# Understand the problem

Find all positive rational that satisfy the equation :

##### Source of the problem

Greece MO 2019, Problem 3

##### Topic

Number Theory

##### Difficulty Level

7/10

##### Suggested Book

Challenges and Thrills of Pre College Mathematics

# Start with hints

Do you really need a hint? Try it first!

Let us the given equation in terms of rational numbers and simplify the equation. Let and where . We have And since we must have and . 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 for some naturals , then there exists a natural such that and where and .

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, , there exists a such that , if we have and continuing like this , is unbounded ,contradiction. So , we must have wich means and from we have So, the solutions come out to be and , where is any positive integer.

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