Mixing
How does PKCS 1.5 solve the insecureness of Textbook RSA?
$begingroup$
As we know, Textbook RSA is not enough safe to use because of the following issues;
Guessable plaintext problem
Small $e$ issue
I want to know how PKCS 1.5 solved these problems.
encryption rsa padding
edited Jan 24 at 8:59
kelalaka
8,43822351
asked Jan 24 at 8:04
DavisDavis
484
$endgroup$
add a comment |
$begingroup$
As we know, Textbook RSA is not enough safe to use because of the following issues;
Guessable plaintext problem
Small $e$ issue
I want to know how PKCS 1.5 solved these problems.
encryption rsa padding
edited Jan 24 at 8:59
kelalaka
8,43822351
asked Jan 24 at 8:04
DavisDavis
484
$endgroup$
add a comment |
$begingroup$
As we know, Textbook RSA is not enough safe to use because of the following issues;
Guessable plaintext problem
Small $e$ issue
I want to know how PKCS 1.5 solved these problems.
encryption rsa padding
edited Jan 24 at 8:59
kelalaka
8,43822351
asked Jan 24 at 8:04
DavisDavis
484
$endgroup$
As we know, Textbook RSA is not enough safe to use because of the following issues;
Guessable plaintext problem
Small $e$ issue
I want to know how PKCS 1.5 solved these problems.
encryption rsa padding
encryption rsa padding
edited Jan 24 at 8:59
kelalaka
8,43822351
asked Jan 24 at 8:04
DavisDavis
484
edited Jan 24 at 8:59
kelalaka
8,43822351
asked Jan 24 at 8:04
DavisDavis
484
edited Jan 24 at 8:59
kelalaka
8,43822351
edited Jan 24 at 8:59
kelalaka
8,43822351
edited Jan 24 at 8:59
kelalaka
8,43822351
8,43822351
asked Jan 24 at 8:04
DavisDavis
484
asked Jan 24 at 8:04
DavisDavis
484
asked Jan 24 at 8:04
DavisDavis
484
484
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
PKCS#1 v1.5 describes a method (formally known as RSAES-PKCS1-v1_5) that turns textbook RSA into a (heuristically) secure encryption scheme for small messages (PKCS#1 v1.5 also describes a signature scheme, which the question and this answer do not consider).
For a $k$-byte ($8k-7$ to $8k$-bit) public modulus part of public key $(N,e)$, the message to be encrypted $M$ is a $mathrm{mLen}$-byte bytestring with $mathrm{mLen}le k-11$. Message $M$ is turned into
$mathrm{EM}=mathtt{0x00}mathbin|mathtt{0x02}mathbin|mathrm{PS}mathbin|mathtt{0x00}mathbin|M$, where $mathrm{PS}$ is drawn as a $k-mathrm{mLen}-3$-byte bytestring consisting of fresh near-uniformly-random non-zero bytes. Then (with big-endian conversion between bytestring and integer left implicit) the ciphertext is $Cgetsmathrm{EM}^ebmod N$. That's textbook RSA encryption of $mathrm{EM}$.
While plaintext $M$ might still be easily guessable, verifying such guess becomes much harder. A brute-force method becomes: try all the possible $mathrm{PS}$ and compute the corresponding $C$, until one matches. That strategy requires $255^{k-mathrm{mLen}-3}/2$ try on average (for each possible $M$). That's next to $2^{63}$ at least, and growing by a factor of $255$ for each byte of message capacity that we remove. That's believed to solve the guessable plaintext problem; but we have no formal proof that it does.
Thanks to $mathtt{0x02}$ at the second byte of $mathrm{EM}$, it holds that $mathrm{EM}>2^{8k-15}$. It follows that for $ege3$, $mathrm{EM}^egg N^2$ for all practical $k$, and that's more than enough to solve a number of small $e$ issues :
- any known extension of the $e^text{th}$ root attack (in this attack against textbook RSA, where $Cgets M^ebmod N$, it is used that for $M<N^{1/e}$, we can compute back $M$ from $C$ as $Mgetssqrt[e]C$ )
Coppersmith's Short Pad Attack (which extends the Franklin-Reiter related messages attack to random padding).
