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I have a system of $8-$equations for $\{x,y,z,t,w,p,r,s\}$ and I want to solve this system. $$ 4 \pi (w+y)+x+y=w+4 \pi (x+z)+z,\\4 \pi r+r+s=4 \pi (s+w-z)+w+z,\\4 \pi (p+s)+r+s=p+4 \pi (r+t)+t,\\4 \pi p+p+t=4 \pi (t+x-y)+x+y,\\p+e^{8 i \pi ^2} (t-x)=y,\\p+y=e^{8 i \pi ^2} (t+x),\\r+s=w+z,\\r+z=s+w $$

I use Mathematica (the code is given below) and the solution is as follows $$x=0,\quad y=0,\quad t=0,\quad w=-\frac{(-1-4 \pi ) z}{4 \pi -1},\quad s=z,\quad r=-\frac{(-1-4 \pi ) z}{4 \pi -1},\quad p=0$$

From the nature of my problem, I expect that at least one of the variables $x$ and $y$ to be nonzero. My question is can I trust this result? Or if I use another method and manually, will I get a different result with one of $x$ and $y$ nonzero?

eqs := {x + y + 4 π (w + y) == w + z + 4 π (x + z),
   r + 4 π r + s == w + 4 π (s + w - z) + z,
   r + s + 4 π (p + s) == p + t + 4 π (r + t),
   p + 4 p π + t == x + 4 π (t + x - y) + y,
   p + E^(8 I π^2) (t - x) == y,
   p + y == E^(8 I π^2) (t + x),
   r + s == w + z,
   r + z == s + w};

Solve[eqs, {x, y, z, t, w, s, r, p}]
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Raffaele
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1 Answers1

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This being a linear system, I have problems to conceive that another solution could exist. In any manner, notice that the results express as functions of $z$ which means that there are somewhere two identical equations.

Just for the "fun" of it, I solved the system with pen and paper using successive eliminations.

Adding $(7)$ to $(8)$, we immediately have $r=w$ and each of them gives $s=z$ (this is where we have the redundancy). Now $(2)$ is satisfied.

Adding $(5)$ to $(6)$, $p=e^{8 i \pi ^2} x$.

Reusing $(6)$, $y=e^{8 i \pi ^2} t$. At this point are left $(1)$, $(3)$ and $(4)$ for $t$, $w$, $x$ and $z$.

From $(4)$, $t=\frac{4 \pi +1}{1-4 \pi }x$ and remain $(1)$ and $(3)$ for $w$, $x$ and $z$.

Adding $(1)$ to $(3)$ gives $x=0$ and $w=-\frac{4 \pi +1}{4 \pi -1}z$.

Finally $z$ stays undefined.

By the end $$x=y=t=p=0 \qquad s=z \qquad r=w=-\frac{4 \pi +1}{4 \pi -1}z$$ which is your result.

Claude Leibovici
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