Is there any tool or technique that allows solving a system of 8 quadratic equations?

Question

I'm looking for a tool, or at least technique, that allows me to solve the following system of equations:

 a*a +  a*b +  a*c +  a*d +  a*e +  a*f +  a*g +  a*h = 0
 b*b +  b*a +  b*d +  b*c +  b*f +  b*e +  b*h +  b*g = 0
 c*c + -c*d +  c*a + -c*b +  c*g + -c*h +  c*e + -c*f = 0
-d*d +  d*c + -d*b +  d*a + -d*h +  d*g + -d*f +  d*e = 1
 e*e + -e*f + -e*g + -e*h +  e*a + -e*b + -e*c + -e*d = 0
-f*f +  f*e +  f*h + -f*g + -f*b +  f*a +  f*d + -f*c = 0
-g*g +  g*h +  g*e +  g*f + -g*c +  g*d +  g*a +  g*b = 0
-h*h +  h*g +  h*f +  h*e + -h*d +  h*c +  h*b +  h*a = 0

For integers a, b, c, d, e, f, g, h. I've tried Wolfram Alpha, but the system is too large, it exceeds the character limit. Is there any alternative tool that can help me solve this particular problem?

Answer 1

The system can be easily solved by applying Gröbner bases. One solution is given by $a=b=c=e=f=g=h=0$ and $d^2+1=0$, so $d=\pm i$. Another one is given by $c=e=f=g=h=0$, and $$d=\frac{1}{2a},\; b=-\frac{2a^2+1}{2a}$$ for nonzero $a$. There are exactly $7$ such types of solutions. None of them is integral. But the system has rational solutions.

Answer 2

The only truly quadratic equation is the $4^{th}$ one, the others are vanishing products of linear factors which can be solved by considering the different subcases.

$$ \begin{align} a (a + b + c + d + e + f + g + h) = 0 \tag{1} \\ b(a + b + c + d + e + f + g + h) = 0 \tag{2} \\ c(a - b + c - d + e - f + g - h) = 0 \tag{3} \\ d(a - b + c - d + e - f + g - h) = 1 \tag{4} \\ e(a - b - c - d + e - f - g - h) = 0 \tag{5} \\ f(a - b - c + d + e - f - g + h) = 0 \tag{6} \\ g(a + b - c + d + e + f - g + h) = 0 \tag{7} \\ h(a + b + c - d + e + f + g - h) = 0 \tag{8} \end{align} $$

Equation $(4)$ implies that $a - b + c - d + e - f + g - h \ne 0$, then $(3)$ implies $c=0$.

• If $\require{cancel}a + b + \color{gray}{\cancel{c}} + d + e + f + g + h = 0$ then equation $(7)$ reduces to $-2g^2=0 \implies g=0$.
  Then equations $(4),(5)$ reduce to $2d(a+e)=1$ and $2e(a+e)=0 \implies e=0, 2ad = 1$.
  The system then becomes: $$ \begin{align} a + b + d + f + h = 0 \\ 2ad = 1 \tag{4'} \\ f(a + d + h) = 0 \tag{6'} \\ h(d + h) = 0 \tag{8'} \end{align} $$

• If $\require{cancel}a + b + \color{gray}{\cancel{c}} + d + e + f + g + h \ne 0$ then equations $(1)-(2)$ imply $a=b=0$.
  The system then becomes: \begin{align} \\ d(- d + e - f + g - h) = 1 \tag{4''} \\ e(- d + e - f - g - h) = 0 \tag{5''} \\ f(d + e - f - g + h) = 0 \tag{6''} \\ g(d + e + f - g + h) = 0 \tag{7''} \\ h(- d + e + f + g - h) = 0 \tag{8''} \end{align}

In both cases, the systems can be solved by further breaking down the different subcases. There don't appear to exist integer solutions, though.