2016 JBMO Problems/Problem 3

Problem

Find all triplets of integers $(a,b,c)$ such that the number

is a power of $2016$ .

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

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

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

2016 JBMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
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