Thanks in advance.

Sandeep.

]]>You can use a fast powering algorithm (number of operations reduces to logarithmic instead of linear) –

To calculate a^k (mod m):

(1) Form a table with three columns and first row: “k” “a” “1”;

(2) at each stage, form a new row as follows:

(a) the entry in column 1 is half the entry in column 1 of the previous row, ignoring fractions;

(b) the entry in column 2 is the square of the entry in column 2 of the previous row (modulo m);

(c) if the entry in column 1 of the previous row is even, the entry in column 3 is the same as the entry in column 3 of the previous row, otherwise it is the product (modulo m) of the entries oin column 2 and column 3 of the previous row.

—

(3) when the entry in column 1 is 0, the entry in column 3 of that row is the required power.

EXAMPLE: To calculate 2^11 (mod 23):

11 2 1

5 4 2

2 16 8

1 3 8

0 9 1

Therefore, 2^11 = 1 (mod 23).

]]>for an unknown element in our ring is not too difficult to analyze, at least in many examples. If there are no solutions at all, then a fortiori there are no solutions in our ideal. So the ideal would not be principal. However, if there is a solution, you would still need to check if there is a solution in the ideal. This can be somewhat trickier.

]]>The definition is

N(J) = |O_F / J |

So in the example,

Z[i]/(5,1+i) = {0}

=> N(J) = |{0}| = 1

I’m afraid I might have overseen this amidst the chaos… sorry..!..

]]>Remember, it is not itself a prime number in general. It just has a prime number factor that doesn’t belong to the list. Given a number that’s not a prime, it is exactly the problem of finding a prime number factor that admits no efficient algorithm.

Recall that if you’re satisfied with an inefficient algorithm, then you can keep finding prime numbers by trying to divide numbers one by one.

]]>vikram mukarji ]]>