RMO Practice Problems - Number Theory #4

Number Theory Level 5

Find the sum of all positive integers $x, y ,z$ such that $x \leq y \leq z$ and $x^{y} + y^{z} = z^{x}$

Details and Assumptions:

If you get the solutions as ( $1,3,10$ ) and ( $2,3,16$ ). Then give your answer as the sum of all $x,y,z$ i.e. $1+3+10+2+3+16=35$ .

Try this set RMO Practice Problems

The answer is 4.

2 solutions

Daniel Ellesar
Sep 25, 2015

For the solution, you need y^z to be < z^x. Therefore if z = 2 you're looking for a square number < 2, if z = 3 a cube number < 3 and so on. With any value of z > 1 the only possible y value is 1 ( it's the only square number < 2 and the only cube number < 3 etc. ) Since y = 1, x = 1, therefore z = 2. Therefore ( 1, 1, 2 ) is the only solution.

It seems like you're assuming $x = 1$ here? I mean, if $z = 2$ and $x = 5$ then there are plenty of values of $y$ such that $y^z < z^x$ .

Patrick Corn - 5 years, 8 months ago

$x \leq y \leq z$ is given

Raushan Sharma - 5 years, 5 months ago

@Surya Prakash please post a solution!

Adarsh Kumar - 5 years, 8 months ago

(0,1,1) is also a solution

Sushant Patel - 4 years ago

Is 0 positive. Please read the question carefully.

Bhaskar Pandey - 3 years, 9 months ago

Mycobacterium Tuberculae
Sep 26, 2015

there is a proof for this theorem but here its indirect application is used

Whats the name of the theorem?

Aaron Jerry Ninan - 3 years, 9 months ago

RMO Practice Problems - Number Theory #4

Try this set RMO Practice Problems

The answer is 4.

2 solutions

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