Posted: May 22nd, 2023

Let n be the integer in the computer problem

In this case n = 618240007109027021

Therefore from the above calculations, we obtain the following;

p = 250387201

Therefore the small prime factors of p-1 are

(2)⁸ (3)⁵ (5)²Â (23) (7)

2) We first select a bound for the factorial.

b = ln (n)

is the acceptable limit.

in this case 43.

Exponent is 43! = 6.0415263063373835637355132068514e+52 (factorial of 43)

then we choose some number to raise to that exponential value.

A number ‘a’ from 2 to n-1 can be used.

In this case, I chose 2

Then that’s a factor of n

If not, raise the bound limit.

P = 250387201

//The function to find the prime factors of n is;

public static BigInteger primefactor(BigInteger n) // n is the integer to factorize

public static BigInteger primefactor(BigInteger n) // n is the integer to factorize

return y.subtract(BigInteger.ONE).gcd(n);}

static void execA()

static void factor(BigInteger n)

public static void main(String[] args) //Main method

}

To factor in a larger value of n, you insert the following method in the code;

static void execB()

execB(); // then call execB() function in main

Factoring n = 642401 using the information;

516107²Â â‰¡ 7 (mod n)

And that

187722²Â â‰¡ 2²Â âˆ™ 7 (mod n)

n = 642401

Therefore factor1 := igcd (516107*187722-2*7, n); // n = 642401

factor1 := 1129

and we obtain factor2 by dividing n by factor1

factor2 := n/factor1;

factor2 := 569

Thus 642401= 1129 * 569 which are indeed primes

given that 2^(n−1) ≡ k²Â _≡ 1 (mod n)

Then we can deduce from “Fermat’s Little theorem” that n is not indeed a prime

number.

Because 2^{(n−1)/2} _≡ ±1 (mod n) and 2^(n−1) ≡ 1 (mod n)

then From the “basic factoring principle”; k²Â â‰¡Â 1²Â (mod n)

and k _≡ ±1 (mod n)

Therefore we can find a nontrivial factor of n using the formula gcd(k-1, n) or gcd(k+1, n)