Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to solve equation of the form $x^y + y^z + z^x = x^z + y^x + z^y$. I grouped like this: $(x^y-y^x) + (y^z -z^y) + (z^x - x^z) = 0$ one of the case is $x^y-y^x = 0; y^z - z^y = 0$ and $z^x - x^z = 0$. Can you discuss either trivial or non-trivial solutions of $(x, y , z)$ in $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{R}$ and $\mathbb{C}$? Thanks in advance.

share|improve this question
18  
What is interesting about this equation? –  Gerry Myerson Apr 15 '12 at 12:50
1  
In $\mathbb R$, you get lots of solutions by setting $x=0$, which leads you to solve $y^z=z^y$ or equivalently $f(y)=f(z)$ where $f(t) = (\log t)/t$. This equation has infinitely many solutions $(y,z)$ where $1\le y<e$ and $e<z$. –  Greg Martin Apr 15 '12 at 18:39
1  
Also for other values of $x$ as well, continuing from those solutions. Here is an animation showing the solutions in $0<y\le 6$, $0 < z \le 6$ for $0 \le x \le 2$. Regions where $F(x,y,z) = x^y + y^z + z^x - (x^z + y^x + z^y) > 0$ are blue, $< 0$ are red. math.ubc.ca/~israel/problems/Fanim.gif –  Robert Israel Apr 16 '12 at 5:17

2 Answers 2

Note that switching any two of $(x,y,z)$ interchanges the two sides of the equation. In particular, the set of solutions is invariant under permutations, and any $x,y,z$ that are not all distinct is a solution (which I will consider a "trivial" solution).

Nontrivial integer solutions include $(0,2,4)$ and $(1,2,3)$ and their permutations. I don't know if there are any other integer solutions, but a computer search didn't find any small ones.

share|improve this answer

At first thought:

I could only think of the trivial triples $(x,y,z)$

$$(0,n,n), (1,n,n), (n,n,0), (n,n,1), (n,0,n), (n,1,n)$$ for any $n \in \mathbb{Z}$

But then further trying $x=2,3,4 \dots$ and $y=z$ also seem to work.

Therefore $$\{ (x,y,z) | x,y,z\in \mathbb{Z}, y=z \} $$ which also includes those when $x=y=z$

By symmetry

$$\{ (x,y,z) | x,y,z\in \mathbb{Z}, x=z \} $$

$$\{ (x,y,z) | x,y,z\in \mathbb{Z}, x=y \} $$

also should work.

share|improve this answer
    
Raman can you generalize the each case in others (R and C) –  gandhi Apr 16 '12 at 6:05
    
@gandhi, yes this idea can be applied to $\mathbb{R}$ and $\mathbb{C}$ as well, but I remember that Diophantine equations allow varibles to be $\mathbb{Z}$ only. (But then I studied this three decades ago) –  Kirthi Raman Apr 16 '12 at 13:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.