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avatar Crypto

Factoring RSA when reusing N

cz flag
Fluctuation10111

Suppose in two RSA instances the same $p,q,N$ are used, but different public keys $a,b$ (and corresponding private keys)

Suppose now we have the two equations

$c_{1}=m^{a} \bmod N$

$c_{2}=m^{b} \bmod N$

Is it possibe to retrieve the original message $m$ with this information?

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rsa
poncho avatar
my flag
poncho
You might want to modify the title - this attack would *not* lead to a factorization of $N$...
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poncho avatar Crypto
my flag
poncho

Is it possible to retrieve the original message $m$ with this information?

Well, this is a standard exercise for beginners (that is, you're supposed to learn from it), and so I won't spell out the answer.

I will give you a hint: if you know $N, c_1 = m^a \bmod N, c_2 = m^b \bmod N$, can you compute the value of $c_3 = m^{a-b} \bmod N$? If so, how could you exploit that?

And, at the end, what condition must exist between $a$ and $b$ to allow you to recover $m$?

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Maarten Bodewes avatar
in flag
Maarten Bodewes
Removed answer so only the hint is left... congrats :)
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SSA avatar
ng flag
SSA
in case he wants to see an example. https://math.stackexchange.com/questions/2730675/decrypt-an-rsa-message-when-its-encrypted-by-same-modulus
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KKK avatar Crypto
in flag
KKK

It is possible to retrieve $m$ if $a$ and $b$ are coprime.

This means $GCD(a,b)=1$. Then, applying the Extended Euclidiean Algorithm we have $$ax+by=GCD(a,b) = 1$$ Therefore, $$c_1^x.c_2^y = m^{ax}.m^{by} = m^{ax+by} = m\bmod N$$

So just like that, we could recover $m$ but this gives us no information about the factorization of $N=pq$.

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