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.

If $a b+c=4$ then $a=\frac{4-c}{b}$ assuming that $b\not=0$ and then $b c+a =4$ which is $b c+\frac{4-c}{b}=4$ so $c(b-\frac{1}{b})+\frac{4}{b} = 4$ so $c= \frac{4-\frac{4}{b}} { b-\frac{1}{b}} = \frac{4}{b+1}$ assuming that $b\not -1,1$ and to the final equation we get that $c a+b=4$ which is $c(\frac{4-c}{b})+b=4$ which is $\frac{4}{b+1}(\frac{4-\frac{4}{b+1}}{b})+b=4$ and solving for $b$ we get that $b=3,\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1-\sqrt{17})$ so $c=1,\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1-\sqrt{17})$ and $a=1,\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1-\sqrt{17})$ so the only solutions are :
$(1,3,1)$ and all its permutations and $(\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1+\sqrt{17}),\frac{1}{2}(-1+\sqrt{17})),(\frac{1}{2}(-1-\sqrt{17}),\frac{1}{2}(-1-\sqrt{17}),\frac{1}{2}(-1-\sqrt{17}))$
