RSA: Is it possible to recover the plaintext given that we have the ciphertext and the public key?
$begingroup$
The $N$ and ciphertext are huge number that is more than 600 digits long.
I'm trying to find the prime factors of $N$ in order to get $p$ and $q$. Using $p$ and $q$ I can get $phi(n)$. And using $phi(n)$ and the public key I would be able to calculate the private key.
So, my question is if it is possible to recover the plaintext given that we have ciphertext and the public key?
If so, is it computationally feasible?
Thank you
rsa
edited 2 hours ago
AleksanderRas
2,0391630
asked 2 hours ago
Maqruis1Maqruis1
61
New contributor
Maqruis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
The $N$ and ciphertext are huge number that is more than 600 digits long.
I'm trying to find the prime factors of $N$ in order to get $p$ and $q$. Using $p$ and $q$ I can get $phi(n)$. And using $phi(n)$ and the public key I would be able to calculate the private key.
So, my question is if it is possible to recover the plaintext given that we have ciphertext and the public key?
If so, is it computationally feasible?
Thank you
rsa
edited 2 hours ago
AleksanderRas
2,0391630
asked 2 hours ago
Maqruis1Maqruis1
61
New contributor
Maqruis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Doing so means essentially breaking the scheme, right? The public key is known (hence the term public), and the modulus $N$ is part of the public key. Thus, what you're asking is whether or not we can find the underlying plaintext corresponding to some given ciphertext, which should not be possible according to our belief on the security of RSA.
$endgroup$
– Daniel
2 hours ago
3
$begingroup$
What kind of digits are that? Bits? Decimals? If they are decimals - which for us humans is kind of the default - then we're probably talking about a 2048 bit key, and RSA 2048 is certainly unbreakable if the key pair generation and decryption procedures are implemented correctly.... Please clarify the type of digits.
$endgroup$
– Maarten Bodewes♦
1 hour ago
add a comment |
$begingroup$
The $N$ and ciphertext are huge number that is more than 600 digits long.
I'm trying to find the prime factors of $N$ in order to get $p$ and $q$. Using $p$ and $q$ I can get $phi(n)$. And using $phi(n)$ and the public key I would be able to calculate the private key.
So, my question is if it is possible to recover the plaintext given that we have ciphertext and the public key?
If so, is it computationally feasible?
Thank you
rsa
edited 2 hours ago
AleksanderRas
2,0391630
asked 2 hours ago
Maqruis1Maqruis1
61
New contributor
Maqruis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The $N$ and ciphertext are huge number that is more than 600 digits long.
I'm trying to find the prime factors of $N$ in order to get $p$ and $q$. Using $p$ and $q$ I can get $phi(n)$. And using $phi(n)$ and the public key I would be able to calculate the private key.
So, my question is if it is possible to recover the plaintext given that we have ciphertext and the public key?
If so, is it computationally feasible?
Thank you
rsa
rsa
edited 2 hours ago
AleksanderRas
2,0391630
asked 2 hours ago
Maqruis1Maqruis1
61
New contributor
Maqruis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
AleksanderRas
2,0391630
asked 2 hours ago
Maqruis1Maqruis1
61
New contributor
Maqruis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
AleksanderRas
2,0391630
edited 2 hours ago
AleksanderRas
2,0391630
edited 2 hours ago
AleksanderRas
2,0391630
2,0391630
asked 2 hours ago
Maqruis1Maqruis1
61
New contributor
Maqruis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Maqruis1Maqruis1
61
asked 2 hours ago
Maqruis1Maqruis1
61
61
New contributor
Maqruis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Maqruis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Maqruis1 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Doing so means essentially breaking the scheme, right? The public key is known (hence the term public), and the modulus $N$ is part of the public key. Thus, what you're asking is whether or not we can find the underlying plaintext corresponding to some given ciphertext, which should not be possible according to our belief on the security of RSA.
$endgroup$
– Daniel
2 hours ago
3
$begingroup$
What kind of digits are that? Bits? Decimals? If they are decimals - which for us humans is kind of the default - then we're probably talking about a 2048 bit key, and RSA 2048 is certainly unbreakable if the key pair generation and decryption procedures are implemented correctly.... Please clarify the type of digits.