RSAES-PKCS1-v1_5 also practically thwarts Hastad's broadcast attack. In this attack against textbook RSA, sending the same message to at least $e$ recipients allows decryption. The random $mathrm{PS}$ makes $mathrm{EM}$ for the same $M$ potentially different. Even for $e=3$, $mathrm{PS}$ at the 8-byte minimum, and $2^{36}$ recipients of the same message $M$, probability that $e$ padded messages $mathrm{EM}$ are identical is lower than one in a million, if I got the math right.
However, many RSAES-PKCS1-v1_5 implementations are vulnerable when used with small $e$, such as $e=3$. That's because implementations often check that the first two bytes of $C^dbmod N$ are $mathtt{0x00}mathbin|mathtt{0x02}$, and/or parse what follows in order to find the first $mathtt{0x00}$, indicating the start of $M$ and allowing recovery of $mathrm{mLen}$. When the result of such check is made available in a way leaking where the failure occurred (by a detailed error code, or timing, or some other side channel) to adversaries able to submit cryptograms for decryption, then Bleichenbacher's padding oracle attack applies. It turns a moderate number of queries to a decrypting entity into an RSA private-key operation for arbitrary argument, allowing decryption of one message, or computing one signature if the same key is used for encryption and signature.
That's not a fatality, though: RSAES-PKCS1-v1_5 padding can be checked in constant time, with a single error code regardless of the particular failure. Another possible strategy is to not check the left 10 bytes; or when $mathrm{mLen}$ is known in advance, make no padding check and simply extract $M$ as the rightmost $mathrm{mLen}$ bytes of $C^dbmod N$.
answered Jan 24 at 10:49
fgrieufgrieu
81.6k7175347
$endgroup$
add a comment |
$begingroup$
The Structure of PKCS#1 v1.5 as follows;
The message $m$ is padded to
x = 0x00 || 0x02 || r || 0x00 || m
and the ciphertext calculated as $c=x^ebmod N$ not by $m^ebmod N$, where $r$ is a random string.
Cube root attack cannot be applied since the padding guarantees that messages are not short.
The random $r$ make the encryption probabilistic so that guessing the message $x$ has a very small probability. Two encryptions of a message $$operatorname{Enc}_{k_{pub}}(m_1) neq operatorname{Enc}_{k_{pub}}(m_1)$$ due to randomization.
- Also, the padding scheme prohibits the malleability property of RSA, that is
$$ operatorname{Enc}_{k_{pub}}(2) cdot c = operatorname{Enc}_{k_{pub}}(2) cdot operatorname{Enc}_{k_{pub}}(m) = operatorname{Enc}_{k_{pub}}(2cdot m)$$
answered Jan 24 at 8:28
kelalakakelalaka
8,43822351
$endgroup$
2
$begingroup$
It is guessing the padded messagexthat has a small probability. Independently: there's more than a single small e issue, only one is covered by this answer.
$endgroup$
– fgrieu
Jan 24 at 11:02
add a comment |
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2 Answers
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2 Answers
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$begingroup$
PKCS#1 v1.5 describes a method (formally known as RSAES-PKCS1-v1_5) that turns textbook RSA into a (heuristically) secure encryption scheme for small messages (PKCS#1 v1.5 also describes a signature scheme, which the question and this answer do not consider).
For a $k$-byte ($8k-7$ to $8k$-bit) public modulus part of public key $(N,e)$, the message to be encrypted $M$ is a $mathrm{mLen}$-byte bytestring with $mathrm{mLen}le k-11$. Message $M$ is turned into
$mathrm{EM}=mathtt{0x00}mathbin|mathtt{0x02}mathbin|mathrm{PS}mathbin|mathtt{0x00}mathbin|M$, where $mathrm{PS}$ is drawn as a $k-mathrm{mLen}-3$-byte bytestring consisting of fresh near-uniformly-random non-zero bytes. Then (with big-endian conversion between bytestring and integer left implicit) the ciphertext is $Cgetsmathrm{EM}^ebmod N$. That's textbook RSA encryption of $mathrm{EM}$.
