$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$

Question

Prove that

$$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$$

Is true for $x$, $y$ and $z$ being positive integers, with $x$ and $y$ being co-prime and $3z<x<y$.

There are 2 cases in this question:

Case 1 - $x$ is odd and $y$ is even (or $x$ is even and $y$ is odd), in this case $z$ is odd.

Case 2 - Both $x$ and $y$ are odd, in this case $z$ is even.

I tried to show some contradiction, we know that for the statement on the RHS to be true, the prime factors must be present in multiples of 3's, however I struggled going any further than that.

Any help would be much appreciated.

Answer 1

It is often a proven method to look for coprime factors.

Suppose $q \neq 3$ is a prime dividing $3z-x$. Then $q$ divides $(3z)^3$, hence $z$. This implies that $q$ divides also $x$. Given the symmetry in $x$ and $y$, $q$ cannot divide $3z-y$ as $x$ and $y$ are coprime. Also $x+y$ is not divisible by $q$ as $y$ isn't.

If $3$ divides $3z-x$ then $3 \mid x$. Hence $3 \nmid y$ as $x$ and $y$ are coprime, and therefor $3 \nmid x+y$ and $3 \nmid 3z-y$.

We conclude that $x+y$, $3z-x$ and $3z-y$ are mutually coprime.

As $(3z)^3=27z^3$ is a third power, there exist positive coprime integers $u,v,w$ such that $z=uvw$ and either:

A:
1. $x+y=9u^3$
2. $3z-x=v^3$
3. $3z-y=w^3$

or

B:
1. $x+y=u^3$
2. $3z-x=9v^3$
3. $3z-y=w^3$

Note that we can skip the case $3z-y=9w^3$ as the Diophantine equation is symmetric in $x$ and $y$.

A: We have $6z-x-y=6uvw-9u^3=v^3+w^3$. Hence, $6uwv=9u^3+v^3+w^3$. As in general $\sqrt[3]{x_1x_2x_3} \le \frac{x_1+x_2+x_3}{3}$, we have $6uvw<3\sqrt[3]{9}.uvw \le 9u^3+v^3+w^3$, so there is no solution.

B: Likewise we have $6uvw<3\sqrt[3]{9}.uvw \le u^3+9v^3+w^3$, so there is also no solution.

We conclude that there exists no positive integers $x,y,z$ such that $x$ and $y$ are coprime and fulfill the Diophantine equation $(3z)^3=3(x+y)(3z-x)(3z-y)$. QED

A final remark: $z$ is even. The case that $x$ is even and $y$ is uneven implies that $(3z-x)(3z-y)$ is an even number, hence $(3z)^3$ is even.

Answer 2

Above equation shown below:

$\left(3z\right)^3\ne 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$

Assume:

$\left(3z\right)^3= 3\left(x+y\right)\left(3z-x\right)\left(3z-y\right)$ ---(1)

For the special case take,

$(x,y,z)=[3m,3n,3(m+n)]$

substituting value of, $(x,y,z)$ in equation (1) we get:

$m^2-mn+n^2=0$ ----(2)

It is well known that equation (2)

has no integer solution's. Hence a contradiction.

Hence our assumption is wrong & there is no

solution for "OP" question for the special case.

Answer 3

Substituting $t=3z$ and analyzing the function $f(x,y,t)=t^3-3(x+y)(t-x)(t-y)$ we get: $$f(x,y,t)=t^3-3(x+y)(t-x)(t-y)=t^3-3(x+y)(t^2-t(x+y)+xy)$$ $$f(x,y,t)=t^3-3t^2(x+y)+3t(x+y)^2-(x+y)^3+(x+y)^3-3(x+y)xy$$ $$f(x,y,t)=(t-x-y)^3+(x+y)(x^2+2xy+y^2-3xy)$$ $$f(x,y,t)=(t-x-y)^3+(x+y)(x^2-xy+y^2)=(t-x-y)^3+x^3+y^3$$ Now define $g(x,y,t)=f(-x,-y,t-x-y)=t^3-x^3-y^3$

Let's asume the function $g$ has an integer solution to $g=0$ and let $a,b,c\in\mathbb{N}$ be such a triplet. It follows that $a^3+b^3=c^3$. This contradicts Fermat's Last Theorem and there can not be such triplet solution!