Math Foundations/C

204: Euclid + the failure of prime factorization for z
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A logical problem
Claim:  $$z=10\uparrow\uparrow10+23$$ cannot be factored into prime numbers.
 * Theorem (Unique Factorization) : Every natural number is uniquely factorizable into primes

We are not happy with such a logical inconsistency! What to do? First let us solidify the claim.

So far
A partial prime factorization $$z=3\times13\times139\times163\times18301\times400109\times27997373\times w$$
 * Question: What does $$w$$ look like?

$$\frac{1}{3}=0.\bar{3}$$

$$\frac{1}{3\times39}=\frac{1}{39}\simeq0.\overline{025641}$$

$$\frac{1}{3\times39\times139}=\frac{1}{5421}\simeq0.00018446781036709\ldots$$☹️

$$\frac{z}{3\times13\times139}=18446781036709\ldots$$ ☹️

Reciprocals
Set $$m=3\times13\times139=5421$$. Then

$$\phi(m)=2\times12\times138=2^4\times3^2\times23=3312$$

$$10^{2^4\times3^2\times23}\mod m \equiv 1$$

$$10^{2^4\times3^2}\mod m \equiv 2536$$

$$10^{2^4\times3^2\times23}\mod m \equiv 1$$

$$10^{2^4\times23}\mod m \equiv 3337$$

$$10^{2^2\times3\times23}\mod m \equiv 1$$

$$10^{2\times3\times23}\mod m \equiv 1$$

$$10^{3\times23}\mod m \equiv 3613$$

$$\therefore$$

$$ \begin{align*} ord(10) &= 2\times3\times23\\ &= 138 \end{align*} $$ in U(m).

Decimal
$$\frac{z}{3\times13\times39}$$=1844678103670909426305109758347

length of the repeating cycle = 138
 * Exercise 204.1 : How many such repeating cycles are there?
 * Exercise 204.2 : What does the last cycle look like?

All the primes
$$3\times13\times139\times163\times18301\times 400109\times27997373$$

$$ \begin{align*} phi(m) &= 2\times12\times138\times162\times18300\times 400108\times27997372=109989425893460184115200\\ &= 2^11 \cdot 3^7 \cdot 5^2 \cdot 13\cdot 23^2 \cdot 61 \cdot 4349 \cdot 538411 \end{align*} $$