# A Problem of Divisibility by 17 from the 1894 Eotvos Competition

### Problem

### Solution 1

Let $u=2x+3y$ and $v=9x+5y.$ Then $4u+v=17(x+y).$ Since $4$ is not divisible by $17,$ $u$ and $v$ are either both divisible by $17$ or both not.

### Solution 2

Let $u=2x+3y$ and $v=9x+5y.$ $9u-2v=17y.$ Since neither $9$ nor $2$ are divisible by $17,$ $u$ and $v$ are either both divisible by $17$ or both not.

### Solution 3

Let $u=2x+3y$ and $v=9x+5y.$ $3v-5u=17x.$ Since neither $3$ nor $5$ are divisible by $17,$ $u$ and $v$ are either both divisible by $17$ or both not.

### Solution 4

One proof: $4v-u=34x+17y.$ Thus, $17|(4v-u).$ However, this is not possible if $17$ divides only one of $\{u,v\},$ for $17|u~\Rightarrow~17|4v~\Rightarrow~17|v$ and $17|v~\Rightarrow~17|4v~\Rightarrow~17|u.$

### Solution 5

$13u = v\text{ (mod }17).$ Now, given that $\gcd(13,17)=1,$ one is zero $\mod~17$ iff the other one is.

### Solution 6

Let $u=2x+3y$ and $v=9x+5y.$ Then $u^2+v^2=17(5x^2+6xy+2x^2).$ $u^2$ and $v^2$ are divisible by $17$ iff $u$ and $v$ are, and either both are or both are not.

### Acknowledgment

This is problem 1894/1 Hungarian Problem Book I by Elvira Rapaport (MAA, 1963).

Solution 1 is by Keith Dawid, Lorenzo Villa; Solutions 2 and 3 are by yours truly; Solution 4 is by Amit Itagi; Solution 5 is by plutoo on twitter; Solution 6 is by Lorenzo Villa.

|Contact| |Up| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny