Posted: May 22nd, 2023

In this case n = 618240007109027021

Therefore from the above calculations, we obtain the following;

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Let n be the integer in the computer problem

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p = 250387201

Therefore the small prime factors of p-1 are

(2)^{8}Â (3)^{5}Â (5)^{2}Â (23) (7)

2)Â 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 â€˜aâ€™ from 2 to n-1 can be used.

In this case, I chose 2

Then thatâ€™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;

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

Factoring n = 642401 using the information;

516107^{2}Â â‰¡ 7 (mod n)

And that

187722^{2}Â â‰¡ 2^{2}Â 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}^{âˆ’}^{1)}Â *â‰¡ k*^{2}Â *_â‰¡Â *1 (modÂ *n*)

Then we can deduce from â€œFermatâ€™s Little theoremâ€ that n is not indeed a prime

number.

Because 2^{(}^{n}^{âˆ’}^{1)}^{/}^{2}Â *_â‰¡ Â±*1 (modÂ *n*) and 2^{(}^{n}^{âˆ’}^{1)}Â *â‰¡Â *1 (modÂ *n*)

then From the â€œbasic factoring principleâ€;Â *k*^{2}Â *â‰¡Â *1^{2}Â (modÂ *n*)

*and k _â‰¡ Â±*1 (modÂ *n*)

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

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