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arXiv 0904.1935math.NT
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满足abc猜想的方程c=a+b的个数

The number of equations c=a+b satisfying the abc-conjecture

Constantin M. Petridi

摘要:

我们证明了,对于一个正整数$c$和任何给定的$\varepsilon$,$0<\varepsilon<1$,方程$c=a+b$,$a<b$,正互质整数$a$和$b$,满足不等式$$c < R(c)^{\frac{\varepsilon}{1+\varepsilon}}R(a)^{\frac{1}{1+\varepsilon}}R(b)^{\frac{1}{1+\varepsilon}},$$,其中R(n)是$n$的根的数目$N(c)$是$c\to\infty$ $$N(c)=(1-\varepsilon)\frac{ϕ(c)}{2}+O\Bigl(\frac{ϕ(c)}{2}\Bigr).$$一个类似的abc猜想不等式$c<R(abc)^{1+\varepsilon}$(没有常数因子)也将被证明。

英文摘要:

We prove that for a positive integer $c$ and any given $\varepsilon$, $0<\varepsilon<1$, the number $N(c)$ of equations $c=a+b$, $a<b$, with positive coprime integers $a$ and $b$, which satisfy the inequality $$c < R(c)^{\frac{\varepsilon}{1+\varepsilon}}R(a)^{\frac{1}{1+\varepsilon}}R(b)^{\frac{1}{1+\varepsilon}},$$ where R(n) is the radical of $n$, is for $c\to\infty$ $$N(c)=(1-\varepsilon)\frac{ϕ(c)}{2}+O\Bigl(\frac{ϕ(c)}{2}\Bigr).$$ An analogue for the abc-conjecture inequality $c<R(abc)^{1+\varepsilon}$ (without a constant factor) will also be proved.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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