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Given $x,y,z,n\in\mathbb{N}$, $x,y,z,n>0$, and $x\neq y$, my conjecture is that

$$ (x+y+z)^n-(x+z)^n-(y+z)^n=0 \Longrightarrow x^n+y^n+z^n-(x+y)^n=0, $$

where one can easily recognize Fermat's equation among the hypotheses.

My ultimate goal is to find the structural connection between these two equations, without supplementary knowledge (e.g. we know that $(x+y+z)^n-(x+z)^n-(y+z)^n=0\Rightarrow n\leq 2$). This means, for instance, to find a way to transform the first equation into the second one, e.g. by an appropriate change of variables, showing that the two equations admit the same solutions. Hence, the conjecture.

However, I tried to prove such statement by means of the triangular inequality and the binomial expansion, also by reductio ad absurdum. But I could not find a way through this, therefore

Can you suggest some idea or technique to prove or disprove such statement, without using the fact that we know that $(x+y+z)^n-(x+z)^n-(y+z)^n=0\Rightarrow n\leq 2$?

This post is clearly linked to this one Literature about the equation $x^m+y^m+z^m=(x+y)^m$.

NOTE: This post is a correction of a previous one, that I deleted because of errors pointed out by a kind user. Sorry for the inconvenience! Thank you!

  • I don't know if it helps, but it's quite easy to show that $\operatorname{rad}(n)\mid z$ – Mastrem Jun 30 '18 at 10:19
  • @Mastrem Thanks for your comment. Only, I am not familiar with the notation. What does it mean? Thanks for your courtesy! –  Jun 30 '18 at 10:30
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    $\operatorname{rad}(n)$ is the radical of $n$, the largest squarefree divisor of $n$. A squarefree number is a number not divisible by any squares. The radical of $n$ is also the product of all distinct primes dividing $n$. For instance, $\operatorname{rad}(4)=2$, $\operatorname{rad}(12)=6$ and $\operatorname{rad}(10)=10$ – Mastrem Jun 30 '18 at 11:30
  • @Mastrem Thanks! And what does the symbol $| z$ mean? –  Jun 30 '18 at 11:45
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    $a\mid b$ means that $a$ divides $b$ – Mastrem Jun 30 '18 at 12:39
  • Ah! Thanks! I didn't know! –  Jun 30 '18 at 12:40
  • What makes you think there is such "structural connection"? The implication may be true without any useful way to convert one equation to the other. For example $x^3+y^4=7 \Longrightarrow x^n+y^n+z^n-(x+y)^n=0$ is true but the equations are not connected in any meaningful way. The implication is simply true because the left side is always false. – Sil Jun 30 '18 at 17:06
  • I am not sure to have understood your observation. In your example, the hypothesis contains two different powers (3 and 4), whereas my problem requires the same power $n$. Please, can you rephrase your comment? Thanks! –  Jun 30 '18 at 17:18
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    Point is that having implication which is true does not necessarily mean there is some deep connection between the two sides of implication. For example I can say if Albert Einstein is alive, then I am $200$ years old. This implication is clearly true, though there is no connection between its two sides. For this reason implications where left side is false are usually quite uninteresting. – Sil Jun 30 '18 at 17:32
  • Note that $1^3+8^3+6^3=(1+8)^3$, so there do exist solutions for $n=3$ – Mastrem Jul 01 '18 at 12:30
  • @Mastrem Sorry for late reply! I wonder if you can explain me how to show that rad$(n)| z$. Thanks!!! –  Jul 23 '18 at 14:10

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Write the implication as $L=0 \Rightarrow R=0$. Then $L+R=\sum_{1 \leq i \leq n-1}\binom{n}{i}z^{n-i}[(x+y)^i-x^i-y^i]$, which is greater than zero for $n>2$ because the term $(x+y)^i-x^i-y^i$ is positive for $i >1$. Thus, $L=0 \Rightarrow R>0$ for $n>2$, without any additional knowledge.

Aravind
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