Difference between revisions of "2016 JBMO Problems/Problem 3"

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(Solution)
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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)</math> then <math>(c-a) = x+y</math> and <math> x,y \in \mathbb{Z}</math>
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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>
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 +
<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>
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'''Case 1:''' If <math>n=0</math>
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<math>2016^n = 2016^0 = 1 \equiv 1 (\mod 2016)</math>
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<math>xy(x+y) = -2</math>
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<math>-2 = (-1)(-1)(+2) = (-1)(+1)(+2)=(+1)(+1)(-2)</math>
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<math>(x,y) \in \{(-1,-1), (-2,1), (-1,2)\}</math>
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<math>(a,b,c) = (k, k+1, k+2)</math> and all cyclic permutations, with <math>k \in \mathbb{Z} </math>
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 +
'''Case 2:''' If <math>n>0</math>
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<math>2016^n \equiv 0 (\mod 2016)</math>
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<math>xy(x+y) + 4 = 2 \cdot 2016^n</math>
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Using module arithmetic, it can be proved that there is no solution.
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'''Method 1:''' Using modulo 9
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 +
<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:
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 +
<math>xy(x+y) + 4 = 0 \equiv (\mod 3)</math>
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 +
<math>xy(x+y) = 2 \equiv (\mod 3)</math>
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<math>x = 1 \equiv (\mod 3)</math> and <math>y = 1 \equiv (\mod 3)</math>
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<math>x,y \in \{ 1(\mod 3), 4(\mod 3), 7(\mod 3) \}</math>
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<math>xy(x+y) \equiv 2 (\mod 9)</math> which is absurd since <math>xy(x+y) \equiv 5 (\mod 9)</math>.
  
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'''Method 2:''' Using modulo 7
  
Case 2: If <math>n>0</math>
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<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 $(a,b,c)$ such that the number

\[N = \frac{(a-b)(b-c)(c-a)}{2} + 2\]

is a power of $2016$.

(A power of $2016$ is an integer of form $2016^n$,where $n$ is a non-negative integer.)

Solution

It is given that $a,b,c \in \mathbb{Z}$

Let $(a-b) = -x$ and $(b-c)=-y$ then $(c-a) = x+y$ and $x,y \in \mathbb{Z}$

$(a-b)(b-c)(c-a) = xy(x+y)$

We can then distinguish between two cases:

Case 1: If $n=0$

$2016^n = 2016^0 = 1 \equiv 1 (\mod 2016)$

$xy(x+y) = -2$

$-2 = (-1)(-1)(+2) = (-1)(+1)(+2)=(+1)(+1)(-2)$

$(x,y) \in \{(-1,-1), (-2,1), (-1,2)\}$

$(a,b,c) = (k, k+1, k+2)$ and all cyclic permutations, with $k \in \mathbb{Z}$

Case 2: If $n>0$ $2016^n \equiv 0 (\mod 2016)$

$xy(x+y) + 4 = 2 \cdot 2016^n$

Using module arithmetic, it can be proved that there is no solution.

Method 1: Using modulo 9

$xy(x+y) + 4 = 0 \equiv (\mod 9)$

$xy(x+y) = 5 \equiv (\mod 9)$

In general, it is impossible to find integer values for $x,y$ to satisfy the last statement.

One way to show this is by checking all 27 cases modulo 9. An other one is the following:

$xy(x+y) + 4 = 0 \equiv (\mod 3)$

$xy(x+y) = 2 \equiv (\mod 3)$

$x = 1 \equiv (\mod 3)$ and $y = 1 \equiv (\mod 3)$

$x,y \in \{ 1(\mod 3), 4(\mod 3), 7(\mod 3) \}$

$xy(x+y) \equiv 2 (\mod 9)$ which is absurd since $xy(x+y) \equiv 5 (\mod 9)$.

Method 2: Using modulo 7

$2016 = 2^5 \cdot 3^2 \cdot 7$

See also

2016 JBMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4
All JBMO Problems and Solutions