Can an RSA (private) modulus contain (hex) zeroes?
$begingroup$
Consider a private modulus that is supposedly 4096 bits. But since it contains two hex pairs of zeroes, i.e. ...:00:...:00... such that the bit length of the decimal value instead becomes 4093 bits, is it actually a 4096 bit key or not?
I hope my question is vaguely intelligible.
rsa
asked 3 hours ago
AdamAdam
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$endgroup$
add a comment |
$begingroup$
Consider a private modulus that is supposedly 4096 bits. But since it contains two hex pairs of zeroes, i.e. ...:00:...:00... such that the bit length of the decimal value instead becomes 4093 bits, is it actually a 4096 bit key or not?
I hope my question is vaguely intelligible.
rsa
asked 3 hours ago
AdamAdam
61
New contributor
Adam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Are you saying that the 2 most significant bytes are 0, or just there happen to be multiple zero substrings in places other than the most significant place? Also, why are your bringing up the decimal representation of the value?
$endgroup$
– Ken Goss
3 hours ago
$begingroup$
What's "the bit length of the decimal value" of an integer? How is 4093 obtained for 4096 and "two hex pairs of zeroes", that is 8 bits? Is "the private modulus" really the public modulus $N$ in an RSA public key $(N,e)$? If so; yes, $N$ may contain zeroes, in binary, hex, and decimal. By convention, the RSA key size is the bit size of $N$, that is the integer $k$ such that $2^{k-1}le N<2^k$. I may turn this into an answer if the question becomes clear.
$endgroup$
– fgrieu
4 mins ago
add a comment |
$begingroup$
Consider a private modulus that is supposedly 4096 bits. But since it contains two hex pairs of zeroes, i.e. ...:00:...:00... such that the bit length of the decimal value instead becomes 4093 bits, is it actually a 4096 bit key or not?
I hope my question is vaguely intelligible.
rsa
asked 3 hours ago
AdamAdam
61
New contributor
Adam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Consider a private modulus that is supposedly 4096 bits. But since it contains two hex pairs of zeroes, i.e. ...:00:...:00... such that the bit length of the decimal value instead becomes 4093 bits, is it actually a 4096 bit key or not?
I hope my question is vaguely intelligible.
rsa
rsa
asked 3 hours ago
AdamAdam
61
New contributor
Adam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
AdamAdam
61
New contributor
Adam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
AdamAdam
61
New contributor
Adam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
AdamAdam
61
asked 3 hours ago
AdamAdam
61
61
New contributor
Adam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Adam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
Are you saying that the 2 most significant bytes are 0, or just there happen to be multiple zero substrings in places other than the most significant place? Also, why are your bringing up the decimal representation of the value?
$endgroup$
– Ken Goss
3 hours ago
$begingroup$
What's "the bit length of the decimal value" of an integer? How is 4093 obtained for 4096 and "two hex pairs of zeroes", that is 8 bits? Is "the private modulus" really the public modulus $N$ in an RSA public key $(N,e)$? If so; yes, $N$ may contain zeroes, in binary, hex, and decimal. By convention, the RSA key size is the bit size of $N$, that is the integer $k$ such that $2^{k-1}le N<2^k$. I may turn this into an answer if the question becomes clear.
$endgroup$
– fgrieu
4 mins ago
add a comment |
$begingroup$
Are you saying that the 2 most significant bytes are 0, or just there happen to be multiple zero substrings in places other than the most significant place? Also, why are your bringing up the decimal representation of the value?
$endgroup$
– Ken Goss
3 hours ago
$begingroup$
What's "the bit length of the decimal value" of an integer? How is 4093 obtained for 4096 and "two hex pairs of zeroes", that is 8 bits? Is "the private modulus" really the public modulus $N$ in an RSA public key $(N,e)$? If so; yes, $N$ may contain zeroes, in binary, hex, and decimal. By convention, the RSA key size is the bit size of $N$, that is the integer $k$ such that $2^{k-1}le N<2^k$. I may turn this into an answer if the question becomes clear.
$endgroup$
– fgrieu
4 mins ago
$begingroup$
Are you saying that the 2 most significant bytes are 0, or just there happen to be multiple zero substrings in places other than the most significant place? Also, why are your bringing up the decimal representation of the value?
$endgroup$
– Ken Goss
3 hours ago
$begingroup$
Are you saying that the 2 most significant bytes are 0, or just there happen to be multiple zero substrings in places other than the most significant place? Also, why are your bringing up the decimal representation of the value?
