For each $m\in \mathbb N$ we define $rad\ (m)=\prod p_i$, where $m=\prod p_i^{\alpha_i}$. abc Conjecture Suppose $\epsilon >0$ is an arbitrary number, then there exist $K$ depinding on $\epsilon$ that for each 3 numbers $a,b,c\in\mathbb Z$ that $gcd (a,b)=1$ and $a+b=c$ then: \[ max\{|a|,|b|,|c|\}\leq K(rad\ (abc))^{1+\epsilon} \] Now prove each of the following statements by using the $abc$ conjecture : a) Fermat's last theorem for $n>N$ where $N$ is some natural number. b) We call $n=\prod p_i^{\alpha_i}$ strong if and only $\alpha_i\geq 2$. c) Prove that there are finitely many $n$ such that $n,\ n+1,\ n+2$ are strong. d) Prove that there are finitely many rational numbers $\frac pq$ such that: \[ \Big| \sqrt[3]{2}-\frac pq \Big|<\frac{2^ {1384}}{q^3} \]
Problem
Source: Iran 2005
Tags: number theory, greatest common divisor, logarithms, geometry, number theory proposed
21.09.2005 23:02
ABC proves only an asymptotic version of Fermat conjecture, not the full Fermat's Last Theorem.
21.09.2005 23:16
More: e) prove an asymptotic version of Catalan's conjecture. f) prove that there are infinitely many Wieferich primes.
27.09.2005 17:52
I think that Omid took a mistake. part $a)$ must be : Prove Last fermat theorem for $n$ bigger than something
09.02.2013 09:55
part a) let $K=f(\epsilon )$.suppose $x^n+y^n=z^n$ for large enough $n$. by abc Conjecture we have $f(\epsilon )(rad(xyz))^{1+\epsilon}\geq max\left \{ x^n,y^n,z^n \right \}\geq (xyz)^{n/3}$. We define $\pi (m)$ be the number of prime divisor of $m$. easy to see there exist at least one $q$ such that $q$ is prime and divides one of the $x,y,z$ and $\sqrt[\pi (x)+\pi (y)+\pi (z)]{k}\geq p_{i}^{n/3-1-\epsilon }$ . so $k\geq q^{(n/3-1-\epsilon).(\pi (x)+\pi (y)+\pi (z)) }\geq q^{(n/3-1-\epsilon )}\geq 2^{(n/3-1-\epsilon )}$but we let $n$ be large enough so we have $\log _{2}f(\epsilon )\geq \infty $ contraction. the part d is not hard by Conjecture and Geometry of Numbers but my solution is very long and I had no time for write it. I cant solve the part c can anybody help me about this part?
12.08.2020 13:46
Does this work? part c) First AFTSOC so there exist infinitely many $n$ s.t $n,n+1,n+2$ are strong and take $n$ sufficiently large. Note that $(n^2)+(4(n+1))=(n+2)^2$ So by abc conjecture we have: $$(n+2)^2\leq f(\epsilon)(rad((n^2 )(4n+4)((n+2)^2)))^{1+\epsilon}=f(\epsilon)(rad((n)(4n+4)(n+2)))^{1+\epsilon}$$(*) (note that by the condition non of the numbers $n,n+1,n+2$ are 2(mod 4) so $n+1$ must be a multiple of 4 thus $n,n+1,n+2$ are pairwise coprime) $n+1$ is even so $rad(4n+4)=rad(n+1)$ We know that : $rad(n)\leq \sqrt{n},rad(n+1)\leq \sqrt{n+1},rad(n+2)\leq \sqrt{n+2}$ (since $n,n+1,n+2$ are strong then) so: $$rad((n)(4n+4)(n+2))\leq ((n)(n+1)(n+2))^{\frac{1}{2}}$$(**) Now by (*) , (**) we have the following inequality which has to be tru for all positive $\epsilon$: $$(n+2)^4\leq f(\epsilon)^2((n)(n+1)(n+2))^{1+\epsilon}$$But $((n)(n+1)(n+2))^{1+\epsilon}\leq (n+2)^{3+3\epsilon}$ so finally we have: $$n+2 \leq f(\epsilon)^2(n+2)^{3\epsilon}$$but the last inequality is obviously wrong for $\epsilon=\frac{1}{4}$ and sufficiently large $n$ so we're done.