Is there a simpler approach to these system of equations?

Question

I recently came across the following system of equations:
$$x + y + z = 1 \\ x^2 + y^2 + z^2 = 2 \\ x^3 + y ^3 + z^3 = 3$$

And I have two questions:

One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Furthermore, is there a way to expand the proof for a more generalized set of equations, that is for this set :

$$x + y + z = 1 \\ x^2 + y^2 + z^2 = 2 \\ ...\\x^n + y^n + z^n = n$$
Two, is there a simpler approach for the prior solution set than substitution? As of now, I'm not getting anywhere with this method. I end up getting into a long-winded series of substitution and isolation that yields something like $z = z$ or $1 = 1$.

Sorry if it's a repeat question; I couldn't exactly find an accurate way to search for the equations.

Answer 1

As lab bhattacharjee has mentioned, you can view $x,y,z$ as roots of a degree 3 polynomial, and thereby obtain:

$$x+y+z=1$$ $$xy+yz+zx=-\frac{1}{2}$$ $$xyz=\frac{1}{6}$$

Vieta's Formulas relates the above expressions to the coefficients of polynomials, from which we can solve for $x,y,z$

In response to your other question:

Note that the first three relations: $$x+y+z=1$$ $$x^2+y^2+z^2=2$$ $$x^3+y^3+z^3=3$$ completely define $x,y,z$ (up to symmetry), as these relations define a degree $3$ polynomial, which clearly has $3$ (not necessarily distinct) roots, which must be $x,y,z$ respeectively.

Thus $x^4+y^4+z^4$, for instance, is already defined as $\frac{31}{6}$, and not $4$, so the general question has no solutions.

The general identities $x_1^k+x_2^k+\dots+x_m^k=n_k$ for some polynomial with degree $m$ and roots $x_1,\dots,x_m$ and their relation to the coefficients of the polynomial are actually well-known as Newton's Identities, so you can actually easily derive the relations that lab bhattacharjee mentioned in his posts.

Answer 2

HINT:

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$

$(x+y+z)^3=x^3+y^3+z^3+3(xy+yz+zx)(x+y+z)-3xyz$

So, we can find $x+y+z,xy+yz+zx,xyz $

Now $x,y,z$ are the roots of $t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz=0$

I think the same pattern can be applied for the higher powers

Answer 3

We will make this into a cubic equation in terms of $z$. We have $x+y = 1-z$ and $x^2 + y^2 = 2-z^2$.

Factor: $$x^3 + y^3 = ( x + y ) ( x^2 - xy + y^2 ) = ( 1-z ) ( 2-z^2 - xy )$$

$$2xy = ( x + y )^2 - ( x^2 + y^2 ) = ( 1-z )^2 - ( 2 - z^2 ) = 2z^2 - 2z - 1$$

So our third equation becomes (multiplying by $2$):

$$( 1 - z ) ( -4z^2 + 2z + 5 ) + 2z^3 = 6$$

Multiplying both sides by $2$ and expanding the polynomial,

$$6z^3 - 6z^2 - 3z -1 = 0$$

This will have $3$ roots (in the complex numbers), which will be solutions for $x$, $y$, and $z$ (by symmetry).

Answer 4

For this particular set of three equations, there is no solution. (And therefore there is no solution if you add more equations either.)

From the first equation $x+y+z = 1$ you can eliminate $z$ as $z = 1 - x - y$. Then the second equation $x^2 + y^2 + z^2 = 2$ becomes $x^2 + y^2 + (1 - x - y)^2 = 2$, which being a quadratic expression in $x$ and $y$, defines a conic section in the $x$-$y$ plane, in this case an ellipse. After this, the third equation $x^3 + y^3 + z^3$ which is equivalent to $x^3 + y^3 + (1 - x - y)^3$ defines another curve in the $x$-$y$ plane, which happens not to intersect with the ellipse.

The ellipse in the middle is the curve $x^2 + y^2 + (1-x-y)^2 = 2$; the curves outside but not touching it are $x^3 + y^3 + (1-x-y)^3 = 3$.

---

Now that we know what to prove (that there is no solution), we can look for an alebraic proof. :-)