Powered Puzzle - 05/05/2013

[showmyiq]

Can you find such natural numbers m and n, that fulfill the equation:

\begin{equation} (2n)^{2n} -1 = m^3 \end{equation}

[showmyiq]

The problem is solved by Sergey Sokolov via facebook

\begin{equation}
(2n)^{2n} -1 = m^3 \\
[(2n)^{n} -1][(2n)^{n} +1] = m^3\\
\text{Let K}= (2n)^{n} -1\\
K(K+2) = m^3
\end{equation}
Now, we can easily see that K is an odd number (it equals an even number + 1).
So, K and K+2 are both odd numbers, therefore m is an odd number too.

As we know from the fundamental theorem of arithmetic known as canonical representation of a positive integer – every positive integer n>1 can be represented in exactly one way as a product of prime powers! For reference follow this link:

http://en.wikipedia.org/wiki/Fundamenta ... arithmetic

\begin{equation}
n= p_1^{\alpha_1}p_2^{\alpha_2} \cdots p_k^{\alpha_k}
= \prod_{i=1}^{k}p_i^{\alpha_i}
\end{equation}
So, let’s substitute m with its canonical representation.
\begin{equation}
m= p_1^{\alpha_1}p_2^{\alpha_2} \cdots p_k^{\alpha_k}
= \prod_{i=1}^{k}p_i^{\alpha_i} \\
m^3= p_1^{3\alpha_1}p_2^{3\alpha_2} \cdots p_k^{3\alpha_k}
= \prod_{i=1}^{k}p_i^{3\alpha_i} \\
\end{equation}
Since m^3 is dividable by K, then K should be a part of this canonical representation.
Let’s define it like that:

\begin{equation}
K= p_1^{\beta_1}p_2^{\beta_2} \cdots p_k^{\beta_k}
= \prod_{i=1}^{k}p_i^{\beta_i} \\\text{, where } {\beta_i} < {\alpha_i}\text{ for every i, and }\beta_i\text{ can be equal to 0}\text{ for every i}
\end{equation}
Now, let’s see what is going on.

\begin{equation}
K+2= \frac {M^3} {K} = \frac {\prod_{i=1}^{k}p_i^{3\alpha_i}} {\prod_{i=1}^{k}p_i^{\beta_i}}
= \prod_{i=1}^{k}p_i^{3\alpha_i -\beta_i} \\
\prod_{i=1}^{k}p_i^{\beta_i} + 2 = \prod_{i=1}^{k}p_i^{3\alpha_i -\beta_i} \\
\end{equation}
Let’s stop here and analyze the results:

Sergey proposed this kind of approach. Since K and K+2 doesn't have common divider, then:

\begin{equation}
K=a^3 \\
K+2 = b^3
\end{equation}

,for some random integers a and b. Since the difference between two adjacent cubes is rising and rising, such numbers are not able to exist! For example:

\begin{equation}
2 = b^3 - a^3 = (b-a)(b^2 + ab + a^2)
\text{ ,but } b^2+ab+a^2 >2 \\
\text {Therefore contradiction!}
\end{equation}

Now I want to continue his idea with canonical representation:

\begin{equation}
\prod_{i=1}^{k}p_i^{\beta_i} + 2 = \prod_{i=1}^{k}p_i^{3\alpha_i -\beta_i} \\
\text{Let's assume that for some integer } {j < k} \text{ , }
\alpha_j>0 \text{ and } \beta_j>0\\
\text{Then we can state that the two products in the equation will be dividable by }p_j\\
\text{Therefore, 2 is dividable by }p_j \text{ as well!}\\
\text{So, if such integer j exists, it should be the power of }p_j=2\\
2^j + 2 = 2^{j+\zeta},\text{ for some integer } \zeta>0 \text{ (because } 3\alpha_i - \beta_i > \beta_i \text{)}\\
K = 2^j \text{ ,which is obviously an even number, but early we said K is an odd}\\
\text{So the only scenario here is }j=0 \text{ and } K=2^0=1\\
(2n)^n -1=0\\
(2n)^n=1 \text{ ,therefore } n=0\\
(2*0)^0 = 0^0
\end{equation}
Now, we just hit two very interesting topics.

1. Is 0 a natural number or not?
2. What is the result of 0^0? Undefined value or 1?

About question 1, check this link: http://en.wikipedia.org/wiki/Natural_number

There is no universal agreement about whether to include zero in the set of natural numbers: some define the natural numbers to be the positive integers {1, 2, 3, ...}, while for others the term designates the non-negative integers {0, 1, 2, 3, ...}.

About question 2, check this link: https://cs.uwaterloo.ca/~alopez-o/math- ... ode14.html

The discussion on 0^0 is very old, Euler argues for 0^0 = 1 since a^0 = 1 for a != 0. The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitschrift für Mathematik und Physik. Consensus has recently been built around setting the value of 0^0 = 1.

Quote: “Some textbooks leave the quantity 0^0 undefined, because the functions x^0 and 0^x have different limiting values when x decreases to 0. But this is a mistake. We must define x^0 = 1 for all x, if the binomial theorem is to be valid when x=0, y=0, and/or x=-y. The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant.”

If you believe that 0^0 equals 1 and that 0 is a natural number, then the solution is solved, m=n=0.
if you don’t believe in some of the two statements – no possible solutions!