>Assume that p is a prime number and n is an arbitrary natural number.>Prove that (1+n)^p - n^p - 1 is divisible by p.Well, /sci/? 14 year old Japanese girls can do it. Surely you're smarter than an 8th grader.
>>16480742How is the Condescending Troll business these days? Does it pay well or do you do it for personal satisfaction? How did you get into the business? What coursework did you need to fail?Scientifically speaking, of course. Thanks.
>>16480742homework bait thread
>>16480742Why should I do this problem? It's on you to give me the incentive, resources and prerequisites to do it. If you're not able to do that, then it is a failing on your part, not mine. Personally, I prefer to specialize in deeper matters like the metaphysical nature of conscious experientia.
>>16480795Kek. Got 'em.
[math] (1+n)^p - n^p - 1 [/math]Suffices to say that divisibility by p implies that [math] (1+n)^p - n^p - 1 [/math] mod p = 0 mod p[math] (1+n)^p [/math] mod p = [math] 1+n^p [/math]Hence:[math] (1+n)^p - n^p - 1 = 1+n^p-n^p - 1 = 0 [/math] mod pWasn't that hard.
>>16480742Do you know the binomial expansion? Use that.
>>16480742Trivial consequence of Fermat's Last Theorem
>>16480975*Little Theorem FUCK
>>16480818lmao
Isn't this as easy as doing a binomial expansion? The n^p and 1 terms are subtracted out, leaving (p C k)n^k. But since p is prime, from the definition of (p C k), it can't be divided out from the coefficient. So every coefficient in the sum is divisible by p, so the whole sum is divisible by p. They teach binomial expansions in burger high school.
>>16482926>Isn't this as easy as doing a binomial expansion?Yes. It appears to be a high school level problem.