While plaintext $M$ might still be easily guessable, verifying such guess becomes much harder. A brute-force method becomes: try all the possible $mathrm{PS}$ and compute the corresponding $C$, until one matches. That strategy requires $255^{k-mathrm{mLen}-3}/2$ try on average (for each possible $M$). That's next to $2^{63}$ at least, and growing by a factor of $255$ for each byte of message capacity that we remove. That's believed to solve the guessable plaintext problem; but we have no formal proof that it does.
Thanks to $mathtt{0x02}$ at the second byte of $mathrm{EM}$, it holds that $mathrm{EM}>2^{8k-15}$. It follows that for $ege3$, $mathrm{EM}^egg N^2$ for all practical $k$, and that's more than enough to solve a number of small $e$ issues :
- any known extension of the $e^text{th}$ root attack (in this attack against textbook RSA, where $Cgets M^ebmod N$, it is used that for $M<N^{1/e}$, we can compute back $M$ from $C$ as $Mgetssqrt[e]C$ )
Coppersmith's Short Pad Attack (which extends the Franklin-Reiter related messages attack to random padding).
RSAES-PKCS1-v1_5 also practically thwarts Hastad's broadcast attack. In this attack against textbook RSA, sending the same message to at least $e$ recipients allows decryption. The random $mathrm{PS}$ makes $mathrm{EM}$ for the same $M$ potentially different. Even for $e=3$, $mathrm{PS}$ at the 8-byte minimum, and $2^{36}$ recipients of the same message $M$, probability that $e$ padded messages $mathrm{EM}$ are identical is lower than one in a million, if I got the math right.
However, many RSAES-PKCS1-v1_5 implementations are vulnerable when used with small $e$, such as $e=3$. That's because implementations often check that the first two bytes of $C^dbmod N$ are $mathtt{0x00}mathbin|mathtt{0x02}$, and/or parse what follows in order to find the first $mathtt{0x00}$, indicating the start of $M$ and allowing recovery of $mathrm{mLen}$. When the result of such check is made available in a way leaking where the failure occurred (by a detailed error code, or timing, or some other side channel) to adversaries able to submit cryptograms for decryption, then Bleichenbacher's padding oracle attack applies. It turns a moderate number of queries to a decrypting entity into an RSA private-key operation for arbitrary argument, allowing decryption of one message, or computing one signature if the same key is used for encryption and signature.
That's not a fatality, though: RSAES-PKCS1-v1_5 padding can be checked in constant time, with a single error code regardless of the particular failure. Another possible strategy is to not check the left 10 bytes; or when $mathrm{mLen}$ is known in advance, make no padding check and simply extract $M$ as the rightmost $mathrm{mLen}$ bytes of $C^dbmod N$.
answered Jan 24 at 10:49
fgrieufgrieu
81.6k7175347
$endgroup$
add a comment |
$begingroup$
PKCS#1 v1.5 describes a method (formally known as RSAES-PKCS1-v1_5) that turns textbook RSA into a (heuristically) secure encryption scheme for small messages (PKCS#1 v1.5 also describes a signature scheme, which the question and this answer do not consider).
For a $k$-byte ($8k-7$ to $8k$-bit) public modulus part of public key $(N,e)$, the message to be encrypted $M$ is a $mathrm{mLen}$-byte bytestring with $mathrm{mLen}le k-11$. Message $M$ is turned into
$mathrm{EM}=mathtt{0x00}mathbin|mathtt{0x02}mathbin|mathrm{PS}mathbin|mathtt{0x00}mathbin|M$, where $mathrm{PS}$ is drawn as a $k-mathrm{mLen}-3$-byte bytestring consisting of fresh near-uniformly-random non-zero bytes. Then (with big-endian conversion between bytestring and integer left implicit) the ciphertext is $Cgetsmathrm{EM}^ebmod N$. That's textbook RSA encryption of $mathrm{EM}$.
