0
$\begingroup$

I found system of equations on internet, I want to practice math and to solve it.

All $x$,$y$ and $z$ are integers.

$$\begin{cases} x+y+z=3 \\ x^3+y^3+z^3=3 \end{cases} $$

Should I start by watching all combinations that sum up to 3, please give me some hint to start.

  • 0
    Are $x,y,z$ integers? – 2017-02-27
  • 0
    We can see that x=1, y=1 and z=1 works.That solves the system. – 2017-02-27
  • 0
    Yes $x,y,z are integers. – 2017-02-27
  • 0
    Positive integers ? – 2017-02-27
  • 0
    @JeanMarie seems to be a basic system of equations from an Algebra course, that usually implies all real numbers. – 2017-02-27
  • 0
    @Heavenly96, the OP already answered that $x,y,z$ are integers (see above comment). So don't correct JeanMarie. – 2017-02-27
  • 0
    Nice find, @DietrichBurde. Thanks! – 2017-02-27

3 Answers 3

1

Consider the identity $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx). $$ Then since $x+y+z=3=x^3+y^3+z^3 $, thus we need to solve $$1-xyz=x^2+y^2+z^2-xy-yz-zx .$$

  • 0
    This is a well-known identity. I used it several times in high school – 2017-02-27
  • 0
    do you remember this: if sum of three numbers is zero, then the some of cubes of them is equal to 3 times their product? – 2017-02-27
  • 0
    @user420255 It seems that not all of us had such an enriched high school experience... :) – 2017-02-27
1

If you are looking for solutions with $x,y,z$ real, then you can just substitue $z=3-x-y$ in the second equation to find $$ y=\frac{-9+6x-x^2\pm\sqrt{x^4-18x^2+32x-15}}{2x-6} \\ z = 3-x-y \\ x\leq -5 \mbox{ or }x=1\mbox{ or }x\geq 3 $$ (The quartic inside the square root is easy to factor because $x=1$ is a double root.)

So that is a set of three curves, one each in quadrants I, II, and IV.

If you are looking for solutions with $x,y,z\in \Bbb Z$ this immediately gives $$ \{x=1,y=1,z=1\}, \{x=-5,y=4,z=4\} $$ and by symmetry you get two others that lie along the curves in the first and second quadrants $$\{x=4,y=-5,z=4\},\{x=4,y=4,z=-5\}$$

So now we need to wonder whether there are any other integer solutions.

For that to happen, a necessary (but perhaps not sufficient) condition is that $$ x^4-18x^2+32x-15=k^2$$ for some integer $k\not\in \{0,3,4\}$.

We can show that this cannot happen, as follows:

For $x\geq 5$, $$ (x^2-1)^2 < x^4-18x^2+32x-15 + 16(x^2-1)^2

0

For a complete solution see this duplicate.So let me just add a comment. There is a stronger conjecture, namely that the Diophantine equation $$ x^3+y^3+z^3=3 $$ only has $(1,1,1)$ and $(4,4,-5)$ and its permutations as integral solutions, see this MO-question. Cassels showed that any integral solution $(x,y,z)$ has to satisfy $$ x\equiv y\equiv z \bmod 9. $$ Furthermore, Oppenheim studied the Diophantine equation $$ x^3+y^3+z^3=x+y+z. $$

  • 0
    OP wants an undergraduate level answer, but yours is quite advanced when you quoted " Oppenheim"'s work from an article. OP probably get lost reading your answer. – 2017-02-27
  • 0
    @DeepSea The duplicate gives already an elementary answer. My answer hence is just a comment, for someone interested in Diophantine equations. The OP may actually perhaps enjoy a bit more background on number theory. The statements are really elementary! – 2017-02-27