<s>[INST] Find all triples $(a, b, c) \in \mathbb{R} : ab + c = bc + a = ca + b = 4$ Any help on this question would be great. Ive found that (1,1,3), (3,1,1), (1,3,1) but am confused on how to find more anymore (if there are any).Any help would be appreciated.
 [/INST] $ab + c = bc + a = ca + b=4$ 
By symmetry any solution for any any one of $a,b,c$ can be a solution for any other.
$ab +c = bc +a$ means $ab - bc = b(a-c) = a-c$  So either $b=1$ or $a=c$.
So by symmetry:
if $a\ne 1$ then $b=c$ and if $b\ne 1$ then $a = c = b$.  So it is not possible that only one of $a,b,c$ is equal to $1$.
So either:
1) All $a,b,c$ equal $1$
2) None of $a,b,c$ equal $1$ and $a=b=c$.
3) Two of $a,b,c$ equal $1$ and the third does not.
....
1) $1*1 + 1 = 4$ is not possible.
2)we get $a^2 + a = 4$ and $a = \frac {-1 \pm \sqrt{17}}{2}$.
3) wolog $a= b = 1$ so $1 + c = c + 1 = c + 1 = 4$.  $c = 3$.
So there are 5 solutions:
And $(a,b,c) = (1,1,3)$, $(1,3,1)$ , $(3,1,1), ( \frac {-1+ \sqrt{17}}2,\frac {-1+ \sqrt{17}}2,\frac {-1+ \sqrt{17}}2)$ or $ ( \frac {-1- \sqrt{17}}2,\frac {-1- \sqrt{17}}2,\frac {-1- \sqrt{17}}2)$
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