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Problem 678

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Fermat-like Equations

If a triple of positive integers $(a,b,c)$ satisfies $a^2+b^2=c^2$, it is called a Pythagorean triple. No triple $(a,b,c)$ satisfies $a^e+b^e=c^e$ when $e\ge 3$ (Fermat’s Last Theorem). However, if the exponents of the left-hand side and right-hand side differ, this is not true. For example, $3^3+6^3=3^5$.

Let $a,b,c,e,f$ be all positive integers, $0<a<b$, $e\ge 2$, $f\ge 3$ and $c^f\le N$. Let $F(N)$ be the number of $(a,b,c,e,f)$ such that $a^e+b^e=c^f$. You are given $F({10}^3)=7$, $F({10}^5)=53$ and $F({10}^7)=287$.

Find $F({10}^{18})$.

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费马式方程

若正整数三元组$(a,b,c)$满足$a^2+b^2=c^2$，则称之为毕达哥拉斯三元组。根据费马大定理，对于任意$e\ge 3$，不存在三元组$(a,b,c)$满足$a^e+b^e=c^e$。不过，若等号左侧和右侧所选择的指数不同，则方程可能有解。例如，$3^3+6^3=3^5$。

令$a,b,c,e,f$为正整数，且$0<a<b$，$e\ge 2$，$f\ge 3$，$c^f\le N$。记$F(N)$为使得$a^e+b^e=c^f$的数组$(a,b,c,e,f)$总数。已知$F({10}^3)=7$，$F({10}^5)=53$，$F({10}^7)=287$。

求$F({10}^{18})$。

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