$endgroup$
– Maarten Bodewes♦
1 hour ago
add a comment |
2
$begingroup$
Doing so means essentially breaking the scheme, right? The public key is known (hence the term public), and the modulus $N$ is part of the public key. Thus, what you're asking is whether or not we can find the underlying plaintext corresponding to some given ciphertext, which should not be possible according to our belief on the security of RSA.
$endgroup$
– Daniel
2 hours ago
3
$begingroup$
What kind of digits are that? Bits? Decimals? If they are decimals - which for us humans is kind of the default - then we're probably talking about a 2048 bit key, and RSA 2048 is certainly unbreakable if the key pair generation and decryption procedures are implemented correctly.... Please clarify the type of digits.
$endgroup$
– Maarten Bodewes♦
1 hour ago
2
2
$begingroup$
Doing so means essentially breaking the scheme, right? The public key is known (hence the term public), and the modulus $N$ is part of the public key. Thus, what you're asking is whether or not we can find the underlying plaintext corresponding to some given ciphertext, which should not be possible according to our belief on the security of RSA.
$endgroup$
– Daniel
2 hours ago
$begingroup$
Doing so means essentially breaking the scheme, right? The public key is known (hence the term public), and the modulus $N$ is part of the public key. Thus, what you're asking is whether or not we can find the underlying plaintext corresponding to some given ciphertext, which should not be possible according to our belief on the security of RSA.
$endgroup$
– Daniel
2 hours ago
3
3
$begingroup$
What kind of digits are that? Bits? Decimals? If they are decimals - which for us humans is kind of the default - then we're probably talking about a 2048 bit key, and RSA 2048 is certainly unbreakable if the key pair generation and decryption procedures are implemented correctly.... Please clarify the type of digits.
$endgroup$
– Maarten Bodewes♦
1 hour ago
$begingroup$
What kind of digits are that? Bits? Decimals? If they are decimals - which for us humans is kind of the default - then we're probably talking about a 2048 bit key, and RSA 2048 is certainly unbreakable if the key pair generation and decryption procedures are implemented correctly.... Please clarify the type of digits.
$endgroup$
– Maarten Bodewes♦
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It might be feasible, or not.
If "digits" had been binary digits or bits, the answer would have been yes. Anything about 600-bit can be factored by GNFS. The public record is for a 768-bit RSA modulus, factored in 2009. 600-bit is within reach of CADO-NFS and Msieve. That's even packaged into factoring as a service for semi-deep-pocketed script kiddies.
It could be that $N$ was poorly chosen and can be factored much more easily than by GNFS. Since the generation algorithm is unspecified, we can't tell. Poor RSA key generators have happened (see e.g. SmartFacts, ROCA, and predictable SSH hosts keys on way too many platforms).
It could also be that the encryption system used is plaintext RSA, where the ciphertext is $C=M^ebmod N$ with $M$ the plaintext. This contrasts secure RSA, where $M$ is obtained from plaintext and randomness using a padding scheme such as OAEP. If $M$ is the plaintext (or a known deterministic function of the plaintext), some attacks much easier than factoring the modulus might be possible:
- if $M$ is in a known small set (names on the class roll, 16-digit credit-card number), it is possible to enumerate possible $M$, perform encryption, and check against the ciphertext.
- if $M=Ucdot V$ with $U<2^u$, $V<2^v$ (which is likely when $Mll2^{u,v}$ ) then there's a Meet-in-the-Midle attack with cost $O(2^{max(u,v)})$ time and $O(2^{min(u,v)})$ memory.
- if $M<sqrt[e]N$ we have the $e^text{th}$ root attack (which can be extended to slightly larger $M$).
answered 1 hour ago
fgrieufgrieu
78.2k7166331
$endgroup$
$begingroup$
Could you post a link for the Meet-in -the-Middle attack?
$endgroup$
– kelalaka
26 mins ago
1
$begingroup$
@kelalaka: see this, item 3
$endgroup$
– fgrieu
21 mins ago
add a comment |
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$begingroup$
It might be feasible, or not.
If "digits" had been binary digits or bits, the answer would have been yes. Anything about 600-bit can be factored by GNFS. The public record is for a 768-bit RSA modulus, factored in 2009. 600-bit is within reach of CADO-NFS and Msieve. That's even packaged into factoring as a service for semi-deep-pocketed script kiddies.
