勒让德猜想的C程序

它说在任何两个连续的自然数(n = 1，2，3，4，5，…)平方之间总是存在一个质数。这称为勒让德猜想。
猜想：猜想是基于不完整信息的命题或结论，该信息未找到证据，即尚未被证明或被证明。

Mathematically,
there is always one prime p in the range $n^2$ to $(n + 1)^2$ where n is any natural number.

for examples-
2 and 3 are the primes in the range $1^2$ to $2^2$.

5 and 7 are the primes in the range $2^2$ to $3^2$.

11 and 13 are the primes in the range $3^2$ to $4^2$.

17 and 19 are the primes in the range $4^2$ to $5^2$.

例子：

Input : 4 
output: Primes in the range 16 and 25 are:
        17
        19
        23

说明：这里4 ² = 16和5 ² = 25
因此，介于16和25之间的质数是17、19和23。

Input : 10
Output: Primes in the range 100 and 121 are:
        101
        103
        107
        109
        113

// CPP program to verify Legendre's Conjecture
// for a given n.
#include 
using namespace std;
  
// prime checking
bool isprime(int n)
{
    for (int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return false;
    return true;
}
  
void LegendreConjecture(int n)
{
    cout << "Primes in the range " << n * n
         << " and " << (n + 1) * (n + 1)
         << " are:" << endl;
  
    for (int i = n * n; i <= ((n + 1) * (n + 1)); i++)
  
        // searching for primes
        if (isprime(i))
            cout << i << endl;
}
  
// Driver program
int main()
{
    int n = 50;
    LegendreConjecture(n);
    return 0;
}

输出：

Primes in the range 2500 and 2601 are:
2503
2521
2531
2539
2543
2549
2551
2557
2579
2591
2593

请参阅有关勒让德猜想的完整文章，以了解更多详细信息！

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