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I have blinded $m = 16$ with $r= 13$, $e = 7$ and $n = 209$. This resulted in $m' = 464$, which led to a $s' = 8$.

Now, I want to unblind this $s'$ by using $s = s' \times r^{-1}$ but im stuck at filling in the $r^{-1}$?

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What's your value for $d$ and how have you computed it? You need to find $x$ such that $x \cdot 13\equiv 1 \pmod{209}$. This $x$ is the value $r^{-1}$ you are seeking. Hint: extended euclid – DrLecter Nov 25 '13 at 20:43
@DrLecter If i calculate it via the extended euclid r$^{-1}$ = 16, is that correct? – ziggy34 Nov 25 '13 at 23:59
Nope. You just have to check if $13\cdot 16 =1 \mod p$, which is not the case (it's $-1$ or equivalently $208$). – DrLecter Nov 26 '13 at 7:51
Just a side question: in practice, would it be safe to let $r$ be a 10-digit prime relative to $n$? – Kar Dec 12 '14 at 20:27
Figure: RSA Blind Signature using maple. RSA Blind Signature using maple Refer to SEJPM answer: Missing & correction: m=16 and q=19 – Arnold Aug 11 at 10:40

1 Answer 1

So let's first recall how blinded RSA works:

Select primes $p=11$ and $q=19$, calculate the modulus $n=pq=209$, choose a public exponent $e=7$. Now calculate the signature exponent $d\equiv e^{-1} \pmod{(p-1)(q-1)}$ ($d=103, (p-1)(q-1)=180$).

Now let's go to the message-dependant processing:
A message $M$ is represented as the integer $m=16$ which is blinded by first choosing a random integer $r=13$ and then calculating $m'=m*r^e \bmod n$ ($m'=46$). Finally the signature operation is carried out using $s'\equiv(m')^d\equiv(m*r^e)^d\equiv m^d*r^{ed}\equiv s*r \pmod n$($s'=8$). Finally calculate $s\equiv s'*r^{-1} \pmod n$. For this you need $r^{-1} \bmod n$($r^{-1}=193$). This yields the signature $s=81$.

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