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For a Linear Algebra problem I need to find $3$ distinct solutions, $(x,y,z)$ to the following system of equations: $$x^2+y^2+z^2=1 \\ x+y+z=\pm 1$$

such that $x,y,z\in \mathbb Z$

I have found $(x,y,z)=(\pm 1, 0, 0) , ({-1\over 3}, {2\over 3}, {2\over 3})$

But I'm having trouble finding a third solution. I'm not sure if it even exists.

Just to be clear, by "distinct" I mean that that $(0, \pm 1, 0)$ and $(0, 0, \pm 1)$ count the same as the first solution.

Borat Sagdiyev
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4 Answers4

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The general solution of the system $x+y+z=1$ and $x^2+y^2+z^2=1$ is given by $$ x_1=\frac{\sqrt{d}+1-x_3}{2},\quad x_2=-\frac{\sqrt{d}+1-x_3}{2} $$ for arbitrary $x_3$, with $d=-(x_3-1)(3x_3+1)$. To find rational solutions, we must find all rational $x_3$ such that $d$ is a rational square.

The other system with $x+y+z=-1$ and $x^2+y^2+z^2=1$ works the same way.

Dietrich Burde
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$$(-\frac{2}{7},\frac{3}{7},\frac{6}{7})$$ is another one.

Peter
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    How did you find it, if you don't mind me asking? – Borat Sagdiyev Nov 17 '16 at 20:19
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    I just tried the first few squares of the primes , whether I can find a represenatition $x^2+y^2+z^2=p^2$, such that $x+y+z=\pm p$ holds as well. Probably many more solutions exist. – Peter Nov 17 '16 at 20:22
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    $(\frac{15}{19},\frac{10}{19},-\frac{6}{19})$ is another one. And $(\frac{12}{13},\frac{4}{13},-\frac{-3}{13})$ is a solution as well. – Peter Nov 17 '16 at 20:28
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Think this question in a geometry problem. So we know that $x^2+y^2+z^2=1$ is a sphere in 3D space, and $x+y+z=1$ is a plane. The solutions is the crosssection region of those two objects, which is a curve. So there are infinite solutions.

Maybe the above argument is not that simple to visualise (sorry if you cannot). In order to prove the above argument, here we do it in a more direct way.

Now we solve the system equation. Easily we know $y=1-z-x$, so we can have $x^2+(1-z-x)^2+z^2+1=0$. Wiht little bit algebra, we can know that $x^2-(1-z)x+z^2-1=0$, solve $x$ out treating $z$ as a free variable, then we can know $x=\frac{1-z\pm\sqrt{(z-1)(-3z-1)}}{2}$, then in order to get the rational solution, we have to have $(z-1)(-3z-1)=p^2$ where $p\in Q$ and also $z^2<=1$, it is easy to see that the there is a specific case meeting this requirement where $(z-1)=1/q^2(-3z-1)$, $q\in Z$. For instance, so $q=1$ will give you $(1,0,0)$, $q=2$ will give you $(3/7,-2/7,6/7)$, $q=3$ will give you $(-1/3,2/3,2/3)$. Following this logic, keep going on, you can have infinite solutions.

duanduan
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    But are there infinitely many rational solutions? – Borat Sagdiyev Nov 17 '16 at 20:28
  • Yes, it will give you infinite solutions. – duanduan Nov 17 '16 at 20:41
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    IMO this is nowhere near clear that infinitely many rational solutions will come up. By the same line of reasoning, the solutions of $x^2+y^2=3$ form a curve so we should have infinitely many rational points, but there are none. – Wojowu Nov 17 '16 at 21:08
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    @Wojowu, you are right for $x^2+y^2=3$ or any other cases like this, however, the crossection curve in this problem has a free variable $z$, so that is why there could be infinite solutions. – duanduan Nov 18 '16 at 00:50
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One hint : let $x=\frac uq$, $y=\frac vq$ and $z=\frac wq$, with $u$, $v$, $w$ and $q$ integers. Your system is equivalent to (with $\epsilon=\pm1$) : $$\left\{\begin{matrix}u+v+w=\epsilon q \\ u^2+v^2+w^2=q^2\end{matrix} \right.\iff \left\{\begin{matrix}u+v+w=\epsilon q \\ uv+vw+wu=0\end{matrix}\right.$$ So you are looking for triplets $(u,v,w)$ of integers such that $$(u,v,w).(v,w,u)=0$$ So any orthogonal matrix of order $3$ should give you solutions to your problem.

For example, looking for a triplet $(2,-3,a)$, you find $a=6$, $q=\epsilon(2-3-6)=-\epsilon7$, your solutions are $x=-\epsilon\frac27$, $y=\epsilon\frac37$ and $z=\epsilon\frac67$...

Nicolas FRANCOIS
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