An interesting property of a particular set of triples - multiplying two and adding the other always gives 1 
Find all triples of real numbers such that multiplying any two in a
  triple and adding the third always gives $1$.

When will this be the case? How can we find all such triples?
So far, I've let the numbers be $a$, $b$ and $c$. 
Therefore, $ab+c$ etc must $= 1$ but how can I restrict the possibilities to find all possible triples? 
I seem to think this has something to do with 1s and 0s, eg. $0,0,1$ or $0,1,1$
Many thanks guys!!
 A: Hint: you need to use more than one condition at once. For example, we must have
$$ab+c=ac+b.$$
What can you deduce from that equation?
A: If $ab+c=ac+b=bc+a=1$, then $a(b-c)+(c-b)=0$. Factoring out the common factor $b-c$, we get $(a-1)(b-c)=0$. Hence, either $a=1$ or $b=c$. Likewise, either $b=1$ or $a=c$, and either $c=1$ or $a=b$.
Suppose that $a=1$. Then, $b+c=bc+1=1$, so $bc=0$. Hence, one of $b$ and $c$ must be $0$ and the other must be $1$, giving the solutions $(1,0,1)$ and $(1,1,0)$.
Likewise, if $b=1$, then similar steps will give the solutions $(0,1,1)$ and $(1,1,0)$.
Finally, if $c=1$, then similar steps will give the solutions $(0,1,1)$ and $(1,0,1)$.
If none of $a$, $b$, and $c$ are equal to $1$, then they must all be equal. This reduces to solving the equation $a^2+a=1$, or $a^2+a-1=0$. By the quadratic formula, the two resulting solutions are $a=\frac{-1+\sqrt{5}}{2}$ and $a=\frac{-1-\sqrt{5}}{2}$.
Hence, the $5$ solutions are $(0,1,1)$, $(1,0,1)$, $(1,1,0)$, $(\frac{-1+\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2})$, and $(\frac{-1-\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2})$. Using $\phi$, the golden ratio, the last $2$ solutions can be written as $(\phi-1,\phi-1,\phi-1)$ and $(-\phi,-\phi,-\phi)$ respectively.
A: You have three equations and three variables to solve for. The three equations are $$ab + c=1$$ $$ac +b=1$$ $$bc + a=1$$
Using the first equation and solving for $c$, we get $$c = 1-ab$$
This now reduces to two equations $$a(1-ab)+b=a+b-a^2b=1$$ $$b(1-ab)+a=a+b-ab^2=1$$
Subtracting the second equation from the first, I get $$-a^2b+ab^2=0$$ This can be factored as $$ab(b-a)=0$$
Which means that there are three possible cases $a=0, b=0,$ and $a=b$.
If $a=0$, it must also be true that $b=c=1$ in order to satisfy the original three equations. Similarly, if $b=0$, it must be true that $a=c=1$. If $a = b \not = 0$, there is the cubic $$2a-a^3=1$$ that must be solved. There are three solutions $$a= \frac{-1 \pm \sqrt{5}}{2}, 1$$
This means that all solutions to the problem are $(0, 1, 1), (1, 0, 1), (1, 1, 0), \left(\frac{-1 \pm \sqrt{5}}{2}, \frac{-1 \pm \sqrt{5}}{2}, \frac{-1 \pm \sqrt{5}}{2}\right)$
