User:GFauxPas/Sandbox/NumberTheory

Problem Set 2
$(2):$


 * $\checkmark (a)$ Prove $\phi(n) \ge \frac {\sqrt n} 2$ and deduce the limit $\lim_{n \to \infty} \phi(n)$

$(4):$


 * $\checkmark (a):$ Assuming that $p$ and $q$ are distinct primes, prove:


 * $p^{q-1} + q^{p-1} \equiv 1 \bmod pq$

Any interest in this? --GFauxPas (talk) 19:34, 23 December 2014 (UTC)

$\checkmark(6):$

Carefully outline how you would solve the equation:


 * $ax + by + cz = d$

in integers, where $\gcd(a,b,c)\vert d$.

$\checkmark (5):$

Let $a > 1$ be a positive integer and $m, n$ arbitrary positive integers. Prove:


 * $\gcd(a^m-1, a^n - 1) = a^{\gcd(m,n)} - 1$

I figured it out. Should I put it up as a page? --GFauxPas (talk) 16:11, 21 December 2014 (UTC)


 * Please do so. It's an interesting, elegant and non-obvious little result. It reminds me a little of the Law of Quadratic Reciprocity which I vaguely remember, to which it may have a link. --prime mover (talk) 16:16, 21 December 2014 (UTC)


 * My proof needs $v^t - 1 \divides v^{tw} - 1$ for $w \in \N$, do we have that or something stronger somewhere? --GFauxPas (talk) 17:24, 21 December 2014 (UTC)


 * The something stronger might be $x^n - y^n \divides x^{nt}-y^{nt}$. Is that a theorem? I seem to remember it being a theorem. --GFauxPas (talk) 19:51, 21 December 2014 (UTC)


 * We have Integer Less One divides Power Less One/Corollary which is what you are after. Then we have Difference of Two Powers which is easily adjusted by substituting $a = x^n$ and $b = y^n$. Job done. --prime mover (talk) 20:16, 21 December 2014 (UTC)

Theorem
Let $\phi$ be the Euler phi function.

Then $\phi$ satisfies the inequality:


 * $\forall n \in \N: \phi\left({n}\right) > \dfrac {\sqrt{n}}{2}$

Proof
For $n = 1$:


 * $\phi\left({1}\right) = 1 > \dfrac 1 2 = \dfrac {\sqrt{1}}{2}$.

the inequality holds by inspection.

Suppose $n = 2^j$ for some $j \ge 0$.

If $j = 0$, then $n = 2^0 = 1$, and the inequality holds, from above.

Otherwise, if $j > 0$,

Conclude that the inequality holds if $n = 2^j$ for any $j \ge 0$.

Now suppose $n \ne 2^j$ and write:


 * $n = 2^t \times \dfrac {n}{2^t}$

for $t \ge 0$, where $2^t$ is the highest power of $2$ that divides $n$.

Then $\dfrac n {2^t}$ has a prime factorization:


 * $\dfrac n {2^t} = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_r^{\alpha_r}$

where:


 * $p_1 < p_2 < \cdots < p_r$ are distinct primes
 * $\alpha_1, \alpha_2, \ldots, \alpha_r$ are (strictly) positive integers.

Note that $p_1 \ne 2$.

Then:

Here's where I stop and think if I can prove a strict inequality or only a weak one. --GFauxPas (talk) 20:15, 23 December 2014 (UTC)


 * Haven't tried it, but can you go straight to $\phi$ of a general prime power and prove that, then use the fact that $\phi$ is multiplicative? I think in this context induction may be a little unwieldy. --prime mover (talk) 21:29, 23 December 2014 (UTC)


 * I was gonna do: pick any prime factor, and induction is on $\alpha_i$, and the base case is Number Less One is Greater than Square Root. Then invoke Positive Real Number Inequalities can be Multiplied. The problem is that $2^t$ makes life harder, as then the lemma does not work, and that is what the $\dfrac 1 2$ comes to fix.


 * Did you have something more elegant in mind? Maybe cases, first odd $n$ and then even $n$? --GFauxPas (talk) 13:43, 24 December 2014 (UTC)


 * Not really, no. Head's all over the place at the moment. Stuff IRL in the way. --prime mover (talk) 17:23, 24 December 2014 (UTC)