Note that if then is even - that is simple, if then we also have ... -.
That implies that is even -by your equation-, so if , has at least 3 divisors not coprime to it (n, 2 and n/2) (*), but then a contradiction!.
Hello everyone!
I'm working on this problem:
Q: Find all such that .
Here's my work:
Consider the following cases:
Case 1: is prime.
In is prime, then . However, if this is supposed to satisfy the above requirement, then we must have which is not possible for any .
Case 2: , with p and q prime.
We have . To satisfy the above equality, we must have which is not possible since 1 is not a prime.
Case 3: , with p prime.
We have . To satisfy the above equality, we must have . Therefore is a solution.
Now, I tried other cases that failed (like , , ). I believe that is the only solution, but my main question is when do I know where to stop with all these different cases?
Any help is appreciated!