# Efficient approach to verifying that the only real solutions to this system of quadratic equations are trivial?

Let $$a,b,c,d,e,f$$ be real numbers. In the course of my research, I've arrived at the following system of equations $$\begin{eqnarray*} a+d+f &=& 0, \\ ad + af + df - b^2 - c^2 - e^2 &=& 0 \\ ae^2 + c^2 d - 2bc - adf &=& 0. \end{eqnarray*}$$

Inserting this into Wolfram, we get some unwieldy expressions, but all of them require that one of these numbers is complex.

Is there an efficient method for proving directly that the only real solution to this system of equations is the trivial one?

• To bolster the computational case: If you set up this system of equations in Mathematica, then the FindInstance command cannot find two real solutions (it does find the trivial one). That's not a mathematical proof but it does mean that Mathematica fails to find any nontrivial solution. (Code: FindInstance[{a+d+f==0,a d+d f+ a f==b^2+c^2+e^2,a e^2+c^2 d==2 b c+a d f},{a,b,c,d,e,f},Reals,2].) Aug 15, 2021 at 0:00
• @WillJagy Thank you for your comment. I'm not following, what is the implication of the failure of homogeneity? Aug 15, 2021 at 0:01
• @Semiclassical Thank you, this is a great check :) Aug 15, 2021 at 0:02
• One approach that's occasionally used for such polynomial systems, though I don't know how useful it would be here, is the sum of squares method. The idea is to show that a solution of this system of equations would imply that $a_1 f_1^2+a_2 f_2^2+\cdot +a_k f_k^2 =0$, where the $f_k$ are linear functions of the 6 variables. But the left-hand side can only be true if all the linear functions vanish identically, which is easy to test for. Aug 15, 2021 at 0:09

Squaring the first equation you get

$$a^2 + d^2 + f^2 + 2ad + 2af + 2df = 0.$$

Substituting in the second one gives

$$\tfrac12 a^2 + b^2 + c^2 + \tfrac12 d^2 + e^2 + \tfrac12 f^2 = 0,$$

but this is possible only when all the variables are equal to $$0$$.

• Very nice! Thank you :) Aug 15, 2021 at 0:10
• This looks to be an example of the sum-of-squares method I indicated in comments. Well-spotted! Aug 15, 2021 at 0:10
• Another way to have seen this result, though it looks less elegant: Eliminating $f$ from the second equation via the first, we have $a^2+c^2+ad+d^2+e^2+b^2=0$. But $a^2+ad+d^2=\frac12 a^2+\frac12 (a+d)^2+\frac12 d^2$, so the equation amounts to a sum of squares and has only trivial solution. Aug 15, 2021 at 0:43