Solving a specific polynomial system of equations of degree $2$

Question

I want to know if this system of equations has a solution within the complex numbers.

And also if it has solutions in the reals.

None of the variables is allowed to be equal to zero.

$$a_3 b_2 + b_3^2 = a_1 + a_2 b_3 + c_3$$ $$c_3 b_3 + b_2^2 = c_1 + a_2 + b_2 + a_3$$ $$b_1 + b_2 b_3 = a_2$$ $$a_1 b_2 + b_1 b_3 = a_2 b_1 + c_1$$ $$c_1 b_3 + b_1 b_2 = a_2 b_1 + a_1$$

Are there solutions or it is overdetermined ?

We have $8$ variables and $5$ equations so it seems it should have a solution set, but I can not find it.

Did I overlook a hidden contradiction ?

Context :

Assuming I made no mistake , they come from solving this question :

Is this 3D algebra $T$ power-associative?

Where I try to find cases that are power-associative.

I tried software but it gave an error.

EDIT

It turns out that by substition $x/q - v = x' , y/s - w = y'$ in the related link

Is this 3D algebra $T$ power-associative?

we can make $b_2 = c_2 = 0$ , $b_1 = 1$.

This greatly simplifies the system !

We arrive at

$$a_3 b_2 + b_3^2 = a_1 + c_3$$ $$c_3 b_3 + b_2^2 = c_1 + b_2 + a_3$$ $$1 + b_2 b_3 = 0$$ $$a_1 b_2 + b_3 = c_1$$ $$c_1 b_3 + b_2 = a_1$$

Plugging it into mathematica gave me these , but I am not sure if they are all solutions and such.

https://www.wolframalpha.com/input?i=a_3+b_2+%2B+b_3%5E2+%3D+a_1+%2B+c_3%2Cc_3+b_3+%2B+b_2%5E2+%3D+c_1+%2B+b_2+%2B+a_3%2C1+%2B+b_2+b_3+%3D+0%2Ca_1+b_2+%2B+b_3+%3D++c_1%2Cc_1+b_3+%2B+b_2+%3D++a_1

It appears everything is a function of $a_1$ so only one degree of freedom and a few branches,

with 2 pairs of valid real solutions ($c_3$ and $a_3$ must be nonzero )

$$a_1≈2.09362, a_3≈2.50401, b_2≈-0.463692, b_3≈2.1566, c_1≈1.18581, c_3≈1.39623$$

$$a_1≈2.09362, a_3≈2.50401, b_2≈4.12858, b_3≈-0.242214, c_1≈8.40149, c_3≈8.30304$$

Answer 1

Yes, there are solutions. For instance, here is one solution:

$$a_1=4, a_2=3, a_3=18, b_1=1, b_2=1, b_3=2, c_1=3, c_3=12.$$

In general, it is straightforward to find solutions, by simple algebra. Simply pick a variable, eliminate it (find an expression for it in terms of the other variables, then plug in all instances of it for that variable), and repeat.

In this manner, we can obtain the following general solution to your system of equations. Pick $b_1,b_2,b_3$ arbitrarily, such that $b_2 b_3\ne 1$. Set

$$\begin{align*} c_1 &= (b_1^2 + b_1 b_2 b_3 - b_1 b_2^2 + b_1^2 b_2 + b_1 b_2^2 b_3 - b_1 b_3)/(b_2 b_3-1)\\ a_1 &= b_3 c_1 + b_1 b_2 - b_1^2 - b_1 b_2 b_3\\ c_3 &= (a_1 + b_1 b_3 + b_2 b_3^2 - b_2^3 + b_2 c_1 + b_1 b_2 + b_2^2 b_3 + b_2^2 - b_3^2)/(b_2 b_3-1)\\ a_3 &= c_3 b_3 + b_2^2 - c_1 - b_1 - b_2 b_3 - b_2\\ a_2 &= b_1 + b_2 b_3 \end{align*}$$

Then if all of $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_3$ are non-zero, then this gives a valid solution to your system of equations. If any of them are zero, or $b_2b_3=1$, then pick a new set of values of $b_1,b_2,b_3$. This gives you infinitely many solutions to your equations.

In the future, if you face any similar problems, give that a try on your own, and you should be able to solve it yourself.

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How did I find this? Following the approach listed above, eliminating variables in the order $a_2,a_3,c_3,a_1,c_1$. For instance, we find

$$a_2 = b_1 + b_2 b_3,$$

then replacing all occurrences of $a_2$ with the RHS gives us a new system of 4 equations in 7 variables. Now solve for $a_3$, finding

$$a_3 = c_3 b_3 + b_2^2 - c_1 - b_1 - b_2 b_3 - b_2.$$

Plug in this RHS to replace all occurrences of $a_3$, and we now have a new system of 3 equations in 6 variables. Repeat, isolating one variable at a time and plugging in, until you obtain a system of 0 equations in 3 variables. That yields the solution I showed above.