Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$

Question

This problem is from a math competition, but I think is wrong:

Find the value of $x+y$ if:

$$\begin{align} xy &= 1 \\ x^2 + y^2 &= 5 \\ x^3+y^3 &= 8 \end{align}$$

Solution (I think is wrong):

$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$

So we have:

$x^3+y^3 = 4(x+y)$

$x^3+y^3 = 8$

Then:

$8 = 4(x+y)$

$x+y = 2$

However if we replace that value in $(x+y)^2$ we have:

$(x+y)^2 = 2^2 =4$

$(x+y)^2 = x^2+y^2+2xy = 5 + 2 = 7$

As you can see $4 \neq 7$, what is happening?

Answer 1

Let us call $a := x + y$ and $b := xy$. Then the proposed system of equations is equivalent to \begin{align*} \begin{cases} b = 1\\\\ a^{2} - 2b = 5\\\\ a^{3} - 3ab = 8 \end{cases} \Longleftrightarrow \begin{cases} b = 1\\\\ a^{2} = 7\\\\ a^{3} - 3a = 8 \end{cases} & \Longleftrightarrow \begin{cases} b = 1\\\\ a = \pm\sqrt{7}\\\\ a = 2 \end{cases} \end{align*} which clearly has no solutions. I think this answers your question about what is happening.

Hopefully this helps!

Answer 2

The following method has the advantage that it can be generalized for similar kinds of problems with more variables.

Consider the equation satisfied by $\;z=x\;$ and $\;z=y\;$ $$ z^2 - (x+y)z + (xy) = (z-x)(z-y) = 0. \tag1 $$ Then for all integer $n\ge 0$ it is true that $$ z^{n+2} - (x+y)z^{n+1} + (xy)z^n = 0. \tag2 $$ Substitute $x$ and $y$ for $z$ in this equation and add the two equations to get $$ (x^{n+2} + y^{n+2}) - (x+y)(x^{n+1} + y^{n+1}) + (x^n + y^n) = 0. \tag3 $$ Define $\;c := x^1+y^1 = x+y \;$ and note that $\;x^0 + y^0 = 2.\;$ Now assume that $$ x\,y = 1, \quad x^2 + y^2 = 5, \quad \text{ and } \quad x^3 + y^3 = 8. \tag4 $$ Use equation $(3)$ with $\;n=1,\;n=0\;$ and assumptions $(4)$ to get $$ 8 - c\,5 + c = 0 \qquad \text{ and } \qquad 5 - c\,c + 2 = 0. \tag5 $$ The equation for $\;n=1\;$ implies that $\;c=2\;$ but this is inconsistent with the equation for $\;n=0.\;$

Thus the assumptions in $(4)$ are inconsistent and there is no solution.

Answer 3

Alternative approach:

Let $z = (x + y)$.
The following equations over-determine the value of $z$.

• E-1: $~x^2 + y^2 = 5.$
• E-2: $xy = 1.$
• E-3: $x^3+y^3 = 8.$

Since $~z^2 = x^2 + y^2 + 2xy,~$
E-1,E-2 collectively imply that
$\displaystyle z^2 = 7 ~\implies~ z ~\in ~\left\{+\sqrt{7}, -\sqrt{7}\right\}.$

So, $\displaystyle ~~\left\{+\sqrt{7}, -\sqrt{7}\right\}~$ are the only candidate values for $z$.

Since $z^3 = x^3 + y^3 + 3xy(z)$,
E-2,E-3 collectively imply that $z^3 = 8 + 3z \implies z^3 - 3z - 8.$

Trying each candidate value in turn gives:

• $\displaystyle \left[+\sqrt{7}\right]^3 - 3\left[+\sqrt{7}\right] = 4\sqrt{7} \neq 8.$

• $\displaystyle \left[-\sqrt{7}\right]^3 - 3\left[-\sqrt{7}\right] = -4\sqrt{7} \neq 8.$

So, neither of the two candidate values collectively implied by E-1 and E_2, satisfy the constraint collectively implied by E-2 and E-3.

