I was given the following determinant to evaluate.
\begin{pmatrix}x^3+1&x^2y&x^2z\\ xy^2&y^3+1&y^2z\\ xz^2&yz^2&z^3+1\\ \end{pmatrix}.
I simplified it to $$x^3+y^3+z^3+1$$
In the question the determinant was given to be equal to 11 and positive integral solution were asked.
One solution from hit and trial was $$(2,1,1)$$ and it's other 3 permutations. Other number would be way too big to satisfy so it is the only possible solution.
But my question is how do we find all integral solutions (both positive and negative) if there are any other?
We have:
$(x^3+y^3+z^3)=10$
For positive solution you must have:
$x, y, z<\lfloor 10^{\frac13} \rfloor=3$
So you only have numbers 1 and 2 which you found. If you want to find more solutions you may use this identity:
$(x+y+z)^3-(x^3+y^3+z^3)=3(x+y)(x+z)(y+z)$
with restriction x, y and z<3 . LHS must be divisible by 3, the only number can be 4 such that:
$4^3-(x^3+y^3+z^3=10)=54$
so we have:
$3(x+y)(x+z)(y+z)=54\Rightarrow (x+y)(x+z)(y+z)=18=2\times 3\times 3=((2+1)(2+1)(1+1)$
That is $(x, y, z)=(2, 1, 1)$
Let $x+y+z=m$ and $(x+y)(x+z)(y+z)=n$, so we have this equation:
$m^3-3n=10$
You can find more numbers for m and n for this equation which suffices the condition such as:
$(m, n)=(-11, -447), (-8, -174), (-5, -45),(-2, -6),(1, -3),(4, 18),(7, 111), (10, 333), (13, 729)\cdot\cdot\cdot $
As can be seen m makes an arithmetic progression and n is a multiple of a power of 3. Then you need to solve the following equations for more solutions:
$(x+y)(x+z)(y+z)=-6$ which gives $(x, y, z)=(4, -3, -3)$
$(x+y)(x+z)(y+z)=18$which as I said gives $(x, y, z)=(2, 1, 1)$
You can see that:
$111=3\times 37$, $333= 3^2\times 37$, $729=27^2=3^6$ $\cdot\cdot\cdot$
Update: Suppose we have:
$n =\alpha\times \beta\times \gamma$
then the equations of the following must be cosistent:
$\begin{cases}x+y=\alpha\\y+z=\beta\\z+x=\gamma\end{cases}$
Now we want to see whether with $m=-5$ and $n=-45$ the system is consistent or not:
$-45=(-1)(5)(9)=(-3)(3)(5)$
for example:
$\begin{cases}x+y=-1\\x+z=5\\y+x=9\end{cases}$
$[(x+y=-1)+z]=-5\Rightarrow z=-4, x+z=5\rightarrow x=9, x+y=-1\rightarrow y=-10\rightarrow y+z=-14 $
where the third equation says $y+z=9$, so the system is not consistent.The same if for factors $\alpha=-3$, $\beta=3$ and $\gamma =5$, so with $(m, n)=(-5, -45)$ there is no integral solutions. Among an infinite number of m there may be some consistent systems.
Particular case where $x+y+z=0$:
In this case we have:
$x^3+y^3+z^3=3xyz$
For example $(x, y, z)=(5, -2, -3)$, we have:
$5^3-3^3-2^3=3 (5) (-2)(-3)=90$
We try to adapt this with our equation, for this we divide both side by $9$:
$(\frac 5{9^{1/3}})^3+(\frac {-2}{9^{1/3}})^3+(\frac{-3} {9^{1/3}})^3=10 $
That is our equation may also have solutions in $\mathbb Q'$(irrational numbers).