User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

I have another proof for the conditions for logarithms to be irrational.

It involves finding the non-trivial solutions to this Diophantine equation:


 * $a^x = b^y$

But I need a name for this equation so I can name the page. Ideas? --GFauxPas (talk) 20:07, 25 November 2014 (UTC)


 * Equality of Integer Powers? &mdash; Lord_Farin (talk) 20:09, 25 November 2014 (UTC)


 * Sure, that works, thanks. Should I emphasize solutions in $a,b$ or solutions in $x,y$? The presentation would be the same, but I feel like one should make one of those two parameters and the other two variables. I'm not sure that distinction makes sense. --GFauxPas (talk) 20:16, 25 November 2014 (UTC)


 * If the presentation is the same, I would try to formulate it using four-tuples. But present it as it suits you, we can always adjust if room for improvement is discovered. &mdash; Lord_Farin (talk) 20:19, 25 November 2014 (UTC)

It's not even required for my major, but I think number theory is the most enjoyable math course I've taken. Every time I solve an exercise or a problem I feel like I'm finding the solution to a puzzle. Are there other maths like that, where each problem feels like a puzzle? --GFauxPas (talk) 20:54, 25 November 2014 (UTC)


 * I know no fields that have this to the same extent that number theory does, but I've always felt that model theory has a nice mix of theory-building and puzzling. I'm more of a theory-builder myself, so others might be able to give further pointers. &mdash; Lord_Farin (talk) 21:30, 25 November 2014 (UTC)

Theorem
Let $a, b, x, y \in \N_{\ge 1}$, where $a, b \ge 2$.

Let $\left({a,b,x,y}\right)$ be a solution to the Diophantine equation:


 * $a^x = b^y$

Then:

If $x$ divides $y$, then $b$ is an integer power of $a$, of the form:


 * $b = a^{y/x}$

If $y$ divides $x$, then $b$ is an integer power of $a$, of the form:


 * $a = b^{x/y}$

Otherwise, the equation has no solution.

Corollary
Let $\left({a, b, x, y}\right)$ be a solution to:


 * $a^x = b^y$

If $x$ and $y$ are relatively prime, then $x = 1$ or $y = 1$.

Proof
Let $\left({a, b, x, y}\right)$ be a solution to:


 * $a^x = b^y$

Suppose $y \ge x$.

By Prime Divides Power, any prime number that divides $a$ must divide $b$, and vice versa.

Using the Fundamental Theorem of Arithmetic, write $a$ and $b$ as their prime factorizations:

where $p_1, p_2, \cdots, p_n$ are distinct primes and the indices are all strictly positive integers.

Then,

Equating powers of both sides:

Because:


 * $\displaystyle y = \frac {A_1}{B_1}x = \frac {A_2}{B_2}x = \ldots = \frac {A_n}{B_n}x$

we see that $y$ and $x$ are in some constant proportion:


 * $(1): \quad y = k x$

for some $k \in \Q$.

Recall $y \ge x$.

Then $k \ge 1$.

By the Division Theorem:


 * $(2): \quad y = x t + r$

for some $t, r \in \Z$, where $0 \le r < x$.

By hypothesis, $x\ge 1$.

Recall $k \ge 1, 0 \le r < x$.

^^ I need to fix this step^^

If we let $k= \frac p q$ and write:


 * $qy = px$


 * $y = tx + r$

am I allowed to conclude $q = 1, t = p, r = 0$? Would it help to make $x \perp y$? --GFauxPas (talk) 14:13, 26 November 2014 (UTC)


 * Found a counterexample. --GFauxPas (talk) 18:30, 26 November 2014 (UTC)

That is, $y = tx$ for some integer $t$.

If $x \ge y$, then the same argument mutatis mutandis would produce $x = qy$ for some integer $q$.

As $\left({a, b, x, y}\right)$ was arbitrary, we conclude $a^x = b^y$ has a solution iff $x$ divides $y$ or $y$ divides $x$

If the former,

And similarly for $b = a^{x/y}$, if $y$ divides $x$.

Proof of Corollary
Suppose $\left({a, b, x, y}\right)$ is a solution to:


 * $a^x = b^y$

Let $x$ and $y$ be relatively prime.

By the main result, $x$ divides $y$ or $y$ divides $x$.

Um, this is obvious, but how do I state the reasoning behind it? --GFauxPas (talk) 12:49, 26 November 2014 (UTC)