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Prove $3(x+y)(x+z)(y+z)\neq a$ cubic when $x,y,z$ are different co-primal positive integers.

I believe you can look at the prime factors of $x,y$ and $z$. As for the equation to equal a cubic, the prime factors must all be present in multiples of $3$'s.

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  • $\begingroup$ Maybe $x+y,\,y+z,\,x+z$ are co-prime too? $\endgroup$ – Alexey Burdin May 29 at 22:34
  • $\begingroup$ Is that necessarily true for all x,y and z? $\endgroup$ – user578923 May 29 at 22:44
  • $\begingroup$ Furthermore, the fact that they are co-prime does not necessarily mean that it cannot equal a cubic. $\endgroup$ – user578923 May 29 at 22:48
  • $\begingroup$ Suppose it's equal to a perfect cube, then we obtain $x+y+z=\frac{(x+y)+(y+z)+(x+z)}{2}$ and hence $x,y,z$ will be half-integers, multiplying by $2$ each of them we get integer $2x,2y,2z$, for them LHS will be a perfect cube. So this constraints are almost equal. $\endgroup$ – Alexey Burdin May 29 at 22:56
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    $\begingroup$ @AlexeyBurdin There are more constraints, like requiring $ s > \max (a, b, c)$ for terms to remain positive. With your approach, I can't get "positive, integral, coprime" values of $x, y, z$. $\endgroup$ – Calvin Lin May 30 at 1:32
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What you're asking to prove is not always true. One specific counter-example, among infinitely many, is

$$x = 19 \tag{1}\label{eq1A}$$

$$y = 324 = 2^2 \times 3^4 \tag{2}\label{eq2A}$$

$$z = 4\text{,}589 = 13 \times 353 \tag{3}\label{eq3A}$$

These are all distinct, co-primal positive integers. You next have

$$x + y = 343 = 7^3 \tag{4}\label{eq4A}$$

$$x + z = 4\text{,}608 = 3^2 \times 2^9 \tag{5}\label{eq5A}$$

$$y + z = 4\text{,}913 = 17^3 \tag{6}\label{eq6A}$$

This gives

$$\begin{equation}\begin{aligned} 3(x + y)(x + z)(y + z) & = 3(343)(4\text{,}608)(4\text{,}913) \\ & = 3(7^3 \times (3^2 \times 2^9) \times 17^3) \\ & = (2^3)^3 \times 3^3 \times 7^3 \times 17^3 \\ & = (8 \times 3 \times 7 \times 17)^3 \\ & = (2\text{,}856)^3 \end{aligned}\end{equation}\tag{7}\label{eq7A}$$

Are there perhaps some other required conditions which were not mentioned?

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  • $\begingroup$ Yes sorry, there is - the sum of x+y>z with the integers increasing in value from x to y to z $\endgroup$ – user578923 May 30 at 10:44
  • $\begingroup$ @user578923 Thanks for the feedback. However, in my counter-example, you have $x + y = 343 \lt z = 4\text{,}589$, with integers increasing in value. Are saying this is a required condition, or a condition that cannot hold, i.e., you require $x + y \ge z$ when you have $x \lt y \lt z$? Please explicitly state the extra requirement in your question text. $\endgroup$ – John Omielan May 30 at 10:50

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