Find all triples (x,m,n) of positive integers satisfying the equation x^m=2^{2n+1}+2^n+1.

Question

anonymous · Answer

$$\large{x^m=2^{2n+1}+2^n+1}~?$$

raden · Answer

yes

anonymous · Answer

Let $u=2^n$. Then you have a quadratic equation in terms of u: $2u^2+u+(1-x^m)=0$. Hope that helps.

raden · Answer

next, what should i do??

anonymous · Answer

I really have no idea. But $\large 1-x^m=(1-x)(x^{m-1}+x^{m-2}y+\ldots +xy^{m-2}+y^{m-1})$. Sooo... Since u is rational, perhaps apply the rational root theorem?

raden · Answer

hmmm, thanks for ur clue... but i think very" difficul for me 
i hope anyone give me more to solve this problem..
thanks @Herp_Derp

anonymous · Answer

dont worry i'll solve it soon ;)

anonymous · Answer

lol ; not that easy...solved it before for m=2 but for m>2 ........

experimentx · Answer

running a code for half and hour ... still no more than two solutions
n=0, n=4

anonymous · Answer

I believe that only answers are $ (x,m,n)=(2^{2k+1}+2^k+1,1,k) \ and \ (23,2,4) $ but my proof is still incomplete......

experimentx · Answer

for n>0 it would be that
for n=0
(2, 2, 0) this is also true. also note that m>1

anonymous · Answer

m,n and k are >0 
positive integers

experimentx · Answer

ah ... it means i didn't read the question well

anonymous · Answer

lol

anonymous · Answer

i have the complete proof now but it takes me an hour to type it....

anonymous · Answer

Well  note that  $x$  is an odd number.

$m=1 $  is an obvious answer so i want to work on  $m>1 $..............

firstly let  $m=2k+1 $  where $k \ge 1 $ we have

$$ 2^{n}(2^{n+1}+1)=x^{m}-1=(x-1)(x^{m-1}+...+1)$$$ (x^{m-1}+x^{m-2}+...+x+1)=(x^{2k}+x^{2k-1}...+x+1) $ is an odd number because $x$ is odd so 
$x-1$ divides  $2^n$  then  $x-1 \ge 2^n$  and  $x\ge 1+2^n$  hence :
$$ 2^{2n+1}+2^{n}+1=x^{m}\geq (2^{n}+1)^{3}=2^{3n}+3\times 2^{2n}+3\times 2^{n}+1 $$and this is a contradiction.....


now let   $m=2k$  then
$ 2^{n}(2^{n+1}+1)=x^{2k}-1=(x^k-1)(x^k+1)=(y-1)(y+1) $  and  $y$  is odd.
$y-1$  and   $y+1$  are even factors so   $ \gcd(y-1,y+1)=2 $
 
exactly one of them divisible by $4$. Hence $n ≥ 3$  and one of these factors is divisible by $2^{n−1}$ but not by $2^{n} $. So we can write

                                     $y = 2^{n−1}s +t$  , $s$ odd  , $t= ±1$.

Plugging this into the original equation we obtain

                         $2^n(2^{n+1}+1)=(2^{n−1}s +t)^2-1=2^{2n−2}s^2+2^n st $

or, equivalently
                                          $2^{n+1}+1=2^{n−2}s^2+ st $

Therefore
                                            $1-st=2^{n−2}(s^2-8) $

For  $t= 1$  this yields  $s^2 − 8 ≤ 0$,  so  $s = 1$, which fails to satisfy original equation.

For  $t=-1$  equation gives  $1+s=2^{n−2}(s^2-8) \ge 2(s^2-8)$  so  $2s^2-s-17 \le 0$  and  $ s \le 3$ we know that  $s$  is odd so  $s=3$. put this in the last equation with  $t=-1 $ gives  $n=4$. back to the original equation with  $n=4$  u have
                              $$x^m=529=23^2 $$
so  $x=23$  and  $m=2$

the only solutions are $ (x,m,n)=(2^{2l+1}+2^l+1, 1,l) \  and  \ (23,2,4) $

experimentx · Answer

man that's a long solution!!

anonymous · Answer

yes... i've learned it for m=2 from a book or something; i cant remember... it helped me a lot

experimentx · Answer

great!!

anonymous · Answer

a little typo

*****which fails to satisfy *last* equation.******

anonymous · Answer

@RadEn 
come here plz

raden · Answer

hoho, yea... wht about u with other solution in inbox (emai)

anonymous · Answer

but those are incorrect...check them... (11,1,1), (37, 1, 2) are incorrect

anonymous · Answer

im pretty sure that there is no other solution for this diophantine equation

raden · Answer

but, for (x,m,n)=(11,1,1)
=> 11^1= 2^(2(1)+1) + 2^1 + 1
11 = 8 + 2 + 1
i think Right side = Left side

anonymous · Answer

oh so thats answers for m=1 i provided a closed form for particular case of m=1

raden · Answer

my friend tell me that the solution has  infinitely, but i cant prove that

anonymous · Answer

i already proved it for u ...m=1 is an obvious solution of this equation...

anonymous · Answer

let  $m=1$  u will have  

$x=2^{2n+1}+2^n+1 $  now for any natural number like k the answer is 

$ (x,m,n)=(2^{2k+1}+2^k+1,1,k) $

anonymous · Answer

that equation becomes to an obvious form with letting m=1

raden · Answer

but, if we look the problem original :
x^m = 2^(2n+1) + 2^n + 1
satisfy for solution (11,1,1), (37,1,2) isnt it?
and the problem tell that all triples (x,m,n) are positive integers

anonymous · Answer

yes thats right and closed form of this solutions is what i write up there...

anonymous · Answer

just let k=1,2,3,...

raden · Answer

so, what conklusy the solutions it,, only one or more??

anonymous · Answer

there are infinite solution

anonymous · Answer

solutions are 

$(x,m,n)=(23,2,4)$

and 

$(x,m,n)=(2^{2k+1}+2^k+1,1,k)$   $k=1,2,3,...$

raden · Answer

ok, thank u very much.. i want  to save this page, n i will study it ...
case closed.. hehe

anonymous · Answer

OK...save the address of page...

raden · Answer

ok.. see u tomorrow :D