a) Find the minimum positive integer $k$ so that for every positive integers $(x, y) $, for which $x/y^2$ and $y/x^2$, then $xy/(x+y) ^k$ b) Find the minimum positive integer $l$ so that for every positive integers $(x, y, z) $, for which $x/y^2$, $y/z^2$ and $z/x^2$, then $xyz/(x+y+z)^l$
Problem
Source: France JBMO TST 2020 Test 2 P2
Tags: number theory
Tato_
10.03.2020 23:30
What means x/y²? x divided by y²?
Al3jandro0000
10.03.2020 23:31
I think it means $x\mid y $
Tato_
10.03.2020 23:48
But if x|y² and y|x², it means x=y, or not?
Math-Shinai
11.03.2020 00:01
no, 12 and 18 for exmaple
Math-Shinai
11.03.2020 00:14
The answer is $k = 3$. Take $x=12, y = 18$ and eliminate the small cases, and expand to prove that $k = 3$ is indeed the answer.
kaede
11.03.2020 08:45
$
x\mid y^{2} ,\ y\mid z^{2} ,z\mid x^{2} \ \ \cdots ( 1)\\
xyz\mid ( x+y+z)^{l} \ \ \cdots ( 2)
$
Note that $ ( x,y,z) =\left( 2^{4} ,2^{2} ,2\right)$ satisfies $ ( 1)$.
From $ ( 2)$, we have $ 2^{7} \mid \left( 2^{3} +2 +1\right) 2^{l}$, which implise $ l\geq 7$.
$ ( 1)$ hold if and only if $ \nu _{p}( x) \leq 2\nu _{p}( y)$, $ \nu _{p}( y) \leq 2\nu _{p}( z)$, and $ \nu _{p}( z) \leq 2\nu _{p}( x)$ for all primes $ p$.
$ ( 2)$ hold if and only if $ \nu _{p}( x) +\nu _{p}( y) +\nu _{p}( z) \leq l\cdot \nu _{p}( x+y+z)$ for all primes $ p$.
Let $ p$ be an arbitrary prime number.
WLOG, $ \min\{\nu _{p}( x) ,\nu _{p}( y) ,\nu _{p}( z)\} =\nu _{p}( z)$.
Then we have $ \nu _{p}( x+y+z) \geq \nu _{p}( z)$.
From $ \nu _{p}( x) \leq 2\nu _{p}( y)$ and $ \nu _{p}( y) \leq 2\nu _{p}( z)$, we have $ \nu _{p}( x) +\nu _{p}( y) \leq 6\nu _{p}( z)$.
So we have $ \nu _{p}( x) +\nu _{p}( y) +\nu _{p}( z) \leq 7\nu _{p}( z)$.
Therefore, the answer is $ \boxed{l=7}$.