Question

\((2n)! + 1 = (2n+1)^k\)

find all solutions over natural numbers

Answer 1

this is a follow up of earlier problem
http://openstudy.com/study#/updates/566af694e4b090814d904aea

Answer 2

\((2n+1)^k=\left(\begin{matrix}k\\ 0\end{matrix}\right)(2n)^01^k+\left(\begin{matrix}k\\ 1\end{matrix}\right)(2n)^11^{k-1}+....\left(\begin{matrix}k \\ n\end{matrix}\right)(2n)^k1^{k-1}\)
\((2n+1)^k=1+\left(\begin{matrix}k\\ 1\end{matrix}\right)(2n)^11^{k-1}+....\left(\begin{matrix}k \\ k\end{matrix}\right)(2n)^k1^{k-1}\)
\((2n+1)^k=1+k\left( ^{k-1}C_0+ ^{k-1}C_1....+^{k-1}C_{k-1} \right)\)
\((2n+1)^k=1+k2^{k-1}\)

now we just put in our equation
\((2n)!+1=(2n+1)^k\)
\((2n)!+1=1+k2^{k-1}\)
\((2n)!=k2^{k-1}\)

now we hav to find solutions to this equation :/

Answer 3

in the 1st line the coefficient of rightmost term -> \( \left(\begin{matrix}k \\ k\end{matrix}\right)\)

Answer 4

seems like i did something wrong
n=0, k=1 satisfies the equation -> \((2n)!=k2^{k-1}\)
but it doesn't satisfy \((2n)!+1=(2n+1)^k\)

Answer 5

Except for that mistake, it is a great start ! Here I have fixed the mistake :

\((2n)!+1 = (2n+1)^k \)

\(\implies (2n)! = (2n+1)^k-1^k\)

factor right hand side using the formula
http://math.stackexchange.com/questions/117660/proving-xn-yn-x-yxn-1-xn-2-y-x-yn-2-yn-1

\(\implies (2n)! = (2n)\sum\limits_{i=0}^{k-1}(2n+1)^i\)

\(\implies (2n-1)! = \sum\limits_{i=0}^{k-1}(2n+1)^i\)

Answer 6

let \(2n+1 = m\), above equation is same as :
\(\implies (m-2)! = \sum\limits_{i=0}^{k-1}m^i \)

Answer 7

Next we may use the earlier proven result : \((m-2)!\) is divisible by \(m-1\)

Answer 8

so this means we only have a limited number of solutions, then?
n = 1, k = 1 is a pretty trivial one

Answer 9

Yes, there are no solutions for \(n\gt 2\)

Answer 10

yeah n=2,k=2 is also a sol

Answer 11

Yes thats all... I cooked up the problem from this :
http://math.stackexchange.com/questions/1195342/when-does-p-1-1-pk-hold

Answer 12

that link has a simpler version of the same problem...

Answer 13

oh now we have to prove the rest :|

Answer 14

it's weird though... the fact that we can freely choose our \(k\) makes this unbelievable

Answer 15

let \(2n+1 = m\), above equation is same as :
\(\implies (m-2)! = \sum\limits_{i=0}^{k-1}m^i \)

If above equation holds, then it also holds under mod \(m-1\) :
\(\implies 0\equiv \sum\limits_{i=0}^{k-1}1^i\pmod{m-1} \)

\(\implies 0\equiv k \pmod{m-1} \)

\(\implies k=t(m-1) \)

Answer 16

so above shows that we cannot choose \(k\) freely...

Answer 17

yes of course, we can boil the restrictions down but I get a similar feeling for this like I do for the FLT.

Answer 18

of course the proof must be simpler for this, but for a newbie they're all the same :)

Answer 19

proof of FLT is very easy for you

Answer 20

i bet you can prove it on ur own if you try

Answer 21

loooooool

Answer 22

no I mean Fermat's Last Theorem

Answer 23

i thought you're talking about little fermat

Answer 24

sorry I forgot they used FLT for little

Answer 25

FLT is beyong the scope of openstudy

Answer 26

*beyond

Answer 27

openstudy? maybe even stackexchange

Answer 28

i think anyone who took abstract algebra can understand the wiles proof

Answer 29

were you able to understand it?

Answer 30

https://www.math.wisc.edu/~boston/869.pdf

Answer 31

i could never get past the first few pages

Answer 32

This book will describe the recent proof of Fermat’s Last Theorem
by Andrew Wiles, aided by Richard Taylor, for graduate
students and faculty with a reasonably broad background in algebra.
It is hard to give precise prerequisites but a first course
in graduate algebra, covering basic groups, rings, and fields together
with a passing acquaintance with number rings and varieties
should suffice.

Answer 33

it's weird to think how much Galois had done by the time he was my age :(

Answer 34

Galois and Abel contributed a lot to algebra and both died young, so unfortunate

Answer 35

While we're at them, Ramanujan too.

Answer 36

though his contribution can only be appreciated by his fellow mathematicians.

Answer 37

ok how do you prove this lol

Answer 38

It follows \(k = t(2n)\) for \(n\gt 2\), substituting this in the given equation : \[(2n)! + 1 = (2n+1)^{t(2n)}\]

Answer 39

Above equation is impossible because the left hand side is always less than the right hand side

Answer 40

\[(2n)!+1 \lt (2n+1)(2n+1)\cdots (2n\text{ times}) = (2n+1)^{2n}\]

Answer 41

that proves no solutions for \(n > 2\)

just need to check the cases \(n=1,2\) manually

Answer 42

nice :)

Answer 43

the entire proof hinges on the result from previous problem :

\((n-2)!\) is divisible by \(n-1\) whenever \(n\ne p+1\) and \(n\gt 5\)

Answer 44

only because of that result is it possible to simplify the problem by considering mod \(m-1\)

Answer 45

I do not see any other way to work this problem
If we are not familiar with that result, I think we will be stuck forever..

Answer 46

can we differentiate factorials? will that be easy and look good?
like \(\large d \frac{ (n)! }{ d(n) }\)

Answer 47

we may use the continuous version of factorial, the gamma function

Answer 48

\[\Gamma(t) = \int\limits_{0}^{\infty}x^{t-1}e^{-x}\,dx\]

Answer 49

This is an extension of the regular factorial
defined for all complex numbers except nonnegatice integers

Answer 50

\(\Gamma(t+1) = t\Gamma(t)\)