<s>[INST] 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.
 [/INST] *

*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 $[3]$,
$$gcd(ab+c,bc+a,ca+b)=gcd(a,b,c)$$
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