Prove, that there is no solution

[lfdahl]

Prove, that the equation:

\[ 2^{ 2^n}+1 = m^3\]

has no solution in natural numbers.

 

[Greg]

Prove, that the equation:

\[ 2^{ 2^n}+1 = m^3\]

has no solution in natural numbers.

Spoiler

$$2^{2^n}+1=m^3\Rightarrow2^{2^n}=m^3-1=(m-1)(m^2+m+1)$$

The parity of $m^2+m+1$ is odd, hence no natural $m$ and $n$ satisfy the original equation.

 

[kaliprasad]

Not a solution but observation

Spoiler

weaker condition.

this should hold for $2^n$ and $2^{2^n}$ is not required.

 

[lfdahl]

Spoiler

$$2^{2^n}+1=m^3\Rightarrow2^{2^n}=m^3-1=(m-1)(m^2+m+1)$$

The parity of $m^2+m+1$ is odd, hence no natural $m$ and $n$ satisfy the original equation.

Hi, greg1313

Spoiler

You´re on the right track, but (cf. kaliprasads comment), you need to take a few steps more in order to reach the solution

 

[kaliprasad]

Prove, that the equation:

\[ 2^{ 2^n}+1 = m^3\]

has no solution in natural numbers.

my solution: I am taking $2^n$ and not $2^{2^n}$

Spoiler

taking m = p+1
$2^n = m^3-1 = (p+1)^3 -1 = p(p^2+3p+3)$
now $gcd(p, p^2+3p+3) = gcd(p,3) $
so
let us take 2 cases
p is co-prime to 3. then it has got 2 factors that are co-primes so product cannot be a power of 2
p is multiple of 3. then product is multiple of 3 so cannot be power of 2
hence proved

 

[Albert1]

my solution: I am taking $2^n$ and not $2^{2^n}$

Spoiler

taking m = p+1
$2^n = m^3-1 = (p+1)^3 -1 = p(p^2+3p+3)$
now $gcd(p, p^2+3p+3) = 3$
so
let us take 2 cases
p is co-prime to 3. then it has got 2 factors that are co-primes so product cannot be a power of 2
p is multiple of 3. then product is multiple of 3 so cannot be power of 2
hence proved

Spoiler

how is $gcd(p,p^2+3p+3)=3\,\, ?$
if $p=2,4--$

 

[topsquark]

I tried it out for all possible n and m and found no solutions. ( (Smoking) A long haul...It took me almost two hours to write it out.)

-Dan

 

[lfdahl]

I tried it out for all possible n and m and found no solutions. ( (Smoking) A long haul...It took me almost two hours to write it out.)

-Dan

Hi, topsquark!

Thankyou for your effort! I must admit: I am glad indeed, that you did not find any solution with natural numbers ... otherwise the suggested solution would be unreliable.

 

[kaliprasad]

Spoiler

how is $gcd(p,p^2+3p+3)=3\,\, ?$
if $p=2,4--$

thanks Albert

Spoiler

sorry It should be $gcd(p,p^2+3p+3)=gtcd(p,3)$

I have done the needful

 

[topsquark]

thanks Albert

Spoiler

sorry It should be $gcd(p,p^2+3p+3)=gtcd(p,3)$

I have done the needful

"gtcd"? Grand Traverse Conservation District?

Never heard of this one. Can I have a hint? Or just a typo?

-Dan

 

[kaliprasad]

"gtcd"? Grand Traverse Conservation District?

Never heard of this one. Can I have a hint? Or just a typo?

-Dan

typo

 

[lfdahl]

Suggested solution:

Spoiler

Assume, that $2^{ 2^n}+1 = m^3$. Then $m$ must be an odd number, and:
$2^{ 2^n} = m^3-1= (m-1)(m^2+m+1)$
Hence, $m-1=2^s$ and $m^2+m+1=2^t$ for some positive integers $s$ and $t$.
Now,$2^{2s}= (m-1)^2=m^2-2m+1$ and $2^t-2^{2s}=3m$.

But $2^t-2^{2s}$ is even and $3m$ is odd, a contradiction.

 

[Albert1]

Suggested solution:

Spoiler

Assume, that $2^{ 2^n}+1 = m^3$. Then $m$ must be an odd number, and:
$2^{ 2^n} = m^3-1= (m-1)(m^2+m+1)$
Hence, $m-1=2^s$ and $m^2+m+1=2^t$ for some positive integers $s$ and $t$.
Now,$2^{2s}= (m-1)^2=m^2-2m+1$ and $2^t-2^{2s}=3m$.

But $2^t-2^{2s}$ is even and $3m$ is odd, a contradiction.

my solution:

Spoiler

$2^{2^n}=m^3-1=(m-1)(m^2+m+1)$
let $A=2^{2^n}$
$B=(m-1)(m^2+m+1)$
then $A$ must be a multiple of 4
both $m$ and $(m^2+m+1)$ must be odd,set $m-1=2a$($a\in N$), $a$ even or odd
(this does not matter,in fact a is even)
if$A=B$ then
$2^{2^n}=(2a)(m^2+m+1)$
this is impossible ,since $a\times (m^2+m+1)$ can not be expressed as a power of $2$
and we have the proof

 

[lfdahl]

Thankyou, kaliprasad and Albert, for your nice solutions!(Yes)