2016 JBMO Problems/Problem 3
Contents
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.
Solution 2
First, we check when . WLOG, let since one of the factors of must be positive. Then, is forced. Let and so we want . Now, we have two cases, one of or is odd and the other is even, or they are both odd and . Case 1: WLOG is odd and is even. Then . Let where is odd. Note that cannot be even since the product cannot be divisible by 8. After dividing the modular equation by 4, we get . Remember that Fermat's LIttle Theorem states that if . There must be a prime that divides , and by CRT, that means that is one less than that . However, must be odd so the only that works is . Thus, this implies that has no divisor of any other prime other than 2, so it must be a power of 2. However, since is odd, it cannot have a factor of 2. We arrive at a contradiction.
Case 2:
See also
2016 JBMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |