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, $0 \lt a \lt b$, $e \ge 2$, $f \ge 3$ 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 $F(10^{18})$.
To solve this problem, it might be easier to deconstruct it and tackle it in smaller steps. Let’s start by considering the bounded ranges of $a, b, c, e, f$.
The constraints given tell us that $c^f \le 10^{18}$. For $f \ge 3$, a simple check shows that $c \le 10^6$ range has to be considered, since $10^6^3 = 10^{18}$. This already narrow down the range for the third base in the equation to check.
For the bases of the smaller powers, that is $a^e$ and $b^e$, the equation suggests the sum of these numbers equals $c^f$. Here, the exponent $e$ is quite small. Thus $a$ and $b$ could span much bigger ranges. However, knowing that $a$ and $b$ are definitely smaller than $c$ and $a < b$, intuitively the ranges of $a$ and $b$ are bound by the cube root of $c^3$, that is $c$ itself. Next, consider how the exponents $e$ and $f$ can vary. From the given conditions, $e \ge 2$ and $f \ge3$. With $e$ and $f$ being positive integers, it is clear that the combinations of all positive integer solutions for $e$ and $f$ are quite limited. Given these considerations, it is now possible to outline an algorithm to solve for $F(10^{18})$: 1. For every possible value of $c$ from 1 to $10^6$, consider pairs of $a$ and $b$ where $0 < a < b < c$. 2. For each pair of $a, b$, check if $a^e + b^e = c^f$ for some integers $e \ge 2$ and $f \ge 3$. 3. Keep track of how many solutions you find. Depending on the programming language and machine used, this brute-force algorithm could be very time-consuming. You could attempt to optimize it using various ploy, such as skipping pairs of $a$ and $b$ that are obviously too large to fit $a^e + b^e = c^f$ or constructing a lookup table to accelerate the calculation. Notice the possibility of high time complexity of the task. A brute force solution might not be practical. Implementing such an algorithm in Python utilizing numpy to vectorize calculations could reduce the time complexity and speed up the solution. However, to find the exact $F(10^{18})$, without any given further information and constraints and due to its complexity, you would not be able to compute the exact number analytically or using a simple formula. For sure it will require complex programming and computational power to calculate the precise value of $F(10^{18})$ given the above conditions.
More Answers:
$2^{\omega(n)}$Matching Digit Sums
Coloured Graphs