21)=7,z=(35,28)=7,LCM(x,y,z)=35.$
If $n=945,a,b,c=9, 756, 910,$ then $x,y,z=9, 35, 189,LCM(x,y,z)=945.$
I checked all $n<1000$ and this is always true. 
(On the other hand, if $2\mid n$ then $LCM(x,y,z)=n$ or $\dfrac n2.$)
 A: First, consider if $ \gcd(a,b,c) \neq 1$. If so, let $p$ be any prime that divides it, and so $p^3 \mid abc = (n-a)(n-b)(n-c)$. Hence, WLOG $p \mid n-a$, which gives us $p\mid n$. Now, we may divide every term by $p$, and still be in the same situation. Henceforth, assume that $\gcd (a,b,c) = 1$.
Let $p$ be an (odd) prime that divides $n$. Let $p^k$ be the largest power of $p$ that divides $n$. Let $p^\alpha, p^\beta, p^\gamma$ be the largest power of $p$ that divides $a, b, c$ respectively. We have $ p^{\min (k, \alpha ) } $ is the largest power of $p$ that divides $x=(n,a)$. Thus, it remains to show that
$$\max[ \min(k, \alpha), \min (k, \beta), \min (k, \gamma) ] = k, $$
which is equivalent to  $$\max( \alpha, \beta, \gamma ) \geq k.$$
Having motivated looking at this animal, we now show that is it true. WLOG, $ \alpha \geq \beta \geq \gamma = 0$ (since $\gcd(a,b,c) = 1$). If $\beta \geq k$, then we are done. Otherwise, assume that $ k > \beta$.
We will work modulo $p^{k+\beta}$. We are given that
$$ abc = n^3 - (a+b+c)n^2 + (ab+bc+ca) n  -abc $$
Since $k > \beta$ hence $ n^3, n^2 \equiv 0 \pmod{p^{k+ \beta} }$. As such, we get
$$ abc \equiv (ab+bc+ca)n - abc \equiv -abc \pmod{p^{k+\beta}} $$
Hence, $ \alpha + \beta \geq k + \beta$, which gives us $\alpha \geq k$.

Note: The reason for $k > \beta$, is because if $ \alpha \geq \beta > k$, then the $cn^2$ term doesn't disappear when taking mod $p^{k + \beta}$.
