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Using RSA with e=13 (encrypting power), d=17 (decrypting power) & n=33 (RSA modulus) I noticed that once I decrypted the encrypted message it would be different then the original message. Why is that??

I used the primes p=11 & q=3 to get the modulus n=33. So the totient Phi(n) = k = 10*2 = 20

By choosing e=13, (d*e)mod(k)=1 d is 17.

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If I encrypt "4"

(4^13)mod(33) = 31

Decrypting "31" to get back "4"

(31^17)mod(33) = 28 (It's not working)

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Though by using e=3 & d=7 it works. Is there a relationship to these numbers??

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  • $\begingroup$ It would help if you would say what you mean by $e,d,n,p,q$. Not all sources use the same letters for the same things. In particular, if $n=pq$, then $q=13$ should be $q=3$. $\endgroup$ – Gerry Myerson Oct 16 '14 at 6:05
  • $\begingroup$ $31^{17}=(-2)^7=-128=-7=4\pmod{11}$, but $28=6\pmod{11}$, so maybe you should check your calculation. $\endgroup$ – Gerry Myerson Oct 16 '14 at 6:09
  • $\begingroup$ You're right. q=13 should be q=3. Just fixed it. $\endgroup$ – Jader J Rivera Oct 16 '14 at 11:35
  • $\begingroup$ Good. Now, did you check your calculation of $31^{17}\pmod{33}$? $\endgroup$ – Gerry Myerson Oct 16 '14 at 11:59
  • $\begingroup$ Yes. And now it somehow works. Thank you. $\endgroup$ – Jader J Rivera Oct 16 '14 at 12:08
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Might as well state that it was just a precision error.

Changing (31) to (-2) "fixes" the problem, since (31)mod(33) == (-2)mod(33).

And ((-2)^17)mod(33) = (-29), where (-29)mod(33) == (4)mod(33).

So it comes down to that one should use more than 8 bytes or use smaller numbers.

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  • $\begingroup$ Or one should calculate $a^b\pmod c$ by some method more clever than first computing $a^b$ and then reducing modulo $c$. It can be done without ever having to deal with any numbers bigger than $c^2$. $\endgroup$ – Gerry Myerson Oct 19 '14 at 9:07

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