Question

\[\large (p-1)! + 1 = p^k\]

Find all solutions such that \(p\) is prime and \(k\in \mathbb{Z^{+}}\)

Answer 1

so we can rearrange it like this:-
\(\Large (p-2)!=\frac{p^k-1}{p-1}\)

Answer 2

any hint ?

Answer 3

previous problem

Answer 4

btw thats a good start!

Answer 5

seems like my way to solve pre goes to no where xD

Answer 6

thats actually is the first step in the solution i have

Answer 7

hmm
(p-1)|(p-2)!

Answer 8

\[(2-1) \nmid (2-2)!\]

\[(5-1) \nmid (5-2)!\]

Answer 9

yeah for p>5 hmm

Answer 10

Okay for p>5 we have (p-1) | (p-2)!
looks good, keep going...

Answer 11

I can't escape the geometric series no matter what hahaha.

Answer 12

-.- eh lol

Answer 13

Yes! writing it as geometric series makes the next step clear

Answer 14

Based on what we came across the other day and using "p" \[\Large \sigma(p^{k-1})= (p-2)!\]

Answer 15

yes this form might give some hint about next step
\[\Large (p-2)!=\sum\limits_{i=0}^{k-1} p^i \]

Answer 16

hmm i would say no solution :|

Answer 17

thats right! there are no solutions for p>5

how did you arrive at that ?

Answer 18

just a conjecture xD

Answer 19

you're almost there... try using the result of previous problem

Answer 20

previous prblem
http://openstudy.com/users/rational#/updates/55096fc6e4b0b8a80199adcb

Answer 21

Because of the factorial we know that for the statement to be true this statement must also be true \[\Large 2^{\frac{p-1}{2}} | \ p^k-1\]

Answer 22

\(\Large (p-2)!=\sum\limits_{i=0}^{k-1} p^i =0 \mod (p-1) \)

Answer 23

BINGO!

Answer 24

kai Is that because there are (p-1)/2 even factors on left hand side factorial ?

Answer 25

\( \Large \sum\limits_{i=0}^{k-1} p^i= \sum\limits_{i=0}^{k-1} 1^i \mod (p-1)\\ \Large =1+1+...+1 \mod (p-1) = 0\mod (p-1) \) hmmmm
note we have k-1 1's

Answer 26

ugh we have k 1's :2

Answer 27

\[\sum\limits_{i=\color{Red}{0}}^{k-1}1 = k\]

Answer 28

k=0 mod p-1
k= n(p-1)

Answer 29

Yes exactly. =) Now since primes greater than 2 are odd, we can write them as this:
\[\Large p^k-1 = (2n+1)^k-1 = \sum_{i=1}^k \binom{k}{i}(2n)^i\] however for this to be divisible by \[\Large 2^{\frac{p-1}{2}}\] then we can't have the exponent on 2 be larger than 1 since the i=1 term of our sum only has a single 2 in it.... Although now I realize this is going to be an incomplete argument since n itself could be even. However for odd n this argument holds so we have to take this a litte further I guess for this way I'm cooking up. hmm... I'll check out ikram's stuff and come back to this in a minute.

Answer 30

typo @rational :2

Answer 31

An extra thing I should have realized earlier, a simple limit to put on k is this, \[\Large p^p > p! \implies p^p> p^{k+1}-p\] \[\Large p^{p-1}+1>p^k\] Hmmm... just some ideas.

Answer 32

hehehe xD

Answer 33

and ikram got k = n(p-1)

we're in a good position to conclude i guess XD

Answer 34

ok so

p^p>p!

p^(p-1)>(p-1)!

p^k>(p-1)!

Answer 35

\[(p-1)! + 1= p^{\color{blue}{k}}\]

\[(p-1)! + 1= p^{\color{blue}{n(p-1)}}\]

however we have
\[(p-1)! \lt p\cdot p\cdots (\text{p-1 times}) = p^{p-1}\]

so there are no solutions for \(\p \gt 5\) in as much as there are no solutions
for \(n \ge 1\)

Answer 36

nice :3
thanks for the question

Answer 37

Nice!