1
$\begingroup$

How many natural pairs $(x,y)\in\Bbb{N^2}$ exist such that $4^x-25^y=39$ holds?

The answer is $1$ the trivial $(x,y)=(3,1)$ is the only one; I've found Catalan's conjecture generalizations for $n=39$ but there should be an elementary solution.

This was given as one of $20$ questions from entrance exam for university of civil engineering which usually have pretty elementary solutions.

If anyone is curious this is how the exam looked like: Exam (it's not in English though)

$\endgroup$
5
$\begingroup$

Since $(2^x-5^y)(2^x+5^y)=39$, $(2^x-5^y,2^x+5^y)=(1,39)$ or $(3,13)$.

So, $(2^x,5^y)=(20,19)$ or $(8,5)$.

It is impossible that $(2^x,5^y)=(20,19)$.

$(x,y)=(3,1)$.

$\endgroup$
1
  • $\begingroup$ No idea how I missed something so trivial...Anyway thanks for helping $\endgroup$ – kingW3 Jun 18 '17 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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