While plaintext $M$ might still be easily guessable, verifying such guess becomes much harder. A brute-force method becomes: try all the possible $mathrm{PS}$ and compute the corresponding $C$, until one matches. That strategy requires $255^{k-mathrm{mLen}-3}/2$ try on average (for each possible $M$). That's next to $2^{63}$ at least, and growing by a factor of $255$ for each byte of message capacity that we remove. That's believed to solve the guessable plaintext problem; but we have no formal proof that it does.
Thanks to $mathtt{0x02}$ at the second byte of $mathrm{EM}$, it holds that $mathrm{EM}>2^{8k-15}$. It follows that for $ege3$, $mathrm{EM}^egg N^2$ for all practical $k$, and that's more than enough to solve a number of small $e$ issues :
- any known extension of the $e^text{th}$ root attack (in this attack against textbook RSA, where $Cgets M^ebmod N$, it is used that for $M<N^{1/e}$, we can compute back $M$ from $C$ as $Mgetssqrt[e]C$ )
Coppersmith's Short Pad Attack (which extends the Franklin-Reiter related messages attack to random padding).
RSAES-PKCS1-v1_5 also practically thwarts Hastad's broadcast attack. In this attack against textbook RSA, sending the same message to at least $e$ recipients allows decryption. The random $mathrm{PS}$ makes $mathrm{EM}$ for the same $M$ potentially different. Even for $e=3$, $mathrm{PS}$ at the 8-byte minimum, and $2^{36}$ recipients of the same message $M$, probability that $e$ padded messages $mathrm{EM}$ are identical is lower than one in a million, if I got the math right.
However, many RSAES-PKCS1-v1_5 implementations are vulnerable when used with small $e$, such as $e=3$. That's because implementations often check that the first two bytes of $C^dbmod N$ are $mathtt{0x00}mathbin|mathtt{0x02}$, and/or parse what follows in order to find the first $mathtt{0x00}$, indicating the start of $M$ and allowing recovery of $mathrm{mLen}$. When the result of such check is made available in a way leaking where the failure occurred (by a detailed error code, or timing, or some other side channel) to adversaries able to submit cryptograms for decryption, then Bleichenbacher's padding oracle attack applies. It turns a moderate number of queries to a decrypting entity into an RSA private-key operation for arbitrary argument, allowing decryption of one message, or computing one signature if the same key is used for encryption and signature.
That's not a fatality, though: RSAES-PKCS1-v1_5 padding can be checked in constant time, with a single error code regardless of the particular failure. Another possible strategy is to not check the left 10 bytes; or when $mathrm{mLen}$ is known in advance, make no padding check and simply extract $M$ as the rightmost $mathrm{mLen}$ bytes of $C^dbmod N$.
answered Jan 24 at 10:49
fgrieufgrieu
81.6k7175347
$endgroup$
add a comment |
$begingroup$
PKCS#1 v1.5 describes a method (formally known as RSAES-PKCS1-v1_5) that turns textbook RSA into a (heuristically) secure encryption scheme for small messages (PKCS#1 v1.5 also describes a signature scheme, which the question and this answer do not consider).
For a $k$-byte ($8k-7$ to $8k$-bit) public modulus part of public key $(N,e)$, the message to be encrypted $M$ is a $mathrm{mLen}$-byte bytestring with $mathrm{mLen}le k-11$. Message $M$ is turned into
$mathrm{EM}=mathtt{0x00}mathbin|mathtt{0x02}mathbin|mathrm{PS}mathbin|mathtt{0x00}mathbin|M$, where $mathrm{PS}$ is drawn as a $k-mathrm{mLen}-3$-byte bytestring consisting of fresh near-uniformly-random non-zero bytes. Then (with big-endian conversion between bytestring and integer left implicit) the ciphertext is $Cgetsmathrm{EM}^ebmod N$. That's textbook RSA encryption of $mathrm{EM}$.
While plaintext $M$ might still be easily guessable, verifying such guess becomes much harder. A brute-force method becomes: try all the possible $mathrm{PS}$ and compute the corresponding $C$, until one matches. That strategy requires $255^{k-mathrm{mLen}-3}/2$ try on average (for each possible $M$). That's next to $2^{63}$ at least, and growing by a factor of $255$ for each byte of message capacity that we remove. That's believed to solve the guessable plaintext problem; but we have no formal proof that it does.
