Hello,
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.
Hm...
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 .
Then, .
But by above argument.
Thus, we are left with .
This is of course true.
The same argument when and .