2016 JBMO Problems/Problem 3
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
is the unique prime factorization.
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 possible cases modulo 9. An other one is the following:
and
which is absurd since .
Method 2: Using modulo 7
In general, it is impossible to find integer values for to satisfy the last statement.
One way to show this is by checking all 15 possible cases 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 |