 # Understand the problem

Find all positive rational $(x,y)$ that satisfy the equation : $yx^y=y+1$
##### Source of the problem
Greece MO 2019, Problem 3
Number Theory
7/10
##### Suggested Book
Challenges and Thrills of Pre College Mathematics

Do you really need a hint? Try it first!

Let us the given equation in terms of rational numbers and simplify the equation. Let $y=\frac{p}{q}$ and $x=\frac{r}{s}$ where $\gcd(p,q)=\gcd(r,s)=1$. We have $p\cdot \left(\frac{r}{s}\right)^{\frac{p}{q}}=p+q \Longleftrightarrow p^q \cdot r^p=(p+q)^q \cdot s^p$ And since $\gcd(p+q,p)=\gcd(r,s)=1$ we must have $p^q=s^p$ and $r^p=(p+q)^q$. Now, we have to find the solutions from these equations.
Observe both the equations are of the form $x^a = y^b$. The idea is that due to the prime factorization theorem, we can specify that $x^a = y^b$ leads to a special form of the x and y. Claim: If $x^a=y^b$ for some $x,y,a,b$ naturals , then there exists a natural $z$ such that $x=z^m$ and $y=z^n$ where $m=\frac{b}{\gcd(a,b)}$ and $n=\frac{a}{\gcd(a,b)}$.
Proof of Claim: Consider the prime factorization theorem of the x and y in $x^a = y^b$. Observe that it implies x and y must have same set of primes by the prime factorization theorem. Let x and y contain the primes $p_1, p_2, …, p_k$. Let $x = \prod_{i=1}^{k} x_i$ and $y = \prod_{i=1}^{k} y_i$. The above equation implies that $a.x_i = b.y_i$. This implies that $y_i = n.c$ and $x_i = m.c$, where c is a natural number. Hence $x = z^m, y = z^n$.

Using the intuitive claim, $p^q=s^p$ , there exists a $z$ such that $p=z^p$ , if $z \neq 1$ we have $p=z^p=z^{z^p}$ and continuing like this , $p$ is unbounded ,contradiction. So , we must have $z=1$ wich means $p=s=1$ and from $r^p=(p+q)^q$ we have $r=(q+1)^q$ So, the solutions come out to be $x=(q+1)^q$ and $y=\frac{1}{q}$, where $q$ is any positive integer.

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