Thanks to $mathtt{0x02}$ at the second byte of $mathrm{EM}$, it holds that $mathrm{EM}>2^{8k-15}$. It follows that for $ege3$, $mathrm{EM}^egg N^2$ for all practical $k$, and that's more than enough to solve a number of small $e$ issues :
- any known extension of the $e^text{th}$ root attack (in this attack against textbook RSA, where $Cgets M^ebmod N$, it is used that for $M<N^{1/e}$, we can compute back $M$ from $C$ as $Mgetssqrt[e]C$ )
Coppersmith's Short Pad Attack (which extends the Franklin-Reiter related messages attack to random padding).
RSAES-PKCS1-v1_5 also practically thwarts Hastad's broadcast attack. In this attack against textbook RSA, sending the same message to at least $e$ recipients allows decryption. The random $mathrm{PS}$ makes $mathrm{EM}$ for the same $M$ potentially different. Even for $e=3$, $mathrm{PS}$ at the 8-byte minimum, and $2^{36}$ recipients of the same message $M$, probability that $e$ padded messages $mathrm{EM}$ are identical is lower than one in a million, if I got the math right.
However, many RSAES-PKCS1-v1_5 implementations are vulnerable when used with small $e$, such as $e=3$. That's because implementations often check that the first two bytes of $C^dbmod N$ are $mathtt{0x00}mathbin|mathtt{0x02}$, and/or parse what follows in order to find the first $mathtt{0x00}$, indicating the start of $M$ and allowing recovery of $mathrm{mLen}$. When the result of such check is made available in a way leaking where the failure occurred (by a detailed error code, or timing, or some other side channel) to adversaries able to submit cryptograms for decryption, then Bleichenbacher's padding oracle attack applies. It turns a moderate number of queries to a decrypting entity into an RSA private-key operation for arbitrary argument, allowing decryption of one message, or computing one signature if the same key is used for encryption and signature.
That's not a fatality, though: RSAES-PKCS1-v1_5 padding can be checked in constant time, with a single error code regardless of the particular failure. Another possible strategy is to not check the left 10 bytes; or when $mathrm{mLen}$ is known in advance, make no padding check and simply extract $M$ as the rightmost $mathrm{mLen}$ bytes of $C^dbmod N$.
answered Jan 24 at 10:49
fgrieufgrieu
81.6k7175347
$endgroup$
PKCS#1 v1.5 describes a method (formally known as RSAES-PKCS1-v1_5) that turns textbook RSA into a (heuristically) secure encryption scheme for small messages (PKCS#1 v1.5 also describes a signature scheme, which the question and this answer do not consider).
For a $k$-byte ($8k-7$ to $8k$-bit) public modulus part of public key $(N,e)$, the message to be encrypted $M$ is a $mathrm{mLen}$-byte bytestring with $mathrm{mLen}le k-11$. Message $M$ is turned into
$mathrm{EM}=mathtt{0x00}mathbin|mathtt{0x02}mathbin|mathrm{PS}mathbin|mathtt{0x00}mathbin|M$, where $mathrm{PS}$ is drawn as a $k-mathrm{mLen}-3$-byte bytestring consisting of fresh near-uniformly-random non-zero bytes. Then (with big-endian conversion between bytestring and integer left implicit) the ciphertext is $Cgetsmathrm{EM}^ebmod N$. That's textbook RSA encryption of $mathrm{EM}$.
While plaintext $M$ might still be easily guessable, verifying such guess becomes much harder. A brute-force method becomes: try all the possible $mathrm{PS}$ and compute the corresponding $C$, until one matches. That strategy requires $255^{k-mathrm{mLen}-3}/2$ try on average (for each possible $M$). That's next to $2^{63}$ at least, and growing by a factor of $255$ for each byte of message capacity that we remove. That's believed to solve the guessable plaintext problem; but we have no formal proof that it does.
