# $x^n+y^n=z^n$ where $x,y,z$ are real numbers

Problem: If $$x^n+y^n=z^n$$ where $x,y$ and $z$ are real numbers $\gt 0$ and n is any integer $\neq 0$ then $$\frac {x}{n} + y >z$$ assuming $x \lt y$

Background: While studying Fermat's theorem I saw that the following are the characteristics of $x,y$ and $z$ $$x+y \gt z\gt y \gt x$$ But while studying further, I also saw the above to be true.

Request: I could prove this when $n \gt 0$ but couldn't do the same for $n \lt 0$. If anyone could prove or disprove both the cases or show me any site where this has been discussed. A detailed proof would be appreciated. I shall share my proof shortly for a review for the case $n \gt0$

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Here is a proof for $n \geq 1$, unfortunately for $n<0$ we get a weaker ienquality.

For all $n \notin(0,1)$ we have

$$(y+\frac{x}{n})^n = y^n(1+\frac{x}{ny})^n > y^n(1+\frac{x}{y})$$

with the last inequality following from Generalized Bernolli

For $n \geq 1$ you also have:

$$y^n+xy^{n-1}> y^n+x^n =z^n\,,$$

and by combining these two you get the desired result.

Maybe someone can fix it in general.

If $n=-1$, your equation is

$$\frac{1}{x}+\frac{1}{y}=\frac{1}{z} \Rightarrow xz+yz+=xy \Rightarrow z= \frac{xy}{x+y}$$

You need $y-x >z$, or equivalently

$$y-x > \frac{xy}{y+x} \Leftrightarrow y^2 > x^2+xy$$

Note that $y=3, x=2$ is a counterexample to this equation, thus $y=3, x=2, z=\frac{6}{5}, n=-1$ is a counterexample to your claim.

In general, if $n=-m<0$ you want to prove that

$$\frac{1}{x^m}+\frac{1}{y^m} =\frac{1}{z^m} \mbox{implies} \, y-\frac{x}{m} >z$$

but since

$$\frac{z^m}{y^m}+\frac{z^m}{x^m}=1\,,$$ if you set $\frac{z^m}{y^m} =\frac{1}{2}+\alpha$ and $\frac{z^m}{x^m} =\frac{1}{2}-\alpha$, the desired inequality becomes

$$\frac{1}{\sqrt[m]{\frac{1}{2}+\alpha}}-\frac{1}{m\sqrt[m]{\frac{1}{2}-\alpha}} >1$$ Since this is increasing in $\alpha$, is easy to see that this is true for all $0 < \alpha < \frac{1}{2}$ if and only if

$$\frac{m-1}{m\sqrt[m]{\frac{1}{2}}} =\lim_{\alpha \to 0^+} \frac{1}{\sqrt[m]{\frac{1}{2}+\alpha}}-\frac{1}{m\sqrt[m]{\frac{1}{2}-\alpha}} \geq 1$$

or equivalently

$$2 \geq (\frac{m}{m-1})^m$$.

But this is not possible, since the RHS is a sequence which is decreasing to $e$. This shows that for each $m$ there exists some $\alpha$ which makes the inequality fail....

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I couldn't get the last portion of your answer for n>0 . Is it possible to just fill up the last few lines? Thanks for your help – Barun Dasgupta Aug 11 '12 at 9:27
My answer for the case n>0 is as follows. Can you kindly verify and comment. Let $\frac{z}{x}=q$ and $\frac{y}{x}=p$. This shows $1=q^n - p^n \Leftrightarrow 1 = (q-p)(q^{n-1}+\cdots+p^{n-1}).$ Now since $q,p>1$,each term in the bracket $(q^{n-1}+\cdots+p^{n-1})$ is greater than $1$ and we have $n$ such terms in the bracket. So $(q^{n-1}+\cdots+p^{n-1})>n$ Which makes $$(q-p)<1/n$$$$\Leftrightarrow q<p+1/n$$ $$\Leftrightarrow z/x<y/x+1/n$$$$\Leftrightarrow z<y+x/n$$ – Barun Dasgupta Aug 11 '12 at 9:46
@BarunDasgupta $$(y+\frac{x}{n})^n > y^n(1+\frac{x}{y})=y^n+xy^{n-1}\geq y^n+x^n=z^n$$ – N. S. Aug 18 '12 at 6:45

$n = -1, x = y = 2, z = 1$ is a solution (I know you require $x < y$; just decrease x by a bit and increase y by a bit.)

Then, we can make $y-x$ arbitrarily close to 0, while z = 1, so this disproves your assertion for $n = -1.$

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But if you increase $y$ and decrease $x$ then it is no longer true that $y-x=0$. – Asaf Karagila Aug 9 '12 at 17:35
But we can make it arbitrarily small, and then it'll still be < 1. For example; y = 2.1, x = $\frac{21}{11}$ and $y-x = \frac{1}{11} + \frac{1}{10},$ which is still < 1. – only Aug 9 '12 at 17:58
True, but observe that this is still not $0$. – Asaf Karagila Aug 9 '12 at 18:13
Sorry; by "y-x = 0", I meant "in this case, y-x = 0, so if you require $x < y$ we can just change it a bit. I'll edit my answer to be more precise. – only Aug 9 '12 at 18:23

1) The estimates pertaining to fermat's equation have been extensively studied. One may see this paper

http://www.fq.math.ca/Scanned/29-1/bialek.pdf

2) Your problem is a theorem of Grunert, and it admits a simple proof.