$endgroup$
– Ken Goss
3 hours ago
$begingroup$
What's "the bit length of the decimal value" of an integer? How is 4093 obtained for 4096 and "two hex pairs of zeroes", that is 8 bits? Is "the private modulus" really the public modulus $N$ in an RSA public key $(N,e)$? If so; yes, $N$ may contain zeroes, in binary, hex, and decimal. By convention, the RSA key size is the bit size of $N$, that is the integer $k$ such that $2^{k-1}le N<2^k$. I may turn this into an answer if the question becomes clear.
$endgroup$
– fgrieu
4 mins ago
$begingroup$
What's "the bit length of the decimal value" of an integer? How is 4093 obtained for 4096 and "two hex pairs of zeroes", that is 8 bits? Is "the private modulus" really the public modulus $N$ in an RSA public key $(N,e)$? If so; yes, $N$ may contain zeroes, in binary, hex, and decimal. By convention, the RSA key size is the bit size of $N$, that is the integer $k$ such that $2^{k-1}le N<2^k$. I may turn this into an answer if the question becomes clear.
$endgroup$
– fgrieu
4 mins ago
add a comment |
1 Answer
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$begingroup$
There is no such thing as a "private modulus". There are the secret primes and the secret exponent, and there is the public modulus. Assuming you are talking about the primes or the secret exponent, are these bits in the most significant (leftmost) position, or are they somewhere in the middle or end? If they are in the most significant position, then they do not count towards the size of the integer. After all, any finite integer at all can be thought of as having an infinite number of leading zeros! If the zeros are somewhere within the integer, they serve to change its value. A random integer is expected to have a few null bytes.
Think about it this way: there is no difference between 0123 and 00123 (both are the same number), but there is a big ten-fold difference between 1230 and 12300, despite both adding a single zero. Of course this is base 10 whereas hex is base 16, but the principle is the same and zero means the same thing.
You can easily calculate the size of your key by taking the ceiling of the binary logarithm of the public modulus. That is, you can calculate $lceillog_2(N)rceil$ for an integer representation of the public modulus $N$.
answered 3 hours ago
forestforest
3,4901338
$endgroup$
$begingroup$
Can't sleep. You need to take the ceiling, so I can look at the open air.
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
Too cryptic? The size of the RSA key is the ceiling of the log :)
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
@MaartenBodewes Good point. Without taking the ceiling, it's not even an integer.
$endgroup$
– forest
2 hours ago
add a comment |
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$begingroup$
There is no such thing as a "private modulus". There are the secret primes and the secret exponent, and there is the public modulus. Assuming you are talking about the primes or the secret exponent, are these bits in the most significant (leftmost) position, or are they somewhere in the middle or end? If they are in the most significant position, then they do not count towards the size of the integer. After all, any finite integer at all can be thought of as having an infinite number of leading zeros! If the zeros are somewhere within the integer, they serve to change its value. A random integer is expected to have a few null bytes.
Think about it this way: there is no difference between 0123 and 00123 (both are the same number), but there is a big ten-fold difference between 1230 and 12300, despite both adding a single zero. Of course this is base 10 whereas hex is base 16, but the principle is the same and zero means the same thing.
You can easily calculate the size of your key by taking the ceiling of the binary logarithm of the public modulus. That is, you can calculate $lceillog_2(N)rceil$ for an integer representation of the public modulus $N$.
answered 3 hours ago
forestforest
3,4901338
$endgroup$
$begingroup$
Can't sleep. You need to take the ceiling, so I can look at the open air.
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
Too cryptic? The size of the RSA key is the ceiling of the log :)
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
@MaartenBodewes Good point. Without taking the ceiling, it's not even an integer.
$endgroup$
– forest
2 hours ago
add a comment |
$begingroup$
There is no such thing as a "private modulus". There are the secret primes and the secret exponent, and there is the public modulus. Assuming you are talking about the primes or the secret exponent, are these bits in the most significant (leftmost) position, or are they somewhere in the middle or end? If they are in the most significant position, then they do not count towards the size of the integer. After all, any finite integer at all can be thought of as having an infinite number of leading zeros! If the zeros are somewhere within the integer, they serve to change its value. A random integer is expected to have a few null bytes.
Think about it this way: there is no difference between 0123 and 00123 (both are the same number), but there is a big ten-fold difference between 1230 and 12300, despite both adding a single zero. Of course this is base 10 whereas hex is base 16, but the principle is the same and zero means the same thing.
You can easily calculate the size of your key by taking the ceiling of the binary logarithm of the public modulus. That is, you can calculate $lceillog_2(N)rceil$ for an integer representation of the public modulus $N$.
answered 3 hours ago
forestforest
3,4901338
$endgroup$
$begingroup$
Can't sleep. You need to take the ceiling, so I can look at the open air.
