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I am trying to crack an unpadded RSA set up for a homework.

I have public key (R,e) and encrypted message E.

R:=1189873858375600120035497781492406260116261348762168101523668479976252918878873543993974091515716923413521166189;
e:=7;
E:=935688997292570230577548538417371291686892187342525833188430568797480485753147343888751799688783739259397891334;

Obviously, the attack that I'm supposed to use has something to do with the size of the public encryption exponent but I'm struggling to see exactly how to use it.

I understand the relationship between e, d and φ(n) and understand that there is an integer k < min{e, d} such that φ(n) = (e ∗ d − 1)/k but I don't understand how I can use that fact to brute force discover φ(n).

It appears that the solution comes from doing some kind of search from 1..k but since d is still unknown, I don't really see how to take advantage of that.

Can someone enlighten me, please?

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    $\begingroup$ I suspect that the smallness of the encryption exponent has nothing to do with it; instead, consider the modulus, which is approximately circa 372 bits long; how can you take advantage of that? $\endgroup$ – poncho Oct 1 '14 at 3:31
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    $\begingroup$ To elaborate a bit on poncho's comment, most real world RSA encryption schemes use small encryption exponents for performance, usually 3, 17 or 65537 - so $e = 7$ may not be of any help! $\endgroup$ – Thomas Oct 1 '14 at 6:43
  • $\begingroup$ If the message is shorter than the length of the modulus divided by the public exponent, you can simply compute the $e$ root, for example via binary search. For real RSA the padding ensures that the message has similar length as the modulus, but for unpadded RSA short messages can happen. $\endgroup$ – CodesInChaos Oct 1 '14 at 8:10
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    $\begingroup$ The actual statement (see Problem 2) does not hint that padding or small exponent are relevant, so I guess Poncho hinted to the expected resolution path. $\;$ Notice that this homework is not quite past its due date + allowance; please do not post a solution before that. $\endgroup$ – fgrieu Oct 1 '14 at 8:26
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    $\begingroup$ 1) Simply running a factoring software sounds a little dull for a homework question. It's also weird that the moduli for Problem 1 are smaller. If you bothered to setup factoring software for Problem 2 you use it to solve problem 1 as well, without exploiting the same message encryption weakness. 2) the short message and small exponent approach doesn't seem to work. $\endgroup$ – CodesInChaos Oct 1 '14 at 9:54
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The question is from Problem 2 in Assignment 6 given here. It looks much like a straight instance of the RSA problem: we want to find $x\in\mathbb N$ such that $x^e=E\bmod R$ and $x<R$, given $R$, $E$, $e=7$, with $E$ expected to be the product of two unknown primes $p,q$ such that $e$ is coprime with $(p-1)\cdot(q-1)$.

It is correct that knowing none of $d$, $\varphi(R)$ or $\lambda(R)$, nor the factorization of $R$, we can't solve the problem by massaging $e\cdot d\equiv1\pmod{\varphi(R)}$ or $e\cdot d\equiv1\pmod{\lambda(R)}$ with $e$ small.

The fact that $e$ is small would help if it turned out that $E$ (or $i\cdot R+E$ for some small $i$) was an exact power of $e$; for this would allow to find $x$ by extracting an $e$-th root. For random parameters, this happens with odds about ${R^{1/e-1}}$, thus decreasing with $e$. But $R$ is so big that theses odds are vanishingly low for $e\ge2$. In fact, any small odd $e>2$ is believed a safe choice in RSA when $x$ is random or properly padded.

The most classic avenue to solve an RSA problem is to factor the public modulus $R$. This is how Part 1 of Assignment 2 was done, for $R$ of 276 bits. In the present problem, $R$ is 367-bit, therefore brute force factorization is feasible (it was feasible in 1995 by means available to academics; see this answer). However it would require a significant effort, and I do not know that Maple's built-in factorization primitive would factor a properly chosen 367-bit RSA modulus in reasonable time.

We should not rule out factorization though: in this course, Alice (which, we are told, has $R,e$ as public key) has in Assignment 4 made unconventional and rather poor choice of key; so there is a chance that $R$ was not properly chosen and can be factored with little effort, using some of the factorization methods likely alluded to in the course.

Hint 1: The problem can be solved with practically any big-number calculator offering modular exponentiation, without even writing an iterative program.

Hint 2: (Hover mouse to see it)

Alice may have put too much emphasis on being ultra-resistant to trial division for a given magnitude of $R$.

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  • $\begingroup$ Hey Fgrieu, I got a similar assignment problem and I can't run the factorization software on a 309 digit long so roughly almost 1000 bits long n along with e being 5, Although I will try this trial division method but would hope if something better way using lattice theory? $\endgroup$ – codeomnitrix Nov 25 '14 at 6:45
  • $\begingroup$ @codeomnitrix: the assignment in this question was intended to be solvable in one iteration of the Fermat factoring method; that is, the simple 4 steps in this answer. That may be worth trying in your case (especially if there is any hint that the factors are very close, or mention of the method); and perhaps the full Fermat factoring. $\endgroup$ – fgrieu Nov 25 '14 at 12:31
  • $\begingroup$ No fgrieu, neither factors are too close nor the (p-1) or (q-1) is smooth i.e. easily factorized. So that option is wiped out. Can you please help me with Coppersmith's method? Thanks in advance.. Please see the question - crypto.stackexchange.com/questions/20450/… $\endgroup$ – codeomnitrix Nov 27 '14 at 12:44
  • $\begingroup$ @codeomnitrix: I scratched my head and failed to see how Coppersmith's method (also here) could help in your other question as it originally stood; but the modified one is very different! $\endgroup$ – fgrieu Nov 27 '14 at 15:51

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