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# 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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