If a is not coprime with 1729, this means that a doesn't belong to the invertible group of Z/1729Z. So there doesn't exist b such asAlso, I'm not sure how to handle the case where a and 1729 are not relatively prime.
I'm supposed to prove that for any integer using Euler's theorem. I'm trying to first prove it under the assumption that , which allows me to apply Euler's theorem. In this case, I showed that . What I want to do is get from here to so I can multiply both sides by and get the result. However, I'm not sure how to fill in that gap. Is there some theorem I can use? Or am I going about this in the wrong way?
Also, I'm not sure how to handle the case where a and 1729 are not relatively prime. Any help there would be appreciated as well.
Let a be a non-coprime number with 1729. Therefore, there is a common factor d, different from 1.
Is there b such as :
If so, we know that is a multiple of 1729 : ,
Since d divides a, it divides . And it divides 1729 too. So it has to divide 1 too, which is not possible unless d=1 -> contradiction.
Is it better this way ?
Okay, that's definitely within the scope of the material I know, but I'm not sure how it relates to what I'm trying to prove. I need to show that for any integer a. The last part of my original post is asking how to show this when a and 1729 aren't coprime. Perhaps I'm just not seeing the connection?
Also, any help with the first part of my post would be greatly appreciated as well.
Let then the maximal order of mod is .
Also . Combining these two results we get that .
This immediately tells us that .
Multiply both sides by : .
The next step is to argue that: , , .
We will prove the first case, the other two are similar.
Note and maximal order of mod is .
Using a similar argument as in the first paragraph we can get .
This tells us that: .
Say that this means or or .
Say that but not the other factors, i.e. and .
But by above argument.
Thus, we are left with .
This is of course true.
The same argument when and .