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I have system of polynomial equations, say,

$$(a+b)x + (c+a)x^2 +(a+2d)x^3 = 0.$$

My idea is that an equivalent condition to solving this system, for $x\neq 0$ and where $x$ is not a root of the equation, would be equivalent to saying

$$a+b = 0,\; c+a = 0,\; a+2d = 0.$$

My idea is to add all the coefficients then solve the resulting diophantine equation,

$$3a + b + c +2d = 0. $$

Would the solution of this system be equivalent to the system of polynomial equations above?

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    do you need to find all values of $x$ for which this is true, or all values of $a,b,c,d$ for which you get equality *for all possible* x? – 2017-01-30
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    @gt6989b the latter. – 2017-01-30

1 Answers 1

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You must solve $$ \begin{cases} a+b &= 0\\ a+c &= 0\\ a+2d &= 0 \end{cases}, $$ which implies $b = c = -a$ and $d = -a/2$. Hence, your general solution will look like $$ \begin{pmatrix}a\\b\\c\\d\end{pmatrix} = \begin{pmatrix}a\\-a\\-a\\-a/2\end{pmatrix} = (a/2) \begin{pmatrix}2\\-2\\-2\\-1\end{pmatrix}. $$ In other words, any linear scale of the vector $(2, -2, -2, -1)$ will do.

Your approach only produces one constraint out of a much more restrictive problem. You can notice that my solution would satisfy your equation, but the solution $(0,0,-2,1)$ would satisfy your equation, but not solve the original problem correctly.