Thanks to $mathtt{0x02}$ at the second byte of $mathrm{EM}$, it holds that $mathrm{EM}>2^{8k-15}$. It follows that for $ege3$, $mathrm{EM}^egg N^2$ for all practical $k$, and that's more than enough to solve a number of small $e$ issues :
- any known extension of the $e^text{th}$ root attack (in this attack against textbook RSA, where $Cgets M^ebmod N$, it is used that for $M<N^{1/e}$, we can compute back $M$ from $C$ as $Mgetssqrt[e]C$ )
Coppersmith's Short Pad Attack (which extends the Franklin-Reiter related messages attack to random padding).
RSAES-PKCS1-v1_5 also practically thwarts Hastad's broadcast attack. In this attack against textbook RSA, sending the same message to at least $e$ recipients allows decryption. The random $mathrm{PS}$ makes $mathrm{EM}$ for the same $M$ potentially different. Even for $e=3$, $mathrm{PS}$ at the 8-byte minimum, and $2^{36}$ recipients of the same message $M$, probability that $e$ padded messages $mathrm{EM}$ are identical is lower than one in a million, if I got the math right.
However, many RSAES-PKCS1-v1_5 implementations are vulnerable when used with small $e$, such as $e=3$. That's because implementations often check that the first two bytes of $C^dbmod N$ are $mathtt{0x00}mathbin|mathtt{0x02}$, and/or parse what follows in order to find the first $mathtt{0x00}$, indicating the start of $M$ and allowing recovery of $mathrm{mLen}$. When the result of such check is made available in a way leaking where the failure occurred (by a detailed error code, or timing, or some other side channel) to adversaries able to submit cryptograms for decryption, then Bleichenbacher's padding oracle attack applies. It turns a moderate number of queries to a decrypting entity into an RSA private-key operation for arbitrary argument, allowing decryption of one message, or computing one signature if the same key is used for encryption and signature.
That's not a fatality, though: RSAES-PKCS1-v1_5 padding can be checked in constant time, with a single error code regardless of the particular failure. Another possible strategy is to not check the left 10 bytes; or when $mathrm{mLen}$ is known in advance, make no padding check and simply extract $M$ as the rightmost $mathrm{mLen}$ bytes of $C^dbmod N$.
answered Jan 24 at 10:49
fgrieufgrieu
81.6k7175347
edited Jan 24 at 14:37
answered Jan 24 at 10:49
fgrieufgrieu
81.6k7175347
answered Jan 24 at 10:49
fgrieufgrieu
81.6k7175347
answered Jan 24 at 10:49
fgrieufgrieu
81.6k7175347
81.6k7175347
add a comment |
add a comment |
$begingroup$
The Structure of PKCS#1 v1.5 as follows;
The message $m$ is padded to
x = 0x00 || 0x02 || r || 0x00 || m
and the ciphertext calculated as $c=x^ebmod N$ not by $m^ebmod N$, where $r$ is a random string.
Cube root attack cannot be applied since the padding guarantees that messages are not short.
The random $r$ make the encryption probabilistic so that guessing the message $x$ has a very small probability. Two encryptions of a message $$operatorname{Enc}_{k_{pub}}(m_1) neq operatorname{Enc}_{k_{pub}}(m_1)$$ due to randomization.
- Also, the padding scheme prohibits the malleability property of RSA, that is
$$ operatorname{Enc}_{k_{pub}}(2) cdot c = operatorname{Enc}_{k_{pub}}(2) cdot operatorname{Enc}_{k_{pub}}(m) = operatorname{Enc}_{k_{pub}}(2cdot m)$$
answered Jan 24 at 8:28
kelalakakelalaka
8,43822351
$endgroup$
2
$begingroup$
It is guessing the padded messagexthat has a small probability. Independently: there's more than a single small e issue, only one is covered by this answer.
$endgroup$
– fgrieu
Jan 24 at 11:02
add a comment |
$begingroup$
The Structure of PKCS#1 v1.5 as follows;
The message $m$ is padded to
x = 0x00 || 0x02 || r || 0x00 || m
and the ciphertext calculated as $c=x^ebmod N$ not by $m^ebmod N$, where $r$ is a random string.
