Finding 3 natural numbers by the rule of Sarrus

[A.Magnus]

Here is a problem I am working on: Using the Rule of Sarrus:
$$\begin{vmatrix}
x & y & z \\
z & x & y \\
y & z & x \\
\end{vmatrix}
=x^3+y^3+z^3-3xyz,$$
find $x, y, z$ such that $x^3+y^3+z^3-3xyz = 315.$

And here is what I have gotten so far: By row and column operations and by factoring out $(x+y+z)$ from th determinant, I was able to reduce the determinant to this:
$$(x+y+z)
\begin{vmatrix}
1 & 0 & 0 \\
z & x-z & y-z \\
y & z-y & x-y \\
\end{vmatrix}
= (x+y+z) \bigg ((x-z)(x-y) - (y-z)(z-y) \bigg) = 315 = 3 \cdot 3 \cdot 5 \cdot 7.$$

I would love to make an equation out of $(x+y+z)$ with a combination of the factors of $3^2, 5$, or $7$ so that I can come up with the $x, y, z$ by trial and error. But the possibilities are simply way too many to be efficient. How should I go forward instead? As always, thank you for your time and gracious help. ~MA

 

[Opalg]

Here is a problem I am working on: Using the Rule of Sarrus:
$$\begin{vmatrix}
x & y & z \\
z & x & y \\
y & z & x \\
\end{vmatrix}
=x^3+y^3+z^3-3xyz,$$
find $x, y, z$ such that $x^3+y^3+z^3-3xyz = 315.$

And here is what I have gotten so far: By row and column operations and by factoring out $(x+y+z)$ from th determinant, I was able to reduce the determinant to this:
$$(x+y+z)
\begin{vmatrix}
1 & 0 & 0 \\
z & x-z & y-z \\
y & z-y & x-y \\
\end{vmatrix}
= (x+y+z) \bigg ((x-z)(x-y) - (y-z)(z-y) \bigg) = 315 = 3 \cdot 3 \cdot 5 \cdot 7.$$

I would love to make an equation out of $(x+y+z)$ with a combination of the factors of $3^2, 5$, or $7$ so that I can come up with the $x, y, z$ by trial and error. But the possibilities are simply way too many to be efficient. How should I go forward instead? As always, thank you for your time and gracious help. ~MA

I think you have made a good start. Next, I would re-write that large bracket by multiplying it out and rearranging it like this: $$(x-z)(x-y) - (y-z)(z-y) = x^2 + y^2 + z^2 -yz-zx-xy = (x+y+z)^2 - 3(yz+zx+xy).$$ Now let $s = x+y+z$ and $\sigma = yz+zx+xy$, so that the equation becomes $$s(s^2-3\sigma) = 3 \cdot 3 \cdot 5 \cdot 7.$$
We have to assume that there is a solution with $x$, $y$ and $z$ integers, otherwise we don't stand a chance. Making that assumption, you can see that $s$ must be a multiple of $3$. (If it isn't, then neither is $s^2$, nor $s^2-3\sigma$, so the product cannot be a multiple of $3$.) On the other hand, $s$ cannot be a multiple of $9$. (If it is, then $s^2-3\sigma$ is a multiple of $3$, so the product is a multiple of $27$, which is not the case.)

This tells you that the first plausible candidate for $s$ is $3 \cdot 5 = 15$. In that case, $s^2 = 225$ and we must have $s^2 - 3\sigma = 3\cdot7 = 21$. That leads to $\sigma = 68.$ So we are looking for three integers with sum $15$ whose products two at a time add up to $68.$ There is still a bit of trial and error involved here, but broadly speaking if the integers are close together then $\sigma$ will be large, whereas if they are very different then $\sigma$ will be small. For example, if $(x,y,z) = (5,5,5)$ then $\sigma = 75$. But if $(x,y,z) = (13,1,1)$ then $\sigma = 27.$

Over to you ... .

 

[A.Magnus]

I think you have made a good start. Next, I would re-write that large bracket by multiplying it out and rearranging it like this: $$(x-z)(x-y) - (y-z)(z-y) = x^2 + y^2 + z^2 -yz-zx-xy = (x+y+z)^2 - 3(yz+zx+xy).$$ Now let $s = x+y+z$ and $\sigma = yz+zx+xy$, so that the equation becomes $$s(s^2-3\sigma) = 3 \cdot 3 \cdot 5 \cdot 7.$$
We have to assume that there is a solution with $x$, $y$ and $z$ integers, otherwise we don't stand a chance. Making that assumption, you can see that $s$ must be a multiple of $3$. (If it isn't, then neither is $s^2$, nor $s^2-3\sigma$, so the product cannot be a multiple of $3$.) On the other hand, $s$ cannot be a multiple of $9$. (If it is, then $s^2-3\sigma$ is a multiple of $3$, so the product is a multiple of $27$, which is not the case.)

This tells you that the first plausible candidate for $s$ is $3 \cdot 5 = 15$. In that case, $s^2 = 225$ and we must have $s^2 - 3\sigma = 3\cdot7 = 21$. That leads to $\sigma = 68.$ So we are looking for three integers with sum $15$ whose products two at a time add up to $68.$ There is still a bit of trial and error involved here, but broadly speaking if the integers are close together then $\sigma$ will be large, whereas if they are very different then $\sigma$ will be small. For example, if $(x,y,z) = (5,5,5)$ then $\sigma = 75$. But if $(x,y,z) = (13,1,1)$ then $\sigma = 27.$

Over to you ... .

Thank you very much for your gracious help and time. Looks like we have a good lead here. Allow me some times to look it over but in the meantime, please accept this posting as my acknowledgment of your help. Thank you again. ~MA