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I tried it and I wonder if it is a valid proof? The question above is about proving irrationality by contradiction for ${z\over{x+y}}=\sqrt[3]{x^3+y^3\over(x+y)^3}$.

Assuming $z^3=x^3+y^3$ is true for some positive coprime integers x,y,z , consider the expression $z\over{x+y}$, which when x,y,z are coprime integers, $z\over{x+y}$ becomes a positive rational number. If $z^3=x^3+y^3$ is true, $z\over{x+y}$ can be written (I)${z\over{x+y}}=\sqrt[3]{x^3+y^3\over(x+y)^3}$ , where right side of the equation also is to be a positive rational number. If it is not rational but irrational, that should be possible to prove by contradiction.

Let (II) ${z\over{x+y}}=\sqrt[3]{x^3+y^3\over(x+y)^3}={r\over s}$ , where $\frac rs$ is a reduced rational number, so r and s are positive coprime integers, r < s.

Develop (II): $\sqrt[3]{{(x+y)(x^2-xy+y^2)}\over{(x+y)(x+y)^2}}$=$\frac rs$

Reduce left side: $\sqrt[3]{{x^2-xy+y^2}\over{(x+y)^2}}$=$\frac rs$

Divide left side: $\sqrt[3]{1-{3xy\over(x+y)^2}}$=$\frac rs$

Raise both sides to three : ${1-{3xy\over(x+y)^2}}$=$r^3\over s^3$

Multiply both sides by $s^3$ : (III) ${s^3-{s^3*3xy\over(x+y)^2}=r^3}$.

Right side of (III) = $r^3$ is an integer.

$\Rightarrow$ Left side is an integer, $\Rightarrow$ ${s^3*3xy\over(x+y)^2}$ is an integer.

x and y are coprime $\Rightarrow$ both x and y are coprime to x+y.

For left side to be an integer, $3s^3$has to be divisible by$(x+y)^2$ . There are two possibilities:

A. 3 is not a factor of $(x+y)^2$, or B. 3 is a factor of $(x+y)^2$.

For A, let $s^3=k*(x+y)^2$ , k is a positive integer. Insert in (III) $\Rightarrow$ $ {k*(x+y)^2}-{{k*(x+y)^2*3xy}\over{(x+y)^2}}=r^3$ ; ${k*(x+y)^2}-k*3xy=r^3$ ; $r^3=k*(x^2-xy+y^2)$ $\Rightarrow$ For k >1, $r^3$ and $s^3$ share a common factor k $\Rightarrow$ r and s share a common factor.

That is a contradiction.

For B , let $3s^3=k_1*(x+y)^2$ ; $k_1$ is a positive integer ; $s^3=k_1*{(x+y)^2\over3}$ ; Insert in (III) $\Rightarrow$ $r^3=k_1*{(x^2-xy+y^2)\over3}$. Since 3 is a factor in x+y, $(x+y)^2\over3$ is an integer. Also $x^2-xy+y^2=(x+y)^2-3xy$ , and both terms on the right side are divisible by 3. $\Rightarrow$ Left side = $x^2-xy+y^2$ is also divisible by 3, and $(x^2-xy+y^2)\over3$ is an integer. $\Rightarrow$For $k_1>1$, $r^3$ and $s^3$ share a common factor $k_1$ . $\Rightarrow$ r and s share a common factor.

That is a contradiction.

$\Rightarrow$For A or B, when k >1 or $ k_1 >1$, $\frac rs$ is irrational $\Rightarrow$${z\over{x+y}}=\sqrt[3]{x^3+y^3\over(x+y)^3}$ is irrational $\Rightarrow$ either the numerator z or one or both terms of the denominator x+y is irrational. All three of x, y and z cannot be positive integers simultaneously when $z^3=x^3+y^3$.

Fermat´s theorem is true for n=3.

It is not necessary to consider $k=1$ (or $ k_1 =1$). The intent is to prove (II) ${\sqrt[3]{x^3+y^3\over(x+y)^3}}={r\over s}$ , when $r\over s$ is a reduced rational number. The proof $\Rightarrow$ $r=\sqrt[3]{k*( x^2-xy+y^2)}$ and $s=\sqrt[3]{k*(x+y)^2}$ $\Rightarrow$ ${\frac rs}=\sqrt[3]{{k*(x^2-xy+y^2)}\over{k*(x+y)^2}}$, which is irrational for certainty, if $k>1$. (II)is identical with ${\frac rs}=\sqrt[3]{{(x+y)*(x^2-xy+y^2)}\over{(x+y)*(x+y)^2}}$ when $k=x+y$ and $x+y >1$. $\Rightarrow$ $r\over s$ is not a reduced rational number $\Rightarrow$ ${z\over{x+y}}={r\over s}=\sqrt[3]{x^3+y^3\over(x+y)^3}$ is irrational.

Is this really a proof? I wonder about the case of B, and also the dismissal of $k=1$ and $ k_1 =1$.

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  • $\begingroup$ Nothing immediately stands out on a first reading... though I do not see why $k=1$ and $k_1 = 1$ are trivial/unnecessary at all. $\endgroup$ – Artimis Fowl Jun 28 '17 at 18:22
  • $\begingroup$ I'd say you won't find a contradiction since $a z^3=x^3+y^3$ has some integer solutions for many $a$ $\endgroup$ – reuns Jun 28 '17 at 20:16
  • $\begingroup$ Thanks A.Fowl. I edited the proof and hope it is understandable.Better of course to prove it for k=1 too but I could not manage that. $\endgroup$ – Ylvali Jul 2 '17 at 14:19
  • $\begingroup$ Thanks to user 1952009 too. But I don´t understand where your equation comes in, and why it prevents contrdiction. I would appreciate very much if you could explain it, as clearly and elementary as possible. $\endgroup$ – Ylvali Jul 2 '17 at 14:27
  • $\begingroup$ If $x^n+y^n=z^n$ with $n$ natural number, it is known do you have always GCD$(x+y , \frac{x^n+y^n}{x+y})$=GCD$(n,x+y)$. Besides, in $II$ you have always $z=rk$ and $x+y=sk$ for an integer $k$. $\endgroup$ – Piquito Jul 2 '17 at 14:44

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