Question

Arithmetic Differential Equation with two independent solutions:

Answer 1

Given primes p and q, we have x which is a 'function' of these two primes. \[x(p,q)\] Suppose now that x satisfies this differential equation \[\large x'=\frac{p^2+q^2}{pq}x\] what are two solutions to this?

The first obvious choice is to consider: \[\large y=p^{kp} \\ \large y' = kp^{kp} \\ \large y' =ky\] and just choose k to be (p^2+q^2)/pq

But what about another solution?

Answer 2

So this is sort of fun, also I don't know if anyone but wio knows what an arithmetic derivative is, so if you're curious check this out: http://en.wikipedia.org/wiki/Arithmetic_derivative

Ok, so what's the other solution I found?

\[\large x= p^qq^p \\ \large x' = p^{q-1}q^{p+1}+p^{q+1}q^{p-1} \\ \large x' = p^qq^p(\frac{q}{p}+\frac{p}{q}) \\ \large x' = x \frac{q^2+p^2}{qp}\]

Now how do we show this is indeed a separate solution?

Answer 3

Two distinct prime factorization.

Answer 4

One way of doing it is to show that if they're equal it gives a contradiction

\[\large y=x \\ \large p^{\frac{q^2+p^2}{qp}p}=q^pp^q \\ \large p^{\frac{q^2+p^2}{q}}=q^pp^q \\ \large p^{q^2+p^2}=q^{pq}p^{q^2} \\ \large p^{q^2}q^{p^2}=q^{pq}p^{q^2} \\ q^p=p^q\]

I guess my question now is, are there other solutions to this differential equation or can we show there aren't others?

Answer 5

\[
\frac{p^2+q^2}{pq }=\frac{p}{q} + \frac{q}{p}
\]These could add up to anything, s there is no reason to think there wouldn't be quite a few solutions.

Answer 6

If a function of primes can be a sum of primes, then it might as well be any number.

Answer 7

Also, using \(p\) as a solution but not \(q\) despite their symmetry is weird.

Answer 8

Oh what do you mean? Can you give an example of a function of primes that works then, I'm not sure I see how?

Answer 9

\[
x = ap^mq^n\implies x' = a'p^mq^n+amp^{m-1}q^n+anp^mq^{n-1} \\
x' =\left( \frac {a'pq}{a} +mq+np\right) \frac{x}{qp}
\]Since it is addition, how would you be certain that :\[
p^2+q^2 \neq \frac{a'pq+amq+anp}{a}
\]

Answer 10

Since \(a\) can be anything, then \(x\) can be many possible values.

Answer 11

Something worth noting: http://en.wikipedia.org/wiki/Primorial

So basically the Primorial is like the Factorial, except it multiplies only primes up to that number as its upper bound. So for example, 10 primorial is:

\[10 \# = 2*3*5*7\]

just a useful function I think to know about.