## Notes on crypto for beginners (with one promise and an update on R-sequence)

With axiom of induction $\forall P [[P(0) \wedge (\forall_{k \in N})(P(k)\rightarrow P(k+1)))]\rightarrow \forall_{n \in N} P(n)]]$, we often connect infinite descent, ie. we find $P(0)$ and implication for mapping consecutive solutions so that we reach contradiction. In simpler crypto cases we often deal with Vieta jumping, in more advanced cases we deal with complex mapping. But now, if we are to e.g. prove that for each positive integer $K$ there exist infinitely many even positive integers which can be written in more than $K$ ways as the sum of two odd primes, then we find it troublesome to find the right mapping. Often times we could use mapping from lemmas like e.g. if $w$ is an integer, and if two coprime positive integers $x$ and $y$ are such that $xy$ is a $w$-th power of a positive integer, then both $x$ and $y$ are $w$-th powers of positive integers. For crypto purposes we could use the basics like the fact: $\pi\left(\sqrt{p_{1}p_{2}\cdots p_{n}}\right) > 2n.$ or that for $p$ prime there exists a prime number $q$ such that for every integer $n$,$n^p-n$  is not divisible by $q$, or standard things from Riemann, Gauss etc. But, the most important point I’d like to make is that the additonal intelligence used is not the descent itself, but only part of the attack on the code allowing to understand it better. Thus, I’d like to focus a bit more on the descent itself. For its use in crypto (and not only), on three things, ie. Axioms of set theory, Hilbert’s axioms and Tarski’s axioms in the context of the descent. This will be addressed in one of the later posts, and hope you will love it.

It will take me a couple of days. Please, be patient. Besides, me and my friend are in the midst of completing a new proof for $n=3$ of the LFT, with a new argumentation. We will try to handle the major question within a couple of months, assuming this argumentation works.

Enclosed work on $p2^n$ blocker conjecture and $R2$ function. art1

BTW, with regards to the LTF, $x^n+y^n=z^n$, we have ${{(z+y)(z-y)}\over{(z+y)}}(z^{n-1}+yz^{n-3}(z+y)+y^3z^{n-5}(z+y)+...+y^{2k-1}z^{n-(2k+1)}(z+y))$, for even $n$, thus ${{z^{n-1}}\over{z+y}}\rightarrow (z+y)|z^{n-1}\rightarrow (z+x)|z^{n-1}$, thus $\forall_{x,y,z \in N, odd} x^n+y^n=z^n$ exist $k,m \in N$ such that ${ (k-m)|(mx-ky)}$.