How to find d, given p, q, and e in RSA?

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滥情空心
滥情空心 2021-02-07 07:05

I know I need to use the extended euclidean algorithm, but I\'m not sure exactly what calculations I need to do. I have huge numbers. Thanks

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  •  名媛妹妹
    2021-02-07 07:32

    Given that, p=11, q=7, e =17, n=77, φ (n) = 60 and d=?

    First substitute values from the formula:-

    ed mod φ (n) =1

    17 d mod 60 = 1

    The next step: – take the totient of n, which is 60 to your left hand side and [e] to your right hand side.

    60 = 17

    3rd step: – ask how many times 17 goes to 60. That is 3.5….. Ignore the remainder and take 3.

    60 = 3(17)

    Step 4: – now you need to balance this equation 60 = 3(17) such that left hand side equals to right hand side. How?

    60 = 3(17) + 9 <== if you multiply 3 by 17 you get 51 then plus 9, that is 60. Which means both sides are now equal.

    Step 5: – Now take 17 to your left hand side and 9 to your right hand side.

    17 = 9

    Step 6:- ask how many times 9 goes to 17. That is 1.8…….

    17 = 1(9)

    Step 7:- Step 4: – now you need to balance this 17 = 1(9)

    17 = 1(9) + 8 <== if you multiply 1 by 9 you get 9 then plus 8, that is 17. Which means both sides are now equal.

    Step 8:- again take 9 to your left hand side and 8 to your right hand side.

    9 = 1(8)

    9 = 1(8) + 1 <== once you reached +1 to balance your equation, you may stop and start doing back substitution.

    Step A:-Last equation in step 8 which is 9 = 1(8) + 1 can be written as follows: 1.= 9 – 1(8)

    Step B:-We know what is (8) by simple saying 8 = 17 – 1(9) from step 7. Now we can re-write step A as:-

    1=9 -1(17 – 1(9)) <== here since 9=1(9) we can re-write as:-

    1=1(9)-1(17) +1(9) <== group similar terms. In this case you add 1(9) with 1(9) – that is 2(9).

    1=2(9)-1(17)

    Step C: – We know what is (9) by simple saying 9 = 60 – 3(17) from step 4. Now we can re-write step B as:-

    1=2(60-3(17) -1(17)

    1=2(60)-6(17) -1(17) <== group similar terms. In this case you add 6(17) with 1(17) – that is 7(17).

    1=2(60)-7(17) <== at this stage we can stop, nothing more to substitute, therefore take the value next 17. That is 7. Subtract it with the totient.

    60-7=d

    Then therefore the value of d= 53.

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