how can u prove that 900720143 is not a prime.?

Question

without any calculator, by algebra.

Answer 1

The number 900720143 can be written as
(30012)^2 -1. so it is not a prime.
[we know (a^2-1) can be divided by a+1 and a-1]

Answer 2

The only way I know of to PROVE a number is prime is to divide it by every known prime lower than its square root, and check for a remainder. Of course, this assumes that you know all the primes less than the square root, which is what makes this problem very difficult.

I don't know that there is an algebraic method other than this iterative technique to find or prove primeness. This is why the search for large prime numbers is so difficult.

Answer 3

all primes > 3 are of the form 6n + 1. If your number were a prime it would have a remainder of 1 when divided by 6. It doesn't, so not prime.

Disregard this hasty response --my error!!!!

It is true that primes are 6n + 1, but not all of 6n + 1 are prime as others have pointed out. It can also be shown that one less than the square of a prime number > 3 is evenly divisible by 24. So if someone has the time, paper, patinnce or a good enough calculator/spreadsheet, square 900720143, subtract 1, and if the result is divisible by 24, it's prime, otherwise not.

Answer 4

There is no way to determine prime numbers other than applying algorithms.

A simple, ancient algorithm for finding all prime numbers up to a specified integer. (Modify limit according to your need)

limit = 1000000
sieve$ = string of the character "P" with length limit

prime = 2
repeat while prime2 < limit
set the character at the index of each multiple of prime (excluding index prime * 1) in sieve$ to "N"
prime = index of the next instance of "P" in sieve$ after index prime
end repeat

print the index of each instance of "P" in sieve$



A fast, modern algorithm for finding all prime numbers up to a specified integer.


// arbitrary search limit
limit ← 1000000

// initialize the sieve
is_prime(i) ← false, i ∈ [5, limit]

// put in candidate primes:
// integers which have an odd number of
// representations by certain quadratic forms
for (x, y) in [1, √limit] × [1, √limit]:
n ← 4x²+y²
if (n ≤ limit) ∧ (n mod 12 = 1 ∨ n mod 12 = 5):
is_prime(n) ← ¬is_prime(n)
n ← 3x²+y²
if (n ≤ limit) ∧ (n mod 12 = 7):
is_prime(n) ← ¬is_prime(n)
n ← 3x²-y²
if (x > y) ∧ (n ≤ limit) ∧ (n mod 12 = 11):
is_prime(n) ← ¬is_prime(n)

// eliminate composites by sieving
for n in [5, √limit]:
if is_prime(n):
// n is prime, omit multiples of its square; this is
// sufficient because composites which managed to get
// on the list cannot be square-free
is_prime(k) ← false, k ∈ {n², 2n², 3n², ..., limit}

print 2, 3
for n in [5, limit]:
if is_prime(n): print n

Answer 5

yes and when n= 4

6n+ 1 = 25

n= 8

6n + 1 = 49

Answer 6

John V :
I don't think that's true. 5 and 11 are examples of primes that don't follow your '6n+1' rule.

Answer 7

You can use the contrapositive of Fermat's little theorem:
If 2^(n-1) <> 1(mod n) then n is composite.
With the aid of PARI, I found
that 2^(900720142) mod(900720143) = 469550677
so 900720143 is not prime.

Answer 8

By algebra? Simple. Just show that group Z_900720143 is cyclic.

Answer 9

try dividing it by a ton of different numbers (by hand, so you don't break the no calculator rule) and then go from there

Answer 10

USE PRIME FACTORISTAION