I'm working on one privacy-preserving cryptographic protocol that both parties try to compute one function with their private inputs.
If one party is trying to guess the other party's inputs then she should solve about 6 nonlinear equations with 6 variables ($a,b,c,w,y,z$).
Since those variables are representing the latitude and longitude of some coordinates, I'm looking for solutions in integers and real numbers. Here is one example:
$$\left\{ \begin{array}{c} ab+wy-2a-2b-2w-2y=-13\\ ay-bw+2b-2a+2w-2y=15\\ bc+yz-2b-2c-2y-2z=-18\\ bz-cy+2c-2b+2y-2z=5\\ ca+zw-2c-2a-2z-2w=-9\\ cw-az+2a-2c+2z-2w=7 \end{array}\right\}$$
but I don't know how can I solve this system of equations. Could you help me?
Here is an attempt to find all real solutions. Let $A$, $B$, $C$, $W$, $Y$, and $Z$ denote $a-2$, $b-2$, $c-2$, $w-2$, $y-2$, and $z-2$, respectively. We have $$AB+WY=-5\,,$$ $$AY-BW=15\,,$$ $$BC+YZ=-10\,,$$ $$BZ-CY=5\,,$$ $$CA+ZW=-1\,,$$ and $$CW-AZ=7\,.$$ Write $P:=A+W\text{i}$, $Q:=B+Y\text{i}$, and $R:=C+Z\text{i}$. Then, $$\bar{P}Q=-5+15\text{i}\,,$$ $$\bar{Q}R=-10+5\text{i}\,,$$ and $$\bar{R}P=-1+7\text{i}\,.$$ From here, it is easily seen that $|P|=\sqrt{10}$, $|Q|=5$, and $|R|=\sqrt{5}$. Letting $P=\sqrt{10}\,u$, where $u\in\mathbb{C}$ such that $|u|=1$, we get $$Q=5\,\left(\frac{-1+3\text{i}}{\sqrt{10}}\right)\,\bar{u}$$ and $$R=\sqrt{5}\,\left(\frac{-1-7\text{i}}{5\sqrt{2}}\right)\,\bar{u}\,.$$ If $u=\cos(t)+\text{i}\sin(t)$, then $$P=\sqrt{10}\,\cos(t)+\text{i}\,\sqrt{10}\,\sin(t)\,,$$ $$Q=5\,\left(\frac{-\cos(t)+3\,\sin(t)}{\sqrt{10}}\right)+\text{i}\,5\,\left(\frac{3\,\cos(t)+\sin(t)}{\sqrt{10}}\right)$$ and $$R=\sqrt{5}\,\left(\frac{-\cos(t)-7\,\sin(t)}{5\sqrt{2}}\right)+\text{i}\,\sqrt{5}\,\left(\frac{-7\,\cos(t)+\sin(t)}{5\sqrt{2}}\right)\,.$$ That is, $$a=A+2=\text{Re}(P)+2=\sqrt{10}\,\cos(t)+2\,,$$ $$b=B+2=\text{Re}(Q)+2=5\,\left(\frac{-\cos(t)+3\,\sin(t)}{\sqrt{10}}\right)+2\,,$$ $$c=C+2=\text{Re}(R)+2=\sqrt{5}\,\left(\frac{-\cos(t)-7\,\sin(t)}{5\sqrt{2}}\right)+2\,,$$ $$w=W+2=\text{Im}(P)+2=\sqrt{10}\,\sin(t)+2\,,$$ $$y=Y+2=\text{Im}(Q)+2=5\,\left(\frac{3\,\cos(t)+\sin(t)}{\sqrt{10}}\right)+2\,,$$ and $$z=Z+2=\text{Im}(R)+2=\sqrt{5}\,\left(\frac{-7\,\cos(t)+\sin(t)}{5\sqrt{2}}\right)+2\,.$$ There are indeed infinitely many real solutions, but they form a circle in $\mathbb{R}^6$ centered at $(a,b,c,w,y,z)=(2,2,2,2,2,2)$ with radius $\sqrt{10+25+5}=2\sqrt{10}$. In other words, $$\begin{bmatrix}a\\b\\c\\w\\y\\z\end{bmatrix}=\begin{bmatrix}2\\2\\2\\2\\2\\2\end{bmatrix}+\begin{bmatrix}10\\-5\\-1\\0\\15\\-7\end{bmatrix}\,\frac{\cos(t)}{\sqrt{10}}+\begin{bmatrix}0\\15\\-7\\10\\5\\1\end{bmatrix}\,\frac{\sin(t)}{\sqrt{10}}$$ for $t\in\mathbb{R}$, which is contained in the $2$-dimensional affine plane $$\begin{bmatrix}2\\2\\2\\2\\2\\2\end{bmatrix}+\text{span}_\mathbb{R}\left\{\begin{bmatrix}10\\-5\\-1\\0\\15\\-7\end{bmatrix},\begin{bmatrix}0\\15\\-7\\10\\5\\1\end{bmatrix}\right\}\,.$$
If you want to find rational solutions, then the task is to solve for rational numbers $p$ and $q$ such that $p^2+q^2=10$. First, note that $(p,q)=(1,3)$ is a solution. Any other solution except $(p,q)=(1,-3)$ must satisfy $q-3=r\,(p-1)$ for some $r\in\mathbb{Q}$. Plug this in to get $$p^2+\big(r\,(p-1)^2+3\big)^2=10\,,$$ or $$(p-1)\,\big(r^2\,(p-1)+6\,r+(p+1)\big)=0\,.$$ Thus, $p=1$ or $$p=\frac{r^2-6\,r-1}{r^2+1}=1-2\,\left(\frac{3r+1}{r^2+1}\right)\,.$$ This means $q=3$ or $$q=3-2\,\left(\frac{r(3r+1)}{r^2+1}\right)\,.$$ Note that $r=-\dfrac13$ already gives $(p,q)=(1,3)$, so there is no need to treat this case separately. The only point that is included in this parametrization is $(p,q)=(1,-3)$. However, the parametrization $$(p,q)=\Biggl(1-2\,\left(\frac{3r+1}{r^2+1}\right),3-2\,\left(\frac{r(3r+1)}{r^2+1}\right)\Biggr)$$ already includes $(p,q)=(1,-3)$ if we take $r\in\mathbb{Q}\cup\{\infty\}$. That is, the rational solutions $(a,b,c,w,y,z)\in\mathbb{Q}^6$ satisfy $$\small\begin{bmatrix}a\\b\\c\\w\\y\\z\end{bmatrix}=\begin{bmatrix}2\\2\\2\\2\\2\\2\end{bmatrix}+\begin{bmatrix}10\\-5\\-1\\0\\15\\-7\end{bmatrix}\,\frac{1}{10}\,\Biggl(1-2\,\left(\frac{3r+1}{r^2+1}\right)\Biggr)+\begin{bmatrix}0\\15\\-7\\10\\5\\1\end{bmatrix}\,\frac{1}{10}\,\Biggl(3-2\,\left(\frac{r(3r+1)}{r^2+1}\right)\Biggr)$$ for some $r\in\mathbb{Q}\cup\{\infty\}$.