20211127, 16:32  #23  
Apr 2020
599 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
14FE_{16} 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
3F1_{16} Posts 
more of this
I am excited about this.
*smile* 
20211210, 17:33  #26 
Aug 2020
79*6581e4;3*2539e3
757_{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". 
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