20211127, 16:32  #23  
Apr 2020
359_{16} Posts 
Quote:
We get k*binomial(p,k) by choosing the set of k elements and then picking the special element, and p*binomial(p1,k1) by first picking the special element and then choosing the other k1 elements from the remaining p1 elements of our set. More famously, binomial(p1,k1)+binomial(p1,k)=binomial(p,k) is easiest to see combinatorially: we partition the binomial(p,k) choices of k elements from the numbers {1,...,p} into two, depending on whether they contain 1 or not. binomial(p1,k1) is the number of choices containing 1, and binomial(p1,k) is the number of choices not containing 1. Last fiddled with by charybdis on 20211127 at 16:35 

20211129, 16:05  #24  
Feb 2017
Nowhere
5988_{10} Posts 
Quote:
The result can be extended slightly. Since p divides binomial(p,k) for 0 < k < p when p is prime, we have the "freshman's dream" polynomial identity in F_{p}[x,y] Repeatedly raising to the p^{th} power, we see that in F_{p}[x,y] for any positive integer n, which shows that binomial(p^{n},k) is divisible by p for 0 < k < p^{n}. Then the above argument shows that for p prime and any positive integer n, Last fiddled with by Dr Sardonicus on 20211129 at 17:39 Reason: Omit repeated word 

20211203, 13:29  #25 
"Matthew Anderson"
Dec 2010
Oregon, USA
3·17·23 Posts 
more of this
I am excited about this.
*smile* 
20211210, 17:33  #26 
Aug 2020
79*6581e4;3*2539e3
1131_{8} Posts 
Thanks a lot for all the replies, it will take me while to get through all that.
By "reduced" I meant, there would be no smaller integer congruent to them mod p. I forgot about that smaller could also mean "negative". 
20220131, 18:35  #27 
Aug 2020
79*6581e4;3*2539e3
601 Posts 
And another one:
An exercise deals with an integer a having order 3 mod p. And it is to be shown that a+1 has order 6 mod p. As a hint they give: . If I accept that, I can prove that a+1 will have order 6. But how do I get from to ? My problem is that I don't know what I can transform the a^3 into. , but does that help? I also tried writing it as which also didn't get me anywhere. A small nudge in the right direction would be appreciated. 
20220131, 19:35  #28 
Apr 2020
857 Posts 
Hint: how does a^31 factorize?

20220201, 11:41  #29  
"Robert Gerbicz"
Oct 2005
Hungary
3047_{8} Posts 
Quote:
The idea is only that, if you need just that order of (a+1) divides 6, then: (a+1)^6==1 mod p, but that is true, because a^2+a+1  (a+1)^61 for every a integer, here a^2+a+1 is divisible by p, so (a+1)^61 is also divisible by p, hence (a+1)^6==1 mod p. You need more, the order is 6, so it can't be 1,2 or 3 [we know that every "good exponent" is divisible by the order, so the order should divide 6]. If the order is odd, then: (a+1)^3==1 mod p in this case (a+1)^31(a^2+a+1)*(a+2)=2 but the left side is divisible by p, because (a+1)^31 and (a^2+a+1) is divisible by p. This would mean that 2 is divisible by p, but p=2 can't be. Now you need to handle only the order=2 case, so you can't have (a+1)^2==1 mod p. I left it for you. 

20220201, 12:24  #30  
Feb 2017
Nowhere
2^{2}×3×499 Posts 
Quote:
Supplementary exercise: If p is prime, n divides p1, and a^n == 1 (mod p), then the order of a (mod p) is exactly n if and only if a^d =/= 1 for any divisor d of n with n/d > 1. Refinement: If and only if a^(n/q) =/= 1 (mod p) for every prime factor q of n 

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