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Question: How many possible values of (a, b, c, d), with a, b, c, d real, are there such that abc = d, bcd = a, cda = b and dab = c?

I tried multiplying all the four equations to get: $$(abcd)^2 = 1$$

Not sure how to proceed on from here. Won't there be infinite values satisfying this equation?

Gummy bears
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  • say $a\neq \pm 1$. Then $abcd=a\times bcd=a^2\neq 1$. And similarly for $b,c,d$. – lulu May 07 '16 at 11:38
  • (note: your last equation rejects the valid solution ${0,0,0,0}$). – lulu May 07 '16 at 11:40
  • @Lulu Sorry but didn't understand your first comment... How does that help in solving? – Gummy bears May 07 '16 at 11:43
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    It shows that, once you exclude the solution ${0,0,0,0}$, all the variables have to be $\pm 1$. You can have all of one sign, or two of each. But that's it. – lulu May 07 '16 at 11:44

3 Answers3

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If one of the variables is $0$ then all of them are...so one solution is $\{0,0,0,0\}$.

Let's now exclude that solution. As the OP points out, multiplying the equations together yields $$(abcd)^3=abcd\implies (abcd)^2=1$$

Now, suppose that $a\neq \pm 1$. We remark that $$abcd=a\times bcd=a\times a=a^2\neq \pm 1$$ which contradicts $(abcd)^2=1$. Thus $a$ must be $\pm 1$. Similarly, each variable must be $\pm 1$. Inspection quickly shows that either all four have the same sign or we have two of each, and we are done.

lulu
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1) If $abcd = 0$, i.e., if one at least among $a,b,c,d$ is zero, all are clearly zero. Thus there is a solution: $$a=b=c=d=0$$

2) If $abcd \neq 0$, let $A=\ln(|a|), B=\ln(|b|), C=\ln(|c]), D=\ln(|d|).$

Taking absolute values and then logarithms of the 4 equations, we obtain the following linear homogeneous system $$\begin{bmatrix}1&1&1&-1\\-1&1&1&1\\1&-1&1&1\\1&1&-1&1 \end{bmatrix}\begin{bmatrix}A\\B\\C\\D \end{bmatrix}=\begin{bmatrix}0\\0\\0\\0 \end{bmatrix}$$

The determinant of the matrix $M$ of the system is $16 \neq 0$, thus the kernel of $M$ is reduced to $$(A,B,C,D)=(0,0,0,0)=(\ln(1),\ln(1),\ln(1),\ln(1))$$ Therefore

$$|a|=1, |b|=1, |c|=1, |d|=1 \ \ \ (1)$$

All solutions of (1) may not be solutions to the initial system (because, by taking absolute values, we possibly have enlarged the set of solutions).

Thus, we have to check the 16 different possible sign combinations for $a,b,c$ and $d$. Doing this, 8 solutions remain: $$(a,b,c,d)=(-1, -1, -1, -1), (-1, 1, -1, 1), (-1, 1, 1, -1), (-1, -1, 1, 1), (1, 1, -1, -1), (1, -1, -1, 1) , (1, -1, 1, -1), (1, 1, 1, 1).$$

Jean Marie
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$$(abcd)^3=abcd \\ \implies (abcd)^2=1\ or, \ abcd=0 \\ \implies abcd=+1,-1,0 $$ If we look carefully, the initial conditions clearly say that we can't take $3$,$-1$s or only one $-1$. from these, one solution is $(a,b,c,d)=(0,0,0,0)$ and for the other ones i.e. for $abcd=1$ we can say , no of pairs are $\frac{4!}{2! \times 2!}=6 \ (2--(+1)\ and \ 2--(-1))$, $(a,b,c,d)=(1,1,1,1),(-1,-1,-1,-1)$. Hence, total $9$ solutions. $\blacksquare$

sougata
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