MHB Finding 3 natural numbers by the rule of Sarrus

A.Magnus
Messages
138
Reaction score
0
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
 
Physics news on Phys.org
MaryAnn said:
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 ... .
 
Opalg said:
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
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...

Similar threads

Back
Top