On Fermat's Last Theorem for odd prime exponents
We show that the Fermat equation $x^p + y^p = z^p$ has no solutions in coprime positive integers $x, y, z$ for any odd prime $p$.
arXiv subjects
Publications and source records attributed to Zenon B. Batang.
We show that the Fermat equation $x^p + y^p = z^p$ has no solutions in coprime positive integers $x, y, z$ for any odd prime $p$.
For coprime positive integers $a, b, c$, where $a+b=c$, $\gcd(a,b,c)=1$ and $1\leq a < b$, the famous $abc$ conjecture (Masser and Oesterlè, 1985) states that for $\varepsilon > 0$, only finitely many $abc$ triples satisfy $c > R(abc)^{1+\varepsilon}$, where $R(n)$ denotes the radical of $n$. We examine the patterns in squarefree factors of binary additive partitions of positive integers to elucidate the claim of the conjecture. With $abc$ hit referring to any $(a, b, c)$ triple satisfying $R(abc)<c$, we show an algorithm to generate hits forming infinite sequences within sets of equivalence classes of positive integers. Integer patterns in such sequences of hits are heuristically consistent with the claim of the conjecture.
The Collatz conjecture asserts that repeatedly iterating $f(x) = (3x + 1)/2^{a(x)}$, where $a(x)$ is the highest exponent for which $2^{a(x)}$ exactly divides $3x+1$, always lead to $1$ for any odd positive integer $x$. Here, we present an arborescence graph constructed from iterations of $g(x) = (2^{e(x)}x - 1)/3$, which is the inverse of $f(x)$ and where $x \not \equiv [0]_3$ and $e(x)$ is any positive integer satisfying $2^{e(x)}x - 1 \equiv [0]_3$, with $[0]_3$ denoting $0\pmod{3}$. The integer patterns inferred from the resulting arborescence provide new insights into proving the validity of the conjecture.