2016 JBMO Problems/Problem 3
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.)
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:
which is absurd since .
Method 2: Using modulo 7
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