Prove that $x^x+y^y=z^z$ doesn't have integer solutions Prove that $x^x+y^y=z^z$ doesn't have integer solutions
To be honest, I don't see any way to start this problem, I tried for hours but it's not as easy as I thought.
Any hints?
As you can see in the comments, there is a solution for natural numbers, the problem is when the set is extended to integers solutions.
*This is a problem for a national olympiad in my country, this olympiad usually puts all the solutions but i didn't found a solution for this problem in the page, i don't know why.
Possible answer:
Property: 
Be $A,B,C$ $\in \Bbb R$ $\ne 0$ such that $A+B=C$. Then exist $U,V \in {A,B,C}$ such that:  $$2|U| \gt 2|V| \ge |U|$$
Proof: Write $C'=-C$, then $A+B+C'=0$. The numbers $A,B,C \ne 0$ by hypothesis, and they can't be the three of the same sign because then they can't add $0$, so there are two of the numbers $A, B, C'$ that have the same sign and the other has a different sign.
Amplifying the equality $A+B+C'= 0$ $\,$ by $-1$ if its necessary, we can assume that there are two of them (which we will call $X,Y$) that are positive and the third negative (we will call it $-Z$, with $Z\gt 0$). Then we have $Z=X+Y$ with $X,Y,Z \gt 0$ and that $X,Y,Z$ are equal to $|A|, |B| , |C|$ in some order.
WLOG we can assume that $X\le Y$ , then:
$$Z=X+Y\le 2Y \lt 2Y+2X = 2Z$$
Obtaining $2Z \gt 2Y \ge Z$ as we wanted.
Solution: Assume that  $$(1): \; x^x+y^y=z^z$$
has integer solutions $\ne 0$, let $(x,y,z) = (a,b,c)$. According to the property, there exists integers $\ne 0$ $(u,v) \in (a,b,c)$ such that $2|u^u| \gt 2|v^v| \ge |u^u|$. Dividing by 2 and taking $\log$ we see that there are non-zero integers $(u,v)$ such that: $$u\log |u| \gt v\log |v| \ge u\log |u| - \log 2$$
That can be written as: $$(2): \;0 \lt u\log |u| - v\log |v| \le \log 2$$
The inequality (2) says that $u\log|u| \gt 0$ or $v\log|v| \lt 0$, but if $v \log |v| \lt 0$ then: $$u \log |u| - v \log |v| = (-v)\log|-v|\;-\;(-u)\log |-u|$$ with $(-v)\log|-v| \gt 0$, so unless we make a substitution, we can assume that $u \log |u| \gt 0$. In this case, necessarily $u \ge 2$, so 
 the second inequality of (2) says that: $$v \log |v| \ge u \log |u| - \log 2 \ge 2 \log 2 - \log 2 \gt 0$$
Matching this with the first equality of (2) we get: $$ log 2 \ge u \log |u| - v \log |v| \gt (v+1)\log |v| - v\log|v| = \log|v| \ge \log2$$ a contradiction. Thus $x^x+y^y=z^z$ doesn't have any integers solution.
 A: A proof for positive integers.
Suppose $x$ and $y$ are positive integers with $x\le y$.
Then $x^x+y^y<2y^y$.  We can show that 
$$ x^x+y^y \le 2y^y < (y+1)^{y+1}. $$
Consider $$f(y) = (y+1)\log(y+1)-y \log y - \log 2.$$
Note $f(1)>0$ and 
$$f'(y) = \log(y+1)  -\log y > 0$$ for all $y>0$.
Hence, $f(y)>0$ for all $y>0$ and so
$$2y^y < (y+1)^{y+1}$$ for all $y>0$. Thus $x^x+y^y<(y+1)^{y+1}$ and so $x^x+y^y=z^z$ has no solutions in positive integers.
A: (Note that the other solutions don't deal with the case of negative integers.) 
$0^0$ is undefined, so work with these being non-zero. If you believe that $0^0 = 0$ or $1$, that can be easily dealt with, and is left as an exercise to the reader.
We consider cases based on how many of them are positive or negative.

If all of the integers are positive, WLOG $ x \leq y < z$ 
Show that for $n \geq 1$, $(n+1)^{n+1} > (n+1) n^n \geq 2 n^n$.   
Hence $x^x + y^y \leq 2 y^y  < (y+1)^{y+1} \leq z^z$, so there are no solutions. 

If one of the integers is negative, it cannot be $z$. WLOG let $ x < 0$.
If $x = -1$, clearly there are no solutions.
If $ x < -1$, the the LHS is not an integer, but the RHS is. Hence, no solutions. 

If exactly 2 of the integers are negative, then those values of $ |n^n| \leq 1$. The remaining value of $n^n$ from the positive integer is either 1 or $ \geq 4$.
Thus, the only possibility for solutions is $\pm x^x \pm y^y = \pm1$.
If $x = -1$ or $ y = -1$, clearly there are no solutions.
Otherwise, $x, y \leq -2$ and $|n^n| \leq \frac{1}{4}$, and thus this has no solutions. 

If all 3 of the integers are negative, if 2 of them are equal, then the third raised to itself is 0, hence no solutions. So all of them are distinct. WLOG, $ x < y < 0$.       
If any of them are $-1$, then we can easily rule out solutions (like the previous cases).   
Then, for $ n \leq -2$,  $\frac{|n^n|} { |(n-1)^{n-1}|} \geq 3 > 2$ in a manner similar to the all-positive case.
As such, we have no solutions to $\pm |x^x| \pm |y^y| = \pm |z^z|$.
Thus, there are no solutions to $x^x + y^y + z^z = 0 $.
A: This is a proof for positive integers. Obviously, $z> x,y$ and furthermore $x\neq y.$ If not, then $2 = \dfrac{z^z}{x^x}\geq\dfrac{(x+1)^{x+1}}{x^x} >x+1$, so $x=1.$ But then $z^z = 2$ has no integer solution. Therefore, assume $x<y<z$ or $z = x+m+n$, $y = x+m$ with $n,m\geq1.$ Then we have the equation: 
$$1 = \left(\dfrac{x}{x+m+n}\right)^{x}\dfrac{1}{(x+m+n)^{m+n}}+\left(\dfrac{x+m}{x+m+n}\right)^{x+m}\dfrac{1}{(x+m+n)^{n}}.$$
Now notice, that $f(a) = \dfrac{a^a}{(a+t)^a} = \dfrac{1}{(1+\frac{t}{a})^a}$ is decreasing in $a$ converging to $e^{-t}.$ In particular, $f(a)\leq f(1)=\dfrac{1}{t+1}.$ Wit this and the obvious bound $x,n,m\geq 1,$ it follows that: 
$$1\leq \dfrac{1}{1+m+n}\cdot \dfrac{1}{9}+\dfrac{4}{(2+n)^2}\cdot \dfrac{1}{3}\leq \dfrac{1}{3}\cdot\dfrac{1}{9}+\dfrac{4}{9}\cdot\dfrac{1}{3} = \dfrac{4}{27},$$
a contradiction. 
