System of equations with 4 unknowns.

Question

I'm trying to solve this system of equations but I'm reaching a dead end.

$$\begin{array}{lcl} xyz & = & x+y+z \ \ \ \ \ \ \ (1)\\ xyt & = & x+y+t \ \ \ \ \ \ \ (2)\\ xzt & = & x+z+t \ \ \ \ \ \ \ (3)\\ yzt & = & y+z+t \ \ \ \ \ \ \ (4)\\ \end{array}$$

So, (1)-(2) gives $$xyz-zyt=xy(z-t)=z-t\Rightarrow xy=1.$$

(3)-(4) gives $$xzt-yzt=zt(x-y)=x-y\Rightarrow zt=1.$$

This system reduces to $$\begin{array}{lcl} xy & = & 1 \ \ \ \ \ \ \ (5)\\ zt & = & 1 \ \ \ \ \ \ \ (6)\\ \end{array}$$

One trivial solution is $x=y=z=t=\pm1,$ but this can't be because then I'd have divided by zero earlier in my simplifications.

How do I solve this one?

Yes, got that from Wolfram Alpha too. I want to know how to algebraically prove those roots. — Parseval, Jun 14 '17 at 07:08
Can the solution be in $\mathbb{C}$? If yes, we also have $(i,i,-i,-i),$ $(-i,-i,i,i),$ $(i,-i,i,-i),$ $(-i,i,-i,i),$ $(i,-i,-i,i),$ $(-i,i,i,-i).$ — Reinhard Meier, Jun 14 '17 at 08:51
@Parseval Use $xy=1$ instead of $z=t$. Plug this into the first two equations and you get $x+y = 0.$ $xy=1$ and $x+y=0$ gives $(x,y)\in{(i,-i),(-i,i)}.$ The third and fourth equation lead to $zt=1$ or $x=y$. We already know that $x\neq y$, thus we have $zt=1$ which in turn gives $(z,t)\in{(i,-i),(-i,i)}.$. Starting with different pairs of equations will give you all the possible combinations of $i$ and $-i$ — Reinhard Meier, Jun 14 '17 at 09:08
@Parseval the missing complex solution using only equations (1), (2), and (3) comes from the last centered equation $(z^2-3)(z^2+1)=0$ which implies $z \in {\pm \sqrt{3}, \pm i}$. — Paolo Leonetti, Jun 14 '17 at 12:23

score 10 · Accepted Answer · 2017-06-14T10:09:36.243

10

Subtracting $(1)$ and $(2)$, you establish

$$xy=1\lor z=t.$$

But plugging $xy=1$ in $(1)$ reduces it to

$$z=x+y+z$$ or $$x=-y,$$ which is not compatible.

Repeating with a circular permutation of the variables, the only possible solutions are trivially with $x=y=z=t$.

edited Jun 14 '17 at 10:09

answered Jun 14 '17 at 07:47

This was a nice one :) ! – Parseval Jun 14 '17 at 07:51
2

Or complex solutions ;-) – M.Herzkamp Jun 14 '17 at 10:44
@M.Herzkamp: you should mention infinite solutions too. – Jun 14 '17 at 12:15

Paolo Leonetti · Answer 2 · 2017-06-14T08:10:22.760

5

We will find all solutions using only (1), (2), and (3).

Making the difference between the first two equations you get $xy=1$ or $x=y$. Let us suppose that $xy=1$. Then $x,y$ have the same sign and are both $\neq 0$. This implies that $$ z=xyz=x+y+z=x+\frac{1}{x}+z \implies x+\frac{1}{x}=0, $$ which is impossibile since $|x+\frac{1}{x}|\ge 2$ for all non-zero reals $x$ (by AM-GM). Therefore $x=y$ ($\neq 1$ and $\neq -1$, otherwise $xy$ would still be $1$).

The first and second equations become (*) $$ z(x^2-1)=2x \text{ and }t(x^2-1)=2x. $$ In particular, $z=t$. This means the only two first equations imply $x=y \notin \{\pm 1\}$ and $z=t$.

Lastly, using (3), we get $xz^2=x+2z$, hecen we reduce to $$ x(z^2-1)=2z\text{ and }z(x^2-1)=2x. $$ If $x=0$ then also $z=0$ and viceversa. Otherwise substituing the first into the second we get $$ z=\frac{2x}{x^2-1}=\frac{4z}{\left(z^2-1\right) \left(\left(\frac{2z}{z^2-1}\right)^2-1\right)}=\frac{4z(z^2-1)}{4z^2-(z^2-1)^2}. $$ Since $z\neq 0$ then $$ 4z^2-(z^2-1)^2=4(z^2-1) \Leftrightarrow (z^2-3)(z^2+1)=0. $$ Hence $|z|=\sqrt{3}$. Thus we have only the trivial solutions.

edited Jun 14 '17 at 08:10

answered Jun 14 '17 at 07:35

Paolo Leonetti

15,423
3
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According to wolfram alpha, the solutions are $(0,\pm\sqrt{3}).$ – Parseval Jun 14 '17 at 07:38
Yes, I can't see anywhere that you wrote that 0 and $-\sqrt{3}$ are solutions. – Parseval Jun 14 '17 at 07:40
I wrote $x=y$ and $z=t$. Also $x=0$ iff $z=0$, otherwise $|z|=\sqrt{3}$ from which $x=z$. – Paolo Leonetti Jun 14 '17 at 07:41
You made it ten times more complicated than necessary. – Jun 14 '17 at 07:55
Except the fact that I never used the last equation. – Paolo Leonetti Jun 14 '17 at 08:00
@PaoloLeonetti: right, not using the last equation makes it ten times more complicated than necessary. – Jun 14 '17 at 08:06

score 4 · Answer 3 · answered Jun 14 '17 at 07:06

4

Two trivial solutions.

Notice that by symmetry, if we let $x=y=z=t=a$, each equation is just $3a=a^3$, so we can have $(x,y,z,t)=(0,0,0,0), (\sqrt{3},\sqrt{3},\sqrt{3},\sqrt{3}), (-\sqrt{3},-\sqrt{3},-\sqrt{3},-\sqrt{3})$.

answered Jun 14 '17 at 07:06

Saketh Malyala

13,637

why is this down voted? – Saketh Malyala Jun 14 '17 at 07:09
Define "by symmetry"
I did not downvote.
– Parseval Jun 14 '17 at 07:09
Each equation is similar, in that it is of the form abc=a+b+c. it contains all 4 variations of this equation picking from the variables (x,y,z,t). and if all the variables are the same, each equation says the same thing – Saketh Malyala Jun 14 '17 at 07:10
its like if a+b=ab, a+c=ac, b+c=bc – Saketh Malyala Jun 14 '17 at 07:11
all 3 equations would be similar to abc – Saketh Malyala Jun 14 '17 at 07:11
1

@SakethMalyala Wasn't my downvote, but Two trivial solutions doesn't count as an answer. – dxiv Jun 14 '17 at 07:13
I'm sorry :( @dxiv – Saketh Malyala Jun 14 '17 at 07:16

score 0 · Answer 4 · answered Jun 14 '17 at 16:29

0

No. $xyz-zyt=xy(z-t)=z-t\implies (\text{$xy=1$ or $z=t$})$.

answered Jun 14 '17 at 16:29

Steven Alexis Gregory

26,769

System of equations with 4 unknowns.

4 Answers4