Why Does RSA Encryption Return the Original Number?

[James...]

Struggling to put a number through this as I keep getting my original number as the encrypted number too.

A = 11
p = 3 q = 5
n = pq = 15
z = (p-1)(q-1) = 2*4 = 8
k = co-prime of z = 7

So,

A=11
n=15
z=8 (Public key)
k=7(Public key)

kj = 1 (mod z)
7j = 1 (mod 8)

for which I am getting j = 9/7 (private key)

Start of encryption...

A^k = E (mod n)
11^7 = E (mod 15)

19487171/15 = 1299144.733...

1299144 * 15 = 19487160

E = 19487171 - 19487160 = 11 (which is what I started with)Tried using the decrypting part anyway and got...

E^j = A (mod n)

11^(9/7) = A (mod 15)

21.8239547419283/15 = 1.45493031612855

1 * 15 = 15

21.8239547419283 - 15 = 6.8239547419283 (which obviously isn't what I started with)

Any help where I am going wrong would be appreciated, I assume it is where mod is brought in as I haven't used that function before 2 hours ago but it may be somewhere else.

Cheers
james

 

[Borek]

for which I am getting j = 9/7 (private key)

Shouldn't all number involved be integers?

 

[James...]

I'm not sure. Thats where I think I have gone wrong though.

Think I messed up at

kj = 1 (mod z)
7j = 1 (mod 8)

trying to work out the private key as I have tried to teach myself how to do this from an example on a website without really knowing how to do it.

James

 

[James...]

Can anyone help with this? Will ask my math/physics teacher tomorrow but wouldn't mind having another go at it first.

Cheers
James

 

[Borek]

I am sure these should be all integers, but I have not played with RSA for several years.

 

[James...]

I think you are right but I am getting fractions when the modulus function is introduced as I'm unsure how it works.

James

 

[Borek]

mod never gives fractions, mod is about finding remainder.

10 mod 3 = 1
12 mod 5 = 2

and so on.

 

[HallsofIvy]

I'm not sure. Thats where I think I have gone wrong though.

Think I messed up at

kj = 1 (mod z)
7j = 1 (mod 8)

Just try some possibilities: 7(1)= 7, 7(2)= 14= 8+ 6, 7(3)= 21= 2(8)+ 5, 7(4)= 28= 3(8)+ 4, 7(5)= 35= 4(8)+ 7, 7(6)= 42= 5(8)+ 2, 7(7)= 49= 6(8)+ 1. 7j= 1 (mod 8) if and only if u= 7 (mod 8). I can't speak for "kj= 1 (mod z)" because I don't know what z is!

trying to work out the private key as I have tried to teach myself how to do this from an example on a website without really knowing how to do it.

James