Answer 4

We can also take a geometric interpretation of the problem. The curves corresponding to the equations

$$ xy \ = \ 1 \ \ \ , \ \ \ x^2 \ + \ y^2 \ = \ 5 \ \ \ , \ \ \ x^3 \ + \ y^3 \ = \ 8 $$

all have symmetry about the line $ \ y \ = \ x \ $ (only the first two, however, has symmetry about the origin). This suggests an approach in which we look at intersections of these curves with lines through the origin $ \ y \ = \ mx \ \ . \ $ (We do not need to consider a "horizontal" or "vertical" line, since the first curve has the coordinates axes as asymptotes.) If there are value(s) of $ \ m \ $ which "work" for all three curves, then we have located their mutual intersections.

With this substitution, we have $$ m·x^2 \ = \ 1 \ \ \ , \ \ \ (1 + m^2)·x^2 \ = \ 5 \ \ \ , \ \ \ (1 + m^3)·x^3 \ = \ 8 \ \ . $$ For the first two curve equations, we obtain $$ (1 + m^2)·\frac{1}{m} \ \ = \ \ 5 \ \ \Rightarrow \ \ m^2 - 5m + 1 \ \ = \ \ 0 \ \ \Rightarrow \ \ m \ \ = \ \ \frac{5 \ \pm \ \sqrt{21}}{2} \ \ . $$ (As $ \ \frac{2}{5 \ +\ \sqrt{21}} \ = \ \frac{2·(5 \ - \ \sqrt{21})}{25 - 21} \ = \ \frac{5 \ - \ \sqrt{21}}{2} \ \ , \ $ we see that these lines are indeed symmetrical about $ \ y \ = \ x \ \ . \ $ These slopes also tell us that the circle and rectangular hyperbola intersect only in the first and third quadrants.)

However, in the third equation, this gives us $$ (1 + m^3)·\left( \frac{1}{m} \right)^{3/2} \ \ \stackrel{?}{=} \ \ 8 \ \ \Rightarrow \ \ (1 + m^3) \ \ \stackrel{?}{=} \ \ 8· m^{3/2} \ \ $$ $$ \Rightarrow \ \ \left( \ 1 \ + \ \left[ \ \frac{5 \ \pm \ \sqrt{21}}{2} \ \right]^3 \ \right) \ \ \stackrel{?}{=} \ \ 8· \left[ \ \frac{5 \ \pm \ \sqrt{21}}{2} \ \right]^{3/2} $$ $$ \Rightarrow \ \ \ 8 \ + \ \ ( \ 5^3 \ \pm \ 3·5^2·\sqrt{21} \ + \ 3·5·21 \ \pm \ 21^{3/2} \ ) \ \ \stackrel{?}{=} \ \ 16·\sqrt2· ( \ 5 \ \pm \ \sqrt{21} \ )^{3/2} $$ $$ \Rightarrow \ \ \ 8 \ + \ \ ( \ 440 \ \pm \ 96·\sqrt{21} \ \ ) \ \ \stackrel{?}{=} \ \ 16·\sqrt2· ( \ 5 \ \pm \ \sqrt{21} \ )^{3/2} $$ [then, squaring both sides] $$ \Rightarrow \ \ \ ( \ 448 \ \pm \ 96·\sqrt{21} \ )^2 \ \ \stackrel{?}{=} \ \ 512· ( \ 448 \ \pm \ 96·\sqrt{21} \ ) \ \ \Rightarrow \ \ \ 448 \ \pm \ 96·\sqrt{21} \ \ \neq \ \ 512 \ \ . $$

So we may conclude that the pairwise intersections of the three curves do not all fall on the same radial lines from the origin, and thus there is no solution to the given system of non-linear equations.

Interestingly, the second and third curves nearly meet tangentially. The line $ \ y \ = \ x \ $ intersects the circle at $ \ x^2 \ = \ \frac52 \ \Rightarrow \ \left(\sqrt{\frac52} \ , \ \sqrt{\frac52} \ \approx \ 1.5811 \right) \ \ $ and intersects the "cubic curve" at $ \ x^3 \ = \ 4 $ $ \Rightarrow \ \left(\sqrt[3]{4} \ , \ \sqrt[3]{4} \ \approx \ 1.5874 \right) \ \ , \ $ so the intersections of these two curves (in the first quadrant -- there is a second "diagonally symmetric" pair in the second and fourth quadrants ) lie on lines with reciprocal slopes close to $ \ 1 \ \ . \ $ (Those slopes are approximately $ \ m \ \approx \ 1.197 \ \ , \ \ m \ \approx \ 0.835 \ \ . \ ) $