[Solved] Finding a prime number that is one greater than a squared number from a given number [closed]

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The sqrt() method may help. Here is an idea of how to use it:

int squareRoot = (int) Math.sqrt((double) input);

Input being the number you want to round. Casting the result to an int will automatically round it. It is optional. I cast the input to a double, but you only need to do so if your input in an int.

Here is an easier way to check if a number is prime:

if (input % 2 == 0){
    input += 1;
} 
    return input;

I reccomend you look at @useSticks answer again. Using the approach he described I created a method that does exactly what I think you want.

Here is a method that finds the square root of a positive number:

 public static int sqrt(double number){

        double g1;

       if(number==0){
            return 0;
        }

        double squareRoot = number/2;

        do{

            g1=squareRoot;
            squareRoot = (g1 + (number/g1))/2;

        }while((g1-squareRoot)!=0);

           return (int) squareRoot;

    }

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solved Finding a prime number that is one greater than a squared number from a given number [closed]