Difference between revisions of "2016 JBMO Problems/Problem 3"
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It is given that <math>a,b,c \in \mathbb{Z} </math> | It is given that <math>a,b,c \in \mathbb{Z} </math> | ||
− | Let <math>(a-b) = -x</math> and <math>(b-c)=-y | + | 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> |
+ | |||
+ | <math>(a-b)(b-c)(c-a) = xy(x+y)</math> | ||
We can then distinguish between two cases: | We can then distinguish between two cases: | ||
− | Case 1: If <math>n=0</math> | + | '''Case 1:''' If <math>n=0</math> |
+ | |||
+ | <math>2016^n = 2016^0 = 1 \equiv 1 (\mod 2016)</math> | ||
+ | |||
+ | <math>xy(x+y) = -2</math> | ||
+ | |||
+ | <math>-2 = (-1)(-1)(+2) = (-1)(+1)(+2)=(+1)(+1)(-2)</math> | ||
+ | |||
+ | <math>(x,y) \in \{(-1,-1), (-2,1), (-1,2)\}</math> | ||
+ | |||
+ | <math>(a,b,c) = (k, k+1, k+2)</math> and all cyclic permutations, with <math>k \in \mathbb{Z} </math> | ||
+ | |||
+ | '''Case 2:''' If <math>n>0</math> | ||
+ | <math>2016^n \equiv 0 (\mod 2016)</math> | ||
+ | |||
+ | <math>xy(x+y) + 4 = 2 \cdot 2016^n</math> | ||
+ | |||
+ | Using module arithmetic, it can be proved that there is no solution. | ||
+ | |||
+ | '''Method 1:''' Using modulo 9 | ||
+ | |||
+ | <math>xy(x+y) + 4 = 0 \equiv (\mod 9)</math> | ||
+ | |||
+ | <math>xy(x+y) = 5 \equiv (\mod 9)</math> | ||
+ | |||
+ | In general, it is impossible to find integer values for <math>x,y</math> to satisfy the last statement. | ||
+ | |||
+ | One way to show this is by checking all 27 cases modulo 9. An other one is the following: | ||
+ | |||
+ | <math>xy(x+y) + 4 = 0 \equiv (\mod 3)</math> | ||
+ | |||
+ | <math>xy(x+y) = 2 \equiv (\mod 3)</math> | ||
+ | |||
+ | <math>x = 1 \equiv (\mod 3)</math> and <math>y = 1 \equiv (\mod 3)</math> | ||
+ | |||
+ | <math>x,y \in \{ 1(\mod 3), 4(\mod 3), 7(\mod 3) \}</math> | ||
+ | |||
+ | <math>xy(x+y) \equiv 2 (\mod 9)</math> which is absurd since <math>xy(x+y) \equiv 5 (\mod 9)</math>. | ||
+ | '''Method 2:''' Using modulo 7 | ||
− | + | <math>2016 = 2^5 \cdot 3^2 \cdot 7</math> | |
== See also == | == See also == |
Revision as of 01:33, 23 April 2018
Problem
Find all triplets of integers such that the number
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
and all cyclic permutations, with
Case 2: If
Using module arithmetic, it can be proved that there is no solution.
Method 1: Using modulo 9
In general, it is impossible to find integer values for to satisfy the last statement.
One way to show this is by checking all 27 cases modulo 9. An other one is the following:
and
which is absurd since .
Method 2: Using modulo 7
See also
2016 JBMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |