Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$.

Question

Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$.

What are all integral solutions of this equation?

I can only find $x=y=z=1$.

Answer 1

Using identity: $(x+y+z)^3 = x^3+y^3+z^3 + 3(x+y)(y+z)(z+x)$, we have:

$3^3 = 3 + 3(x+y)(y+z)(z+x) \Rightarrow (x+y)(y+z)(z+x) = 8$. From this you should be able to deduce the answer.

Answer 2

Using the Power Mean Inequality we have that:

$$ 1=\frac{1}{3}(x^3 + y^3 + z^3) \leq \left[\frac{1}{3}(x+y+z)\right]^3 =1$$

Equality only holds when $x=y=z=1$.

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Let's replace $z \mapsto -z$ and try to solve the system of equations for $x,y,z \geq 0$:

$$ x^3 + y^3 = 3 + z^3 \text{ and } x + y = z + 3 $$

We can still use the power mean inequality (in only two of the varaibles) and use the linear relation

$$ \frac{1}{3}z^3 < \frac{1}{3}(3 + z^3 ) = \frac{1}{3}(x^3 + y^3 ) \leq \left[\frac{1}{3}(x+y)\right]^3 = \left[\frac{1}{3}(z+3)\right]^3 < \frac{1}{24}(z+3)^3 $$

Maybe if I take cube roots the left and right sides are easier to compare. I also rounded $3^3 \approx 27$

$$ z < \frac{1}{2}(z + 3)$$

This means $0 \leq z < 3$. Can it be that $0^3 + 3 = 3, 1^3 + 3 = 4, 2^3 + 3 = 11$ are the sums of two cubes?

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If two of the 3 numbers are negative, the starting point should be:

$$ x^3 + y^3 = -3 + z^3 \text{ and } x + y = z - 3 $$

Then using the power mean inequality we can attempt to deduce a range on $z$

$$ \frac{1}{3}(-3 + z^3 ) < \frac{1}{24}(z-3)^3 $$

This one is harder to estimate. After fiddling around, we see that $z \leq 0$:

$$ z < \frac{1}{2}\sqrt[3]{(z-3)^3 + 24} < \frac{1}{2}z $$

So there are definitely no solutiuons of this kind.

Answer 3

From the identity $$(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x)$$ we obtain $8 = (x+y)(y+z)(z+x)$. It follows that $(3−x)(3−y)(3− z) = 8$. On the other hand,

$$(3−x)+(3−y)+(3−z)−3(x+y+z) = 6,$$

implying that either $|3−x|, |3−y|, |3−z|$ are all even, or exactly one of them is even. In the first case, we get $|3 −x| = |3−y| = |3−z| = 2$, yielding $x, y, z \in \{1, 5\}$. Because $x+y+z = 3$, the only possibility is $x = y = z = 1$.

In the second case, one of $|3−x|, |3−y|, |3−z|$ must be $8$, say $|3−x| = 8$, yielding $x \in \{−5, 11\}$ and $|3−y| = |3−z| = 1$, from which $y, z \in \{2, z\}$. Taking into account that $x+y +z = 3$, the only possibility is $x = −5$ and $y = z = 4$. In conclusion, the desired triples are

$$(1, 1, 1), (−5, 4, 4), (4,−5, 4), \text{ and }(4, 4,−5).$$