If $a,b,c \in \mathbb N$ and $\gcd(a^2-1,b^2-1,c^2-1)=1$, prove that $\gcd(ab+c,bc+a,ca+b)=\gcd(a,b,c)$. The whole problem is in the title. I'm out of ideas and can't think of a way to solve this, so I'd like to get some help. Thanks.
 A: *

*if $n$ divides $a$, $b$ and $c$ then it divides $(ab+c,bc+a,ca+b)$. Hence
  $$gcd(a,b,c)\le gcd(ab+c,bc+a,ca+b) \quad [1]$$

*if $n$ divides $ab+c$, $bc+a$ and $ca+b$, then it divides
  $$c(ca+b)-(bc+a)=a(c^2-1)$$
  $$b(bc+a)-(ab+c)=c(b^2-1)$$
  $$a(ab+c)-(ca+b)=b(a^2-1)$$
Hence $$gcd(ab+c,bc+a,ca+b)\le gcd(a(c^2-1),c(b^2-1),b(a^2-1))\quad [2]$$
Now, we know that $$gcd(xy,z)\le gcd(x,z).gcd(y,z)$$
and $gcd(a,a^2-1)=1$ (the same for $b$ and $c$) (So when you develop all terms you can ignore any gcd that contains $x$ and $x^2-1$)
Hence $$gcd(a(c^2-1),c(b^2-1),b(a^2-1))\le gcd(a,b,c).gcd(a^2-1,b^2-1,c^2-1)=gcd(a,b,c)$$
Hence (from $[2]$)  $$gcd(ab+c,bc+a,ca+b)\le gcd(a,b,c) \quad [3]$$  

Now from $[1]$ and $[3]$,
$$gcd(ab+c,bc+a,ca+b)=gcd(a,b,c)$$
A: Let $r=ab+c,\ s=bc+a,\ t=ca+b.$ That $\gcd(a,b,c)|\gcd(r,s,t)$ is immediate. Note from the definitions that
$$a=s-bc, \\ b= t-ac, \\ c=r-ab. \tag{1}$$
Substitution of two of the right sides of $(1)$ for the corresponding left-side letters of  the product term of each of the three definitions of $r,s,t$ gives the relations
$$r=st-acs-bct+c(abc+1), \\ s=rt-acr-abt+a(abc+1),\\ t=rs-bcr-abs+b(abc+1).\tag{2}$$
Now suppose $d$ divides each of $r,s,t.$ Then from $(2)$ we have that $d$ divides each of 
$a(abc+1)\ ,b(abc+1),\ c(abc+1).$ Now in each of these products the two factors are coprime. For example any prime $p$ dividing both $a$ and $abc+1$ would have to divide $1$, not possible. It follows that, provided we can show that $d$ fails to divide $abc+1$ nontrivially, then we can conclude that $d$ divides each of $a,b,c$, thus that $\gcd(r,s,t)|\gcd(a,b,c)$ to finish the proof.
Now note that, for any independent choices of $\varepsilon_i= \pm 1$, we have that a common divisor  $d$ of $r,s,t$ divides the quantity 
$$\varepsilon_1 c (ab+c)+\varepsilon_2 a(bc+a)+\varepsilon_3 b (ca+b).$$
This last may be rearranged to
$$(\varepsilon_1+\varepsilon_2+\varepsilon_3)(abc+1) + \\
\varepsilon_1 (c^2-1)+\varepsilon_2 (a^2-1)+\varepsilon_3 (b^2-1).$$
Here is where the assumption $\gcd(a^2-1,b^2-1,c^2-1)=1$ plays in. If it happened that $d|(abc+1),$ then by choosing specific values for the $\varepsilon_i=\pm 1$ and adding, we can show that $d$ divides each of $2(a^2-1),\ 2(b^2-1), 2(c^2-1)$, so that by assumption we arrive at $d=2$ or $d=1$. Now if $abc+1$ is even, then each of $a,b,c$ is odd, which implies each of $a^2-1,b^2-1,c^2-1$ is even, against the assumption their gcd is 1. So in fact we cannot have $d=2$, and have shown $d=1$ is implied by $d|abc+1.$ This completes the argument, as detailed above.
