two questions involving $x^3+y^3+z^3-3xyz$ factorization

Question

(1) Given that $x^3+y^3+z^3=3xyz+1$, determine the minimum of $x^2+y^2+z^2$. I know that Lagrange multiplier can solve this but I believe there is a way out using the factorisation: $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$$

(2) Determine all the positive integral triplet $(x,y,z)$ such that $x^3+y^3+z^3-3xyz=p$ for prime $p\geq 3$.

So according to the above factorisation, we need $x+y+z=p$ and $x^2+y^2+z^2=1+xy+yz+xz$. Can anyone give me an idea or hint to continue? Thanks

Answer 1

In the first part, the minimum is $1$.

Let $\sigma_1 = x + y + z$ and $\sigma_2 = xy + xz + yz$. Then we have $$\sigma_1^2 - 3\sigma_2 = x^2 + y^2 + z^2 - xy - xz - yz = \frac{1}{4}[(2x - y -z)^2 + 3(y - z)^2] \geq 0.$$

Next, using your suggested factorization, we have $$1 = x^3 + y^3 + z^3 - 3xyz = \sigma_1 (\sigma_1^2 - 3\sigma_2).$$ Taking into account the previous inequality, this shows that $\sigma_1 > 0$.

Subject to these conditions, we are to minimize $$x^2 + y^2 + z^2 = \sigma_1^2 - 2\sigma_1 = \frac{2}{3}(\sigma_1^2 - 3\sigma_1) + \frac{1}{3}\sigma_1^2 = \frac{2}{3}\frac{1}{\sigma_1} + \frac{1}{3}\sigma_1^2.$$

The function $f(t) = \frac{2}{3t} + \frac{t^2}{3}$ has a minimum of $1$ for $t > 0$, attained at $t = 1$. Thus $x^2 + y^2 + z^2 \geq 1$.

Conversely, this minimum is attained, for example, when $x = 1, y = 0, z = 0.$

Edit. Here is a solution to the second part. We have $$\sigma_1 = x + y + z = p$$ and $$1 = x^2 + y^2 + z^2 - xy - xz - yz = \sigma_1^2 - 3\sigma_2 = p^2 - 3\sigma_2,$$ so $\sigma_2 = \frac{p^2 - 1}{3}$. Since $\sigma_2$ is an integer, this shows that $p > 3$.

Now $x,y,z$ are the real roots of some cubic polynomial $$f(t) = t^3 -\sigma_1 t^2 + \sigma_2 t + q = t^3 - pt^2 + \frac{p^2 - 1}{3} t + q.$$

But $f(t)$ has its local maximum at $t = \frac{p-1}{3}$ and its local minimum at $t = \frac{p+1}{3}$. Unless $f(t)$ has a double root at $\frac{p-1}{3}$ or $\frac{p+1}{3}$, one of the integers $x, y, z$ must lie in the interval $((p-1)/3,(p+1)/3)$. But this is impossible since $p$ is not a multiple of $3$.

The remaining cases are as follows, and they do provide solutions:

(a) When $p$ is congruent to $1$ modulo 3, the numbers $(p-1)/3$, $(p-1)/3$, $(p+2)/3$.

(b) When $p$ is congruent to $2$ modulo 3, the numbers $(p-2)/3$, $(p+1)/3$, $(p+1)/3$.

Answer 2

since $$(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=1\Longrightarrow x+y+z>0$$ and $$x^2+y^2+z^2-xy-yz-xz=\dfrac{1}{x+y+z}$$ so $$x^2+y^2+z^2-\dfrac{1}{2}[(x+y+z)^2-(x^2+y^2+z^2)]=\dfrac{1}{x+y+z}$$ so $$\dfrac{3}{2}(x^2+y^2+z^2)=\dfrac{1}{2}(x+y+z)^2+\dfrac{1}{x+y+z}$$ use AM-GM inequality we have $$\dfrac{1}{2}(x+y+z)^2+\dfrac{1}{x+y+z}=\dfrac{1}{2}(x+y+z)^2+\dfrac{1}{2(x+y+z)}+\dfrac{1}{2(x+y+z)}\ge 3\cdot\dfrac{1}{2}$$ so $$x^2+y^2+z^2\ge 1$$