A University of Cambridge Math interview question

Make the substitution $P=x_{1}x_{2}x_{3}x_{4}$ . Our system of equations simplify to:

$x_{i}+\frac{P}{x_{i}}=2$ , where $i=1,2,3,4$

Clearing denominators,we can write the equation above as

${x_{i}}^{2}-2x_{i}+P=0$

Using the quadratic formula, we find that $x_{i}=1 \pm \sqrt{1-P}$ , so $x_{i}$ can only take $2$ forms.

Therefore, for a certain set of solution $(x_{1},x_{2},x_{3},x_{4})$ which satisfies the systems of equation above, it must take the form of $(a,a,a,a),(a,a,a,b)$ or $(a,a,b,b)$ . Don't worry about permutation for now, we will take care of that later.

First case: $(a,a,a,a)$

We get $a^{3}+a-2=0$ , which simplifies to

$(a-1)(a^{2}+a+2)=0$ .

It is easy to see that the only real root is $a=1$ , therefore our first set of solution is $(1,1,1,1)$

Second case: $(a,a,a,b)$

We obtain two equations $a+a^{2}b=2$ and $b+a^{3}=2$ by substitution.

Equating the L.H.S. of both equations yield

$a+a^{2}b=b+a^{3}$ .

Moving all the terms to one side and factorizing yields

$(b-a)(1-a^{2})=0$

$(b-a)(1-a)(1+a)=0$

$a=b$ or $a=1$ or $a=-1$

If $b=a$ , we get back the same solution in first case.

If $a=1$ , substitution into $a+a^{2}b=2$ yields

$1+b=2 \Rightarrow b=1$ , which gives us again the first solution.

If $a=-1$ , substitution into $a+a^{2}b=2$ yields

$b-1=2 \Rightarrow b=3$ .

Therefore, we found our second set of solution, that is $(-1,-1,-1,3)$ and its permutation.

Third case: $(a,a,b,b)$

We get the equations $b+a^{3}=2$ and $a+a^{2}b=2$ by substitution.

Equating the L.H.S. of both equations give:

$b+a^{3}=a+a^{2}b$

$(b-a)(1-a^{2})=0$ but this is the same as our second case!

We therefore conclude that there exists only five sets of solution to the systems of equation above, namely $(1,1,1,1)$ and the permutations of $(3,-1,-1,-1)$

Here is a slightly different approach. If $\{p,q,r,s\} = \{1,2,3,4\}$ then, by subtracting two of the equations, $(x_p - x_q)(1 - x_rx_s) = 0 \;.$ If no two of $x_1,x_2,x_3,x_4$ are equal, then $x_rx_s = 1$ for all $r \neq s$ . But this would imply that $x_1 = x_2 = x_3 = x_4$ , which we have said is impossible. Thus at least two of the $x_i$ must be equal.

Suppose that the four values are $a,a,b,c$ . Then the equations become $a + abc \; = \; b + a^2c \; = \; c + a^2b \; = \; 2$ Subtracting the last two gives $(b-c)(1 - a^2) \; = \; 0 \;.$ If $a^2 \neq 1$ , then $b=c$ . But then $a + ab^2 = b + a^2b = 2$ , so that $(a-b)(1-ab) = 0$ , If $ab = 1$ then $a+b = 2$ , so that $a=b=1$ . If $a=b$ then $0 = a^3 + a - 2 = (a-1)(a^2 + a + 2)$ , and so $a=b=1$ . Either way, we have $a=1$ , which is not possible.

Thus we must have $a^2 = 1$ .

• If $a=1$ we deduce that $b+c=2$ and $bc = 1$ , so that $b=c=1$ .
• If $a=-1$ we deduce that $b+c=2$ and $bc = -3$ , so that $\{b,c\} = \{-1,3\}$ .

Thus the only possibilities are $(1,1,1,1)$ and $(-1,-1,-1,3)$ (plus permutations), giving us $\boxed{5}$ solutions.