Prove that phi(a^n - 1) is divisible by n

[xax]

Main Question or Discussion Point

As the title says, for any a>1 and n>0

 

[robert Ihnot]

Set $$a^n=1+M$$. Consequently $$a^n\equiv 1ModM$$ and is the smallest n for a>1, n>0. It also follows from the last equation that a belongs to the reduced residue group of order $$\phi{M}$$. Consequently by LaGrange's Theorem, n is a divisor of the order of the group.

 

[al-mahed]

Set $$a^n=1+M$$. Consequently $$a^n\equiv 1ModM$$ and is the smallest n for a>1, n>0. It also follows from the last equation that a belongs to the reduced residue group of order $$\phi{M}$$. Consequently by LaGrange's Theorem, n is a divisor of the order of the group.

You cannot say that for sure, or I am wrong?

 

[robert Ihnot]

al-mahed: You cannot say that for sure, or I am wrong?

I can say what I meant for sure. However, the way I wrote it, "And is the smallest n for a>1,
n>0, which you have put in red, this, I see, is confusing. It is not what I am asserting, but is repeating the conditions given by xax, "As the title says, for any a>1 and n>0."

So that since n >0, and a>1, which is a very minimal restriction, we know we have a positive value here which I deliberately set: $$a^n=1+M$$ This is the smallest value of n which can be used to achieve the required non-trivial equation: $$a^n\equiv 1ModM$$ since the left side is, in fact, just 1 more than M.

Thanks for the comment, it shows someone is reading this.

 

[xax]

robert, why does m = n and $$a^n=1+m$$? This is exactly what I need to prove?
I understood why $$a^n=1+M$$, but did not understand why M=n.
Thanks

 

[robert Ihnot]

M does not equal n. M=(a^n-1). Thus the order of the group $$\phi{M}=\phi(a^n-1)$$. n is the order of the element a.

 

[xax]

you've saved me again robert, thanks.

 

[robert Ihnot]

What you have to do sometimes is look at problems. Say, 2^4 =16. So we set the modulus at 15. Thus we have 2^4 ==1 Mod 15.

Now phi(15)= phi(5)*phi(3) = 4x2 = 8. That means that the reduced residue group of elements relatively prime to 15 are 8 in number. They are the elements that form the multiplicative group. (3 for example is not in this group since there is no solution to 3X==1 Mod 15. So 3 has no inverse.)

In fact, the eight elements are: 1, 2, 4, 7, 8, 11, 13, 14. Each one of these group elements has an inverse. For example: 1*1==1 Mod 15, 2*8=16==1 Mod 15, 4*4==1 Mod 15, 7*13 = 91==1 Mod 15, and so forth.

The theorem then states that the order of the element, which is 4, divides the order of the group which is 8.