Prove that none of the numbers $2^{2^n}+ 1$, $n = 0, 1, 2, \dots$ is a perfect cube.
Problem
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Tags: geometry, 3D geometry, modular arithmetic, algebra, factorization, difference of cubes, special factorizations
fractals
19.04.2013 21:29
Reduce mod seven and find that it is never $-1,0,1$, but cubes are $-1,0,1$ mod seven, done.
countyguy
19.04.2013 22:01
djb86 wrote: Prove that none of the numbers $2^{2^n}+ 1$, $n = 0, 1, 2, \dots$ is a perfect cube. Please excuse me if this is bad math. I am not up to the level of this forum.
Let's make an equation: $2^{2^n}+1=x^3$.
$2^{2^n}=x^3-1$
Now, let's apply difference of cubes factorization.
$2^{2^n}=(x-1)(x^2+x+1)$
OK, $2^n$ will always be a positive integer if $n$ is a nonnegative integer, and $2^{2^n}$ is a positive integer if $2^n$ is, so that narrows what we have to worry about. $x$ must be an integer, which means that both of our factors are going to be integers. They must be powers of two, because otherwise we have unwanted prime factors, as the product must be a power of $2$ to match the left side.
Case 1: Both factors are even powers of $2$.
For $x-1$ to be even, $x$ must be odd. If $x$ is odd, $x^2$ must also be odd. $1$ is also odd. This means that the right factor will require us to add $3$ odd numbers, which will lead to an odd sum. This proves that we cannot have two even factors, let alone two even powers of $2$, let alone an integer solution.
Case 2: $x-1=1$
In this case, the only possible value of $x$ is $x=2$. Plugging it in, we get $(2-1)(2^2+2+1)=(1)(7)=7$. This $2^{2^n}$ is clearly not $7$ if you plug possible values of $n$ in, so we still do not have an integer solution.
Case 3: $x^2+x+1=1$
In this case, we can solve it as a quadratic.
$x^2+x+1=1$
$x^2+x=0$
$x(x+1)=0$
$x=0$ or $-1$.
If we plug either of these values into the first factor, we get negative values. The right side is positive, and a negative multiplied by a positive is negative, so $x^3-1$ is negative. As we showed earlier, $2^{2^n}$ is going to always be positive in a solution, and we only have negative values, so none will work. We have considered all possible cases and none worked, so we have proved that none of the numbers $2^{2^n}+ 1$, $n = 0, 1, 2, \dots$ is a perfect cube.
EDIT: Sniped, the solution before me is much quicker and doesn't take forever to write, so fractals wins.
mentalgenius
19.04.2013 22:06
countyguy, your solution is also pretty good; it's just you did a bunch of extra steps. As soon as you factored the RHS, you could have just been like: Since $x^2 + x + 1 = x(x+1) + 1$, it must always be odd. (Instead of doing this sub-factorization, you could also have just done casework with 2 cases for $x$: odd and even. Hence, to have the RHS a power of 2, $x$ must be 1 or zero. However, neither of these works, so there are $\boxed{\text{no solutions.}}$
mavropnevma
05.01.2014 00:30
In fact the Fermat numbers $F_n = 2^{2^n}+ 1$, $n = 0, 1, 2, \dots$ cannot be perfect powers $k$ for any integer $k\geq 2$. For $n\geq 1$, $F_n = (F_{n-1} -1)^2 +1$, so $F_n$ cannot be a perfect square. This rules out $k$ even. For $k$ odd, if $F_n = a^k$, then $2^{2^n} = F_n - 1 = a^k-1 = (a-1)(a^{k-1}+\cdots + a + 1)$. Since $a$ must be odd, it follows $a^{k-1}+\cdots + a + 1 > 1$ must be odd, absurd.
cobbler
05.01.2014 00:48
This is probably why Fermat thought he had primes.
ssilwa
05.01.2014 01:17
The result also follows by $\pmod 9$.
bobthesmartypants
05.01.2014 01:40
Extension: prove that $3^{2^n}+1$ for non-negative integers $n$ cannot be a perfect power other than $n=0$ where $3^{2^n}+1=4=2^2$.
The solution is simpler than you think.
ssilwa
05.01.2014 01:52
${3^{2^n}}+1 \equiv 2 \pmod 4$,contradiction.
bobthesmartypants
05.01.2014 02:27
Correct. In fact, taking $\pmod{4}$ takes care of all general $x^{2^n}+1$ for all odd $x$. But what about $3^{2^n}-1$? Can it ever be a perfect square? (perfect power is attainable; for example, $n=1$ yields $3^{2^n}-1=8=2^3$).
ssilwa
05.01.2014 02:49
Just $\pmod 3$. Also, please stop posting trivial problems.... here is a non trivial problem for you: http://www.artofproblemsolving.com/Forum/viewforum.php?f=151
bobthesmartypants
05.01.2014 03:09
None less trivial than the original problem. If you keep on complaining that my problems are trivial, take your time and solve another one of my trivial problems so we can keep the marathon going: bobthesmartypants wrote:
A $9\times 9\times 9$ space is filled with unit cubes that behave as follows:
Every second, if a $1\times 1\times 1$ space is not filled, and exactly three out of six adjacent cells are filled, then that space becomes filled.
Does there exist a placing of $80$ unit cubes somewhere in the space such that the entire space becomes filled in a finite amount of time?
Found in the Does there Exist Marathon
ssilwa
05.01.2014 08:25
@above: My question was not intended to be perceived as an attack to anything, specially to your mathematical abilities... to which you feel to respond to as you did... anyway, the purpose was to imply that if I ever need instructional help from you, I will seek you. In the mean time, you can post your problems on the thread as new topics if you wish..
MerlinLegion07
01.09.2022 15:05
We have this : $2^{2^n}+ 1=t^3$ $\implies 2^{2^n}=t^{3}-1$ $2^{2^n}=(t-1)((t-1)^{2}+3t)$ case 1: (t-1) - even $\implies$ (t-1)^{2}+3(t-1)+3 - odd $\implies$ no solution case 2: t-1=1 $\implies$ (t-1)^{2}+3(t-1)+3 - odd $\implies$ no solution