Cube root attack cannot be applied since the padding guarantees that messages are not short.
The random $r$ make the encryption probabilistic so that guessing the message $x$ has a very small probability. Two encryptions of a message $$operatorname{Enc}_{k_{pub}}(m_1) neq operatorname{Enc}_{k_{pub}}(m_1)$$ due to randomization.
- Also, the padding scheme prohibits the malleability property of RSA, that is
$$ operatorname{Enc}_{k_{pub}}(2) cdot c = operatorname{Enc}_{k_{pub}}(2) cdot operatorname{Enc}_{k_{pub}}(m) = operatorname{Enc}_{k_{pub}}(2cdot m)$$
answered Jan 24 at 8:28
kelalakakelalaka
8,43822351
$endgroup$
2
$begingroup$
It is guessing the padded messagexthat has a small probability. Independently: there's more than a single small e issue, only one is covered by this answer.
$endgroup$
– fgrieu
Jan 24 at 11:02
add a comment |
$begingroup$
The Structure of PKCS#1 v1.5 as follows;
The message $m$ is padded to
x = 0x00 || 0x02 || r || 0x00 || m
and the ciphertext calculated as $c=x^ebmod N$ not by $m^ebmod N$, where $r$ is a random string.
Cube root attack cannot be applied since the padding guarantees that messages are not short.
The random $r$ make the encryption probabilistic so that guessing the message $x$ has a very small probability. Two encryptions of a message $$operatorname{Enc}_{k_{pub}}(m_1) neq operatorname{Enc}_{k_{pub}}(m_1)$$ due to randomization.
- Also, the padding scheme prohibits the malleability property of RSA, that is
$$ operatorname{Enc}_{k_{pub}}(2) cdot c = operatorname{Enc}_{k_{pub}}(2) cdot operatorname{Enc}_{k_{pub}}(m) = operatorname{Enc}_{k_{pub}}(2cdot m)$$
answered Jan 24 at 8:28
kelalakakelalaka
8,43822351
$endgroup$
The Structure of PKCS#1 v1.5 as follows;
The message $m$ is padded to
x = 0x00 || 0x02 || r || 0x00 || m
and the ciphertext calculated as $c=x^ebmod N$ not by $m^ebmod N$, where $r$ is a random string.
Cube root attack cannot be applied since the padding guarantees that messages are not short.
The random $r$ make the encryption probabilistic so that guessing the message $x$ has a very small probability. Two encryptions of a message $$operatorname{Enc}_{k_{pub}}(m_1) neq operatorname{Enc}_{k_{pub}}(m_1)$$ due to randomization.
- Also, the padding scheme prohibits the malleability property of RSA, that is
$$ operatorname{Enc}_{k_{pub}}(2) cdot c = operatorname{Enc}_{k_{pub}}(2) cdot operatorname{Enc}_{k_{pub}}(m) = operatorname{Enc}_{k_{pub}}(2cdot m)$$
answered Jan 24 at 8:28
kelalakakelalaka
8,43822351
edited Jan 24 at 11:05
answered Jan 24 at 8:28
kelalakakelalaka
8,43822351
answered Jan 24 at 8:28
kelalakakelalaka
8,43822351
answered Jan 24 at 8:28
kelalakakelalaka
8,43822351
8,43822351
2
$begingroup$
It is guessing the padded messagexthat has a small probability. Independently: there's more than a single small e issue, only one is covered by this answer.
$endgroup$
– fgrieu
Jan 24 at 11:02
add a comment |
2
$begingroup$
It is guessing the padded messagexthat has a small probability. Independently: there's more than a single small e issue, only one is covered by this answer.
$endgroup$
– fgrieu
Jan 24 at 11:02
2
2
$begingroup$
It is guessing the padded message
x that has a small probability. Independently: there's more than a single small e issue, only one is covered by this answer.$endgroup$
– fgrieu
Jan 24 at 11:02
$begingroup$
It is guessing the padded message
x that has a small probability. Independently: there's more than a single small e issue, only one is covered by this answer.$endgroup$
– fgrieu
Jan 24 at 11:02
add a comment |
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