🧵 Untitled Thread
Anonymous at Mon, 18 Nov 2024 14:51:09 UTC No. 16480742
>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.
Anonymous at Mon, 18 Nov 2024 15:42:37 UTC No. 16480795
>>16480742
How 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.
Anonymous at Mon, 18 Nov 2024 15:56:44 UTC No. 16480810
>>16480742
homework bait thread
Anonymous at Mon, 18 Nov 2024 16:03:17 UTC No. 16480818
>>16480742
Why 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.
🗑️ Anonymous at Mon, 18 Nov 2024 16:15:04 UTC No. 16480824
>>16480742
Apply binomial expansion on (n+1)^p
Except for k = 0 and k = p terms, all other terms in C(p, k) are divisible by p, which thankfully are substracted away by n^p and 1.
p does not have to be prime
Anonymous at Mon, 18 Nov 2024 16:15:07 UTC No. 16480825
>>16480795
Kek. Got 'em.
Anonymous at Mon, 18 Nov 2024 17:13:43 UTC No. 16480887
[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 p
Wasn't that hard.
Anonymous at Mon, 18 Nov 2024 18:03:42 UTC No. 16480969
>>16480742
Do you know the binomial expansion? Use that.
Anonymous at Mon, 18 Nov 2024 18:06:28 UTC No. 16480975
>>16480742
Trivial consequence of Fermat's Last Theorem
Anonymous at Mon, 18 Nov 2024 18:07:29 UTC No. 16480976
>>16480975
*Little Theorem FUCK
🗑️ Anonymous at Mon, 18 Nov 2024 23:34:19 UTC No. 16481431
>>16480742
(1+n)^p - n^p - 1 =
(1 + p*n + ... + p*n^(p-1) + n^p) - n^p - 1 =
p*n + ... + p*n^(p-1) =
Sum of p!/(k!(p-k)!)*n^(p-k) where k goes from 1 to p-1
Factoring p away from each coeficient:
p!/(k!(p-k)!) =
p * (p-1)!/(k!(p-k)!)
Since k goes from 1 to p-1:
Factorial of k! doesn't contain p.
Factorial of (p-k)! doesn't contain p.
Since p is prime it can only be divided by p,
therefore all coeficients are multiples of p,
therefore all sumands are multiples of p,
therefore the sumation is multiple of p,
therefore the sumation is divisible by p.
🗑️ Anonymous at Mon, 18 Nov 2024 23:35:21 UTC No. 16481433
>>16480742
(1+n)^p - n^p - 1 =
(1 + p*n + ... + p*n^(p-1) + n^p) - n^p - 1 =
p*n + ... + p*n^(p-1) =
Sum of p!/(k!(p-k)!)*n^(p-k) where k goes from 1 to p-1
Factoring p away from each coefficient:
p!/(k!(p-k)!) =
p * (p-1)!/(k!(p-k)!)
Since k goes from 1 to p-1:
Factorial of k! doesn't contain p.
Factorial of (p-k)! doesn't contain p.
Since p is prime it can only be divided by p,
therefore all coefficient are multiples of p,
therefore all summands are multiples of p,
therefore the summation is multiple of p,
therefore the summation is divisible by p.
🗑️ Anonymous at Mon, 18 Nov 2024 23:36:57 UTC No. 16481436
>>16480742
(1+n)^p - n^p - 1 =
(1 + p*n + ... + p*n^(p-1) + n^p) - n^p - 1 =
p*n + ... + p*n^(p-1) =
Sum of p!/(k!(p-k)!)*n^(p-k) where k goes from 1 to p-1
Factoring p away from each coefficient:
p!/(k!(p-k)!) =
p * (p-1)!/(k!(p-k)!)
Since k goes from 1 to p-1:
Factorial of k! doesn't contain p.
Factorial of (p-k)! doesn't contain p.
Since p is prime it can only be divided by p,
therefore all coefficients are multiples of p,
therefore all summands are multiples of p,
therefore the summation is multiple of p,
therefore the summation is divisible by p.
Anonymous at Tue, 19 Nov 2024 15:29:48 UTC No. 16482402
>>16480818
lmao
Anonymous at Tue, 19 Nov 2024 19:17:45 UTC No. 16482926
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.
Anonymous at Wed, 20 Nov 2024 04:34:04 UTC No. 16485881
>>16482926
>Isn't this as easy as doing a binomial expansion?
Yes. It appears to be a high school level problem.
Anonymous at Fri, 22 Nov 2024 13:21:41 UTC No. 16489101
>>16488058
Shouldn't it be 0<= k <=p ? Also, k isn't necessarily a natural number. You even contradict yourself in your expansion
Anonymous at Sat, 23 Nov 2024 05:26:58 UTC No. 16490208
>>16489101
She's modified her solution to hopefully be more clear.
Anonymous at Sat, 23 Nov 2024 09:39:43 UTC No. 16490282
Bruh> 16480887 is just using the theorem to prove itself
Anonymous at Sat, 23 Nov 2024 09:44:44 UTC No. 16490288
>>16480975
More simple than that. It's a trivial consequence of the binomial theorem.
Anonymous at Sat, 23 Nov 2024 09:47:05 UTC No. 16490289
[math]p \mid {p \choose k}, k=1,\ldots ,p-1[/math]
[math](1+n)^p = \sum_{k=0}^p {p \choose k} n^k = p(\ldots) + 1 + n^p[/math]