Test Tomorrow.....Help!!!....Fermat's Little Theroemby trap101 Tags: test, theroem, tomorrowhelpfermat 

#1
Feb2712, 08:46 PM

P: 327

Prove that if "p" is an odd prime, then
a) 1^{p1}+2^{p1}+.....+(p1)^{p1} "is congruent" to 1(mod p) attempt: Well I know from Fermat's little theorem, that all the constants on the left hand side will reduce to 1's: 1 + 1 + 1...........1 (is congruent to) 1 (mod p). I know there is just a small step to conclude this, but I can't figure it out. b) 1^{p}+2^{p}+.....+(p1)^{p} (congruent to) 0 (mod p) (Hint: it may be useful to show: 1+2+....+(p1) = p(p1) / 2 Kinda lost with this one sad thing is I got help with it today...smh. 



#2
Feb2712, 08:59 PM

P: 29

Just count the ones =)




#3
Feb2712, 09:03 PM

P: 327





#4
Feb2712, 09:04 PM

P: 29

Test Tomorrow.....Help!!!....Fermat's Little Theroem 



#5
Feb2712, 09:16 PM

P: 327

Ok, well if I add up all the ones I would have some sort of sum 1(p1) which would leave me with (p1) congruent to ____ (mod p), but the only "something" i could fathom would be "1"




#6
Feb2712, 09:19 PM

P: 29

I don't think you understand the concept of congruence. Go back to the definition. What does it mean for a to be congruent to b mod p?




#7
Feb2712, 09:22 PM

P: 327

a = pk + b for some constant k, or pab




#8
Feb2712, 09:23 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877





#9
Feb2712, 09:27 PM

P: 29





#10
Feb2712, 09:29 PM

P: 327

well adding 1 to 1 will give 0. But I'm not seeing it with respect to congruence. 



#11
Feb2712, 09:30 PM

P: 327

university level course..... 



#12
Feb2712, 09:30 PM

P: 29





#13
Feb2712, 09:34 PM

P: 327

as far as congruence, my understanding is this: it is the relationship between a divisor and the number being divided, in which the remainder that is obtained can be compared with another number divided by the same divisor that obtains a similar remainder.




#14
Feb2712, 09:43 PM

P: 327

Well that's great....Just so I'm on firm ground with the question: 1 + 1 + 1...........1 (is congruent to) ______ (mod p) adding up all those one's will give me: (p1) (congruent to) ______ (mod p) 



#15
Feb2712, 09:49 PM

P: 29





#16
Feb2712, 09:54 PM

P: 327

I think I just realized it:
So going back to the fact that I will have a sum of (p1), p(p1) can be written as p(1) + (1).....in other words, my "remainder" is 1 and so (p1) "is congruent to" 1 (mod p) sometimes I get so possessed into thinking of the congruence as an equals sign and start carrying things over from side to side. Thanks by the way. 



#17
Feb2712, 09:55 PM

P: 29




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