It could be that $N$ was poorly chosen and can be factored much more easily than by GNFS. Since the generation algorithm is unspecified, we can't tell. Poor RSA key generators have happened (see e.g. SmartFacts, ROCA, and predictable SSH hosts keys on way too many platforms).
It could also be that the encryption system used is plaintext RSA, where the ciphertext is $C=M^ebmod N$ with $M$ the plaintext. This contrasts secure RSA, where $M$ is obtained from plaintext and randomness using a padding scheme such as OAEP. If $M$ is the plaintext (or a known deterministic function of the plaintext), some attacks much easier than factoring the modulus might be possible:
- if $M$ is in a known small set (names on the class roll, 16-digit credit-card number), it is possible to enumerate possible $M$, perform encryption, and check against the ciphertext.
- if $M=Ucdot V$ with $U<2^u$, $V<2^v$ (which is likely when $Mll2^{u,v}$ ) then there's a Meet-in-the-Midle attack with cost $O(2^{max(u,v)})$ time and $O(2^{min(u,v)})$ memory.
- if $M<sqrt[e]N$ we have the $e^text{th}$ root attack (which can be extended to slightly larger $M$).
answered 1 hour ago
fgrieufgrieu
78.2k7166331
$endgroup$
$begingroup$
Could you post a link for the Meet-in -the-Middle attack?
$endgroup$
– kelalaka
26 mins ago
1
$begingroup$
@kelalaka: see this, item 3
$endgroup$
– fgrieu
21 mins ago
add a comment |
$begingroup$
It might be feasible, or not.
If "digits" had been binary digits or bits, the answer would have been yes. Anything about 600-bit can be factored by GNFS. The public record is for a 768-bit RSA modulus, factored in 2009. 600-bit is within reach of CADO-NFS and Msieve. That's even packaged into factoring as a service for semi-deep-pocketed script kiddies.
It could be that $N$ was poorly chosen and can be factored much more easily than by GNFS. Since the generation algorithm is unspecified, we can't tell. Poor RSA key generators have happened (see e.g. SmartFacts, ROCA, and predictable SSH hosts keys on way too many platforms).
It could also be that the encryption system used is plaintext RSA, where the ciphertext is $C=M^ebmod N$ with $M$ the plaintext. This contrasts secure RSA, where $M$ is obtained from plaintext and randomness using a padding scheme such as OAEP. If $M$ is the plaintext (or a known deterministic function of the plaintext), some attacks much easier than factoring the modulus might be possible:
- if $M$ is in a known small set (names on the class roll, 16-digit credit-card number), it is possible to enumerate possible $M$, perform encryption, and check against the ciphertext.
- if $M=Ucdot V$ with $U<2^u$, $V<2^v$ (which is likely when $Mll2^{u,v}$ ) then there's a Meet-in-the-Midle attack with cost $O(2^{max(u,v)})$ time and $O(2^{min(u,v)})$ memory.
- if $M<sqrt[e]N$ we have the $e^text{th}$ root attack (which can be extended to slightly larger $M$).
answered 1 hour ago
fgrieufgrieu
78.2k7166331
$endgroup$
$begingroup$
Could you post a link for the Meet-in -the-Middle attack?
$endgroup$
– kelalaka
26 mins ago
1
$begingroup$
@kelalaka: see this, item 3
$endgroup$
– fgrieu
21 mins ago
add a comment |
$begingroup$
It might be feasible, or not.
If "digits" had been binary digits or bits, the answer would have been yes. Anything about 600-bit can be factored by GNFS. The public record is for a 768-bit RSA modulus, factored in 2009. 600-bit is within reach of CADO-NFS and Msieve. That's even packaged into factoring as a service for semi-deep-pocketed script kiddies.
It could be that $N$ was poorly chosen and can be factored much more easily than by GNFS. Since the generation algorithm is unspecified, we can't tell. Poor RSA key generators have happened (see e.g. SmartFacts, ROCA, and predictable SSH hosts keys on way too many platforms).
It could also be that the encryption system used is plaintext RSA, where the ciphertext is $C=M^ebmod N$ with $M$ the plaintext. This contrasts secure RSA, where $M$ is obtained from plaintext and randomness using a padding scheme such as OAEP. If $M$ is the plaintext (or a known deterministic function of the plaintext), some attacks much easier than factoring the modulus might be possible:
- if $M$ is in a known small set (names on the class roll, 16-digit credit-card number), it is possible to enumerate possible $M$, perform encryption, and check against the ciphertext.
