# How many real, rational and complex solutions has this system of equations?

Let the system \left\{\begin{aligned} a+b+c &= 3\\ a^2+b^2+c^2 &=5\\ a^3+b^3+c^3 & =12 \end{aligned}\right.

How many real, rational and complex solutions has it?

I read System of three variables of simultaneous equations and found $e_1 = 3, e_2 = 2, e_3 = 1$. Then I do not understand the reason but I get the polynomial $t ^ 3-3t ^ 2 + 2t-1$. The question is how many , not what are the solutions, so at this point I do not know what to do ... or if I'm in the right way. Can you help me please?

• So, how many real, rational and complex solutions are there for the polynomial $t^3-3t^2+2t-1$? – Jack D'Aurizio Oct 22 '17 at 17:57
• You just have to see if the equation $t^3-3t^2+2t-1=0$ has three roots over the rationals, over the reals and over the complexes. I'll start you off. It does have three roots over the complexes. $\ddot\smile$ – Lord Shark the Unknown Oct 22 '17 at 17:58
• You way is right. Now, show that your equation has an unique real root. – Michael Rozenberg Oct 22 '17 at 17:58
• For the rational part, you may apply the rational root theorem. For the remaining part, it is enough to compute the discriminant of such polynomial, or locate its stationary points. – Jack D'Aurizio Oct 22 '17 at 17:58
• en.wikipedia.org/wiki/Rational_root_theorem – Fan Oct 22 '17 at 18:10

Note $$3^2=(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=5+2(ab+bc+ca).$$ from this, one has $$ab+bc+ca=2.$$ Also \begin{eqnarray} 3^3&=&(a+b+c)^3=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\ &=&12+3ab(a+b+c)+3bc(a+b+c)+3ca(a+b+c)-3abc\\ &=&12+3(ab+bc+ca)(a+b+c)-3abc\\ &=&30-3abc \end{eqnarray} from which, one has $$abc=1.$$ Thus $a,b$ and $c$ are the roots of the following equation $$t^3-3t^2+2t-1=0. \tag{1}$$ So the coefficients of (1) are $$a_3=1,a_2=-3,a_1=2,a_0=-1$$ and hence the discriminant is $$\Delta=18a_3a_2a_1a_0-4a_2^3a_0+a_2^2a_1^2-4a_3a_1^3-27a_3^2a_0^2=-23<0$$ and hence (1) has one real root and two complex roots.

• But what is the nature of the roots? – Lord Shark the Unknown Oct 22 '17 at 18:24
• @LordSharktheUnknown, see the update. – xpaul Oct 22 '17 at 18:41

It has 6 solutions

Here they are

$$a_1 = 2.32472$$ $$b_1 = 0.56228·i+0.337641$$ $$c_1 = 0.337641−0.56228·i$$ $$a_2 = 2.32472$$ $$b_ 2= 0.337641−0.56228·i$$ $$c_2 = 0.56228·i+0.337641$$ $$a_3 = 0.337641−0.56228·i$$ $$b_3 = 0.56228·i+0.337641$$ $$c_3 = 5.118e−18·i+2.32472$$ $$a_4 = 0.337641−0.56228·i$$ $$b_4 = 1.11022e−16·i+2.32472$$ $$c_4 = 0.56228·i+0.337641$$ $$a_5 = 0.56228·i+0.337641$$ $$b_5 = 0.337641−0.56228·i$$ $$c_5 = 1.69412e−18·i+2.32472$$

• I don't believe $c_3\ddot\smile$. – Lord Shark the Unknown Oct 22 '17 at 18:18
• @LordSharktheUnknown Why don't you believe $c_3$ – Agile_Eagle Oct 22 '17 at 18:19
• It should be the same as $b_6$, yet its imaginary part is completely different! – Lord Shark the Unknown Oct 22 '17 at 18:20
• Sorry for that it has 5 solutions – Agile_Eagle Oct 22 '17 at 18:22
• It really does have six solutions for $(a,b,c)\ \ddot\frown$. – Lord Shark the Unknown Oct 22 '17 at 18:23