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
Too cryptic? The size of the RSA key is the ceiling of the log :)
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
@MaartenBodewes Good point. Without taking the ceiling, it's not even an integer.
$endgroup$
– forest
2 hours ago
add a comment |
$begingroup$
There is no such thing as a "private modulus". There are the secret primes and the secret exponent, and there is the public modulus. Assuming you are talking about the primes or the secret exponent, are these bits in the most significant (leftmost) position, or are they somewhere in the middle or end? If they are in the most significant position, then they do not count towards the size of the integer. After all, any finite integer at all can be thought of as having an infinite number of leading zeros! If the zeros are somewhere within the integer, they serve to change its value. A random integer is expected to have a few null bytes.
Think about it this way: there is no difference between 0123 and 00123 (both are the same number), but there is a big ten-fold difference between 1230 and 12300, despite both adding a single zero. Of course this is base 10 whereas hex is base 16, but the principle is the same and zero means the same thing.
You can easily calculate the size of your key by taking the ceiling of the binary logarithm of the public modulus. That is, you can calculate $lceillog_2(N)rceil$ for an integer representation of the public modulus $N$.
answered 3 hours ago
forestforest
3,4901338
$endgroup$
There is no such thing as a "private modulus". There are the secret primes and the secret exponent, and there is the public modulus. Assuming you are talking about the primes or the secret exponent, are these bits in the most significant (leftmost) position, or are they somewhere in the middle or end? If they are in the most significant position, then they do not count towards the size of the integer. After all, any finite integer at all can be thought of as having an infinite number of leading zeros! If the zeros are somewhere within the integer, they serve to change its value. A random integer is expected to have a few null bytes.
Think about it this way: there is no difference between 0123 and 00123 (both are the same number), but there is a big ten-fold difference between 1230 and 12300, despite both adding a single zero. Of course this is base 10 whereas hex is base 16, but the principle is the same and zero means the same thing.
You can easily calculate the size of your key by taking the ceiling of the binary logarithm of the public modulus. That is, you can calculate $lceillog_2(N)rceil$ for an integer representation of the public modulus $N$.
answered 3 hours ago
forestforest
3,4901338
edited 1 hour ago
answered 3 hours ago
forestforest
3,4901338
answered 3 hours ago
forestforest
3,4901338
answered 3 hours ago
forestforest
3,4901338
3,4901338
$begingroup$
Can't sleep. You need to take the ceiling, so I can look at the open air.
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
Too cryptic? The size of the RSA key is the ceiling of the log :)
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
@MaartenBodewes Good point. Without taking the ceiling, it's not even an integer.
$endgroup$
– forest
2 hours ago
add a comment |
$begingroup$
Can't sleep. You need to take the ceiling, so I can look at the open air.
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
Too cryptic? The size of the RSA key is the ceiling of the log :)
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
@MaartenBodewes Good point. Without taking the ceiling, it's not even an integer.
$endgroup$
– forest
2 hours ago
$begingroup$
Can't sleep. You need to take the ceiling, so I can look at the open air.
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
Can't sleep. You need to take the ceiling, so I can look at the open air.
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
Too cryptic? The size of the RSA key is the ceiling of the log :)
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
Too cryptic? The size of the RSA key is the ceiling of the log :)
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
@MaartenBodewes Good point. Without taking the ceiling, it's not even an integer.
$endgroup$
– forest
2 hours ago
$begingroup$
@MaartenBodewes Good point. Without taking the ceiling, it's not even an integer.
$endgroup$
– forest
2 hours ago
add a comment |
Adam is a new contributor. Be nice, and check out our Code of Conduct.
Adam is a new contributor. Be nice, and check out our Code of Conduct.
Adam is a new contributor. Be nice, and check out our Code of Conduct.
Adam is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Are you saying that the 2 most significant bytes are 0, or just there happen to be multiple zero substrings in places other than the most significant place? Also, why are your bringing up the decimal representation of the value?
$endgroup$
– Ken Goss
3 hours ago
$begingroup$
What's "the bit length of the decimal value" of an integer? How is 4093 obtained for 4096 and "two hex pairs of zeroes", that is 8 bits? Is "the private modulus" really the public modulus $N$ in an RSA public key $(N,e)$? If so; yes, $N$ may contain zeroes, in binary, hex, and decimal. By convention, the RSA key size is the bit size of $N$, that is the integer $k$ such that $2^{k-1}le N<2^k$. I may turn this into an answer if the question becomes clear.
$endgroup$
– fgrieu
4 mins ago