如何找出前N个斐波那契数中符合w³-1形式的数？

No, 0 is the only Fibonacci number that can be written in the form (w^3 -1). Here's why:

Manual Check of Small Fibonacci Numbers #

Looking at the first few Fibonacci numbers:

• (0 = 1^3 -1) (valid, (w=1))
• (1): (1+1=2) isn't a perfect cube
• (2): (2+1=3) isn't a perfect cube
• (3): (3+1=4) isn't a perfect cube
• (5): (5+1=6) isn't a perfect cube
• (8): (8+1=9) isn't a perfect cube
• All larger Fibonacci numbers grow exponentially, and their value plus 1 never results in a perfect cube.

Mathematical Proof #

Our equation is (Fib(n) = w^3 -1), or rearranged: (w^3 - Fib(n) = 1).

By Mihăilescu's Theorem (previously Catalan's Conjecture), the only pair of positive perfect powers that differ by exactly 1 are (9=3^2) and (8=2^3) (since (3^2 - 2^3 =1)). For our equation to hold, (w^3) would need to be one of these perfect powers, but:

• If (w^3=8), then (Fib(n)=7), which isn't a Fibonacci number.
• If (w^3=9), that's not a perfect cube.

This rules out any possible solutions except when (w=1), which gives (Fib(n)=0). Number theory research confirms there are no larger solutions to (Fib(n)+1 =w^3).

Notes on Your Code #

Your current code works correctly for the range of Fibonacci numbers that fit in a long (up to the 90th term). Beyond that, long will overflow, so if you need to check larger values, switch to BigInteger. Here's a modified version:

public static List<BigInteger> generateFibonacciNumbers(int N) {
    List<BigInteger> fibonacciNumbers = new ArrayList<>();
    BigInteger a = BigInteger.ZERO, b = BigInteger.ONE;
    fibonacciNumbers.add(a);
    if (N > 1) fibonacciNumbers.add(b);

    for (int i = 2; i < N; i++) {
        BigInteger c = a.add(b);
        fibonacciNumbers.add(c);
        a = b;
        b = c;
    }
    return fibonacciNumbers;
}

public static boolean isW3Minus1(BigInteger num) {
    if (num.compareTo(BigInteger.ZERO) < 0) return false;
    BigInteger target = num.add(BigInteger.ONE);
    BigInteger low = BigInteger.ONE;
    BigInteger high = target;

    while (low.compareTo(high) <= 0) {
        BigInteger mid = low.add(high).divide(BigInteger.TWO);
        BigInteger cube = mid.multiply(mid).multiply(mid);
        int comparison = cube.compareTo(target);
        
        if (comparison == 0) return true;
        else if (comparison < 0) low = mid.add(BigInteger.ONE);
        else high = mid.subtract(BigInteger.ONE);
    }
    return false;
}

内容的提问来源于stack exchange，提问作者mary