- if $M=Ucdot V$ with $U<2^u$, $V<2^v$ (which is likely when $Mll2^{u,v}$ ) then there's a Meet-in-the-Midle attack with cost $O(2^{max(u,v)})$ time and $O(2^{min(u,v)})$ memory.
- if $M<sqrt[e]N$ we have the $e^text{th}$ root attack (which can be extended to slightly larger $M$).
answered 1 hour ago
fgrieufgrieu
78.2k7166331
$endgroup$
It might be feasible, or not.
If "digits" had been binary digits or bits, the answer would have been yes. Anything about 600-bit can be factored by GNFS. The public record is for a 768-bit RSA modulus, factored in 2009. 600-bit is within reach of CADO-NFS and Msieve. That's even packaged into factoring as a service for semi-deep-pocketed script kiddies.
It could be that $N$ was poorly chosen and can be factored much more easily than by GNFS. Since the generation algorithm is unspecified, we can't tell. Poor RSA key generators have happened (see e.g. SmartFacts, ROCA, and predictable SSH hosts keys on way too many platforms).
It could also be that the encryption system used is plaintext RSA, where the ciphertext is $C=M^ebmod N$ with $M$ the plaintext. This contrasts secure RSA, where $M$ is obtained from plaintext and randomness using a padding scheme such as OAEP. If $M$ is the plaintext (or a known deterministic function of the plaintext), some attacks much easier than factoring the modulus might be possible:
- if $M$ is in a known small set (names on the class roll, 16-digit credit-card number), it is possible to enumerate possible $M$, perform encryption, and check against the ciphertext.
- if $M=Ucdot V$ with $U<2^u$, $V<2^v$ (which is likely when $Mll2^{u,v}$ ) then there's a Meet-in-the-Midle attack with cost $O(2^{max(u,v)})$ time and $O(2^{min(u,v)})$ memory.
- if $M<sqrt[e]N$ we have the $e^text{th}$ root attack (which can be extended to slightly larger $M$).
answered 1 hour ago
fgrieufgrieu
78.2k7166331
edited 55 mins ago
answered 1 hour ago
fgrieufgrieu
78.2k7166331
answered 1 hour ago
fgrieufgrieu
78.2k7166331
answered 1 hour ago
fgrieufgrieu
78.2k7166331
78.2k7166331
$begingroup$
Could you post a link for the Meet-in -the-Middle attack?
$endgroup$
– kelalaka
26 mins ago
1
$begingroup$
@kelalaka: see this, item 3
$endgroup$
– fgrieu
21 mins ago
add a comment |
$begingroup$
Could you post a link for the Meet-in -the-Middle attack?
$endgroup$
– kelalaka
26 mins ago
1
$begingroup$
@kelalaka: see this, item 3
$endgroup$
– fgrieu
21 mins ago
$begingroup$
Could you post a link for the Meet-in -the-Middle attack?
$endgroup$
– kelalaka
26 mins ago
$begingroup$
Could you post a link for the Meet-in -the-Middle attack?
$endgroup$
– kelalaka
26 mins ago
1
1
$begingroup$
@kelalaka: see this, item 3
$endgroup$
– fgrieu
21 mins ago
$begingroup$
@kelalaka: see this, item 3
$endgroup$
– fgrieu
21 mins ago
add a comment |
Maqruis1 is a new contributor. Be nice, and check out our Code of Conduct.
Maqruis1 is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
Doing so means essentially breaking the scheme, right? The public key is known (hence the term public), and the modulus $N$ is part of the public key. Thus, what you're asking is whether or not we can find the underlying plaintext corresponding to some given ciphertext, which should not be possible according to our belief on the security of RSA.
$endgroup$
– Daniel
2 hours ago
3
$begingroup$
What kind of digits are that? Bits? Decimals? If they are decimals - which for us humans is kind of the default - then we're probably talking about a 2048 bit key, and RSA 2048 is certainly unbreakable if the key pair generation and decryption procedures are implemented correctly.... Please clarify the type of digits.
$endgroup$
– Maarten Bodewes♦
1 hour ago