ganeshie8
  • ganeshie8
\((2n)! + 1 = (2n+1)^k\) find all solutions over natural numbers
Mathematics
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ganeshie8
  • ganeshie8
bubblegum.
  • bubblegum.
\((2n+1)^k=\left(\begin{matrix}k\\ 0\end{matrix}\right)(2n)^01^k+\left(\begin{matrix}k\\ 1\end{matrix}\right)(2n)^11^{k-1}+....\left(\begin{matrix}k \\ n\end{matrix}\right)(2n)^k1^{k-1}\) \((2n+1)^k=1+\left(\begin{matrix}k\\ 1\end{matrix}\right)(2n)^11^{k-1}+....\left(\begin{matrix}k \\ k\end{matrix}\right)(2n)^k1^{k-1}\) \((2n+1)^k=1+k\left( ^{k-1}C_0+ ^{k-1}C_1....+^{k-1}C_{k-1} \right)\) \((2n+1)^k=1+k2^{k-1}\) 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 :/
bubblegum.
  • bubblegum.
in the 1st line the coefficient of rightmost term -> \( \left(\begin{matrix}k \\ k\end{matrix}\right)\)

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bubblegum.
  • bubblegum.
seems like i did something wrong n=0, k=1 satisfies the equation -> \((2n)!=k2^{k-1}\) but it doesn't satisfy \((2n)!+1=(2n+1)^k\)
ganeshie8
  • ganeshie8
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 \(\implies (2n)! = (2n)\sum\limits_{i=0}^{k-1}(2n+1)^i\) \(\implies (2n-1)! = \sum\limits_{i=0}^{k-1}(2n+1)^i\)
ganeshie8
  • ganeshie8
let \(2n+1 = m\), above equation is same as : \(\implies (m-2)! = \sum\limits_{i=0}^{k-1}m^i \)
ganeshie8
  • ganeshie8
Next we may use the earlier proven result : \((m-2)!\) is divisible by \(m-1\)
ParthKohli
  • ParthKohli
so this means we only have a limited number of solutions, then? n = 1, k = 1 is a pretty trivial one
ganeshie8
  • ganeshie8
Yes, there are no solutions for \(n\gt 2\)
bubblegum.
  • bubblegum.
yeah n=2,k=2 is also a sol
ganeshie8
  • ganeshie8
Yes thats all... I cooked up the problem from this : http://math.stackexchange.com/questions/1195342/when-does-p-1-1-pk-hold
ganeshie8
  • ganeshie8
that link has a simpler version of the same problem...
ParthKohli
  • ParthKohli
oh now we have to prove the rest :|
ParthKohli
  • ParthKohli
it's weird though... the fact that we can freely choose our \(k\) makes this unbelievable
ganeshie8
  • ganeshie8
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 \(m-1\) : \(\implies 0\equiv \sum\limits_{i=0}^{k-1}1^i\pmod{m-1} \) \(\implies 0\equiv k \pmod{m-1} \) \(\implies k=t(m-1) \)
ganeshie8
  • ganeshie8
so above shows that we cannot choose \(k\) freely...
ParthKohli
  • ParthKohli
yes of course, we can boil the restrictions down but I get a similar feeling for this like I do for the FLT.
ParthKohli
  • ParthKohli
of course the proof must be simpler for this, but for a newbie they're all the same :)
ganeshie8
  • ganeshie8
proof of FLT is very easy for you
ganeshie8
  • ganeshie8
i bet you can prove it on ur own if you try
ParthKohli
  • ParthKohli
loooooool
ParthKohli
  • ParthKohli
no I mean Fermat's Last Theorem
ganeshie8
  • ganeshie8
i thought you're talking about little fermat
ParthKohli
  • ParthKohli
sorry I forgot they used FLT for little
ganeshie8
  • ganeshie8
FLT is beyong the scope of openstudy
ganeshie8
  • ganeshie8
*beyond
ParthKohli
  • ParthKohli
openstudy? maybe even stackexchange
ganeshie8
  • ganeshie8
i think anyone who took abstract algebra can understand the wiles proof
ParthKohli
  • ParthKohli
were you able to understand it?
ganeshie8
  • ganeshie8
https://www.math.wisc.edu/~boston/869.pdf
ganeshie8
  • ganeshie8
i could never get past the first few pages
ganeshie8
  • ganeshie8
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.
ParthKohli
  • ParthKohli
it's weird to think how much Galois had done by the time he was my age :(
ganeshie8
  • ganeshie8
Galois and Abel contributed a lot to algebra and both died young, so unfortunate
ParthKohli
  • ParthKohli
While we're at them, Ramanujan too.
ParthKohli
  • ParthKohli
though his contribution can only be appreciated by his fellow mathematicians.
ParthKohli
  • ParthKohli
ok how do you prove this lol
ganeshie8
  • ganeshie8
It follows \(k = t(2n)\) for \(n\gt 2\), substituting this in the given equation : \[(2n)! + 1 = (2n+1)^{t(2n)}\]
ganeshie8
  • ganeshie8
Above equation is impossible because the left hand side is always less than the right hand side
ganeshie8
  • ganeshie8
\[(2n)!+1 \lt (2n+1)(2n+1)\cdots (2n\text{ times}) = (2n+1)^{2n}\]
ganeshie8
  • ganeshie8
that proves no solutions for \(n > 2\) just need to check the cases \(n=1,2\) manually
bubblegum.
  • bubblegum.
nice :)
ganeshie8
  • ganeshie8
the entire proof hinges on the result from previous problem : \((n-2)!\) is divisible by \(n-1\) whenever \(n\ne p+1\) and \(n\gt 5\)
ganeshie8
  • ganeshie8
only because of that result is it possible to simplify the problem by considering mod \(m-1\)
ganeshie8
  • ganeshie8
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..
bubblegum.
  • bubblegum.
can we differentiate factorials? will that be easy and look good? like \(\large d \frac{ (n)! }{ d(n) }\)
ganeshie8
  • ganeshie8
we may use the continuous version of factorial, the gamma function
ganeshie8
  • ganeshie8
\[\Gamma(t) = \int\limits_{0}^{\infty}x^{t-1}e^{-x}\,dx\]
ganeshie8
  • ganeshie8
This is an extension of the regular factorial defined for all complex numbers except nonnegatice integers
ganeshie8
  • ganeshie8
\(\Gamma(t+1) = t\Gamma(t)\)

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