Difference between revisions of "2016 JBMO Problems/Problem 3"
(→See also) |
(→Solution) |
||
| Line 8: | Line 8: | ||
== Solution == | == Solution == | ||
| + | It is given that <math>a,b,c \in \mathbb{Z} </math> | ||
| + | Let <math>(a-b) = -x</math> and <math>(b-c)=-y)</math> then <math>(c-a) = x+y</math> and <math> x,y \in \mathbb{Z}</math> | ||
| + | |||
| + | We can then distinguish between two cases: | ||
| + | |||
| + | Case 1: If <math>n=0</math> | ||
| + | |||
| + | |||
| + | Case 2: If <math>n>0</math> | ||
== See also == | == See also == | ||
Revision as of 02:05, 23 April 2018
Problem
Find all triplets of integers 
such that the number
![\[N = \frac{(a-b)(b-c)(c-a)}{2} + 2\]](http://latex.artofproblemsolving.com/4/9/6/496f7460105427895e3c88cb3d99ffadfb21bc41.png)
is a power of 
.
(A power of is an integer of form 
,where 
is a non-negative integer.)
Solution
It is given that 
Let 
and 
then 
and 
We can then distinguish between two cases:
Case 1: If 
Case 2: If 
See also
| 2016 JBMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
