How do you find log of a Z* number?
0
$begingroup$
I understand modular arithmetic on a base level.
I was wondering how can one find log(base 5)20 in Z * base23.
I'm very confused about what Z* represents as I've been trying to figure it out for hours. Is it just a mod of a number? How would you compute it by hand? It is very hard to find any information on that. If you can possible give a simple example with a whole number, because I have no idea how you can calculate log in Z*.
modular-arithmetic cryptography
asked Dec 8 '18 at 2:16
CuriousCurious
11
$endgroup$
|
show 3 more comments
0
$begingroup$
I understand modular arithmetic on a base level.
I was wondering how can one find log(base 5)20 in Z * base23.
I'm very confused about what Z* represents as I've been trying to figure it out for hours. Is it just a mod of a number? How would you compute it by hand? It is very hard to find any information on that. If you can possible give a simple example with a whole number, because I have no idea how you can calculate log in Z*.
modular-arithmetic cryptography
asked Dec 8 '18 at 2:16
CuriousCurious
11
$endgroup$
2
$begingroup$
I think you are asking about the discrete logarithm: en.wikipedia.org/wiki/Discrete_logarithm In general, it is hard to compute, which is why it's important in cryptography.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 2:20
4
$begingroup$
My best guess is that Z*base23 was intended to mean $(mathbb Z/23)^*$, i.e., the multiplicative group of non-zero elements of the 23-element field. So the log in the question would mean the $n$ (unique modulo 22) such that $5^nequiv20pmod23)$. Although @EthanBolker is right about this being hard to compute in general, this particular problem could be solved by just trying one $n$ after another.
$endgroup$
– Andreas Blass
Dec 8 '18 at 2:44
$begingroup$
@AndreasBlass Excellent job of decryption!
$endgroup$
– saulspatz
Dec 8 '18 at 2:57
$begingroup$
@saulspatz Thanks, but I see (now that it's too late to edit my comment) that I didn't do a very good job of including necessary braces in TeX. That congruence should be $5^nequiv20pmod{23}$.
$endgroup$
– Andreas Blass
Dec 8 '18 at 3:01
$begingroup$
@AndreasBlass is 5^n ≡ 20 (mod23) the answer? I'm sorry about the format, I'm new to the website and still struggle to do it properly. The Z I was mentioning is exactly the one you showed. Is there any helpful resources besides Wikipedia that can help me understand what is going on? At least you helped me clarify that Z is basically the mod, but I'm still very confused about how I can compute this by hand.
$endgroup$
– Curious
Dec 8 '18 at 3:06
|
show 3 more comments
0
0
0
$begingroup$
I understand modular arithmetic on a base level.
I was wondering how can one find log(base 5)20 in Z * base23.
I'm very confused about what Z* represents as I've been trying to figure it out for hours. Is it just a mod of a number? How would you compute it by hand? It is very hard to find any information on that. If you can possible give a simple example with a whole number, because I have no idea how you can calculate log in Z*.
modular-arithmetic cryptography
asked Dec 8 '18 at 2:16
CuriousCurious
11
$endgroup$
I understand modular arithmetic on a base level.
I was wondering how can one find log(base 5)20 in Z * base23.
I'm very confused about what Z* represents as I've been trying to figure it out for hours. Is it just a mod of a number? How would you compute it by hand? It is very hard to find any information on that. If you can possible give a simple example with a whole number, because I have no idea how you can calculate log in Z*.
modular-arithmetic cryptography
modular-arithmetic cryptography
asked Dec 8 '18 at 2:16
CuriousCurious
11
asked Dec 8 '18 at 2:16
CuriousCurious
11
asked Dec 8 '18 at 2:16
CuriousCurious
11
asked Dec 8 '18 at 2:16
CuriousCurious
11
asked Dec 8 '18 at 2:16
CuriousCurious
11
11
2
$begingroup$
I think you are asking about the discrete logarithm: en.wikipedia.org/wiki/Discrete_logarithm In general, it is hard to compute, which is why it's important in cryptography.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 2:20
4
$begingroup$
My best guess is that Z*base23 was intended to mean $(mathbb Z/23)^*$, i.e., the multiplicative group of non-zero elements of the 23-element field. So the log in the question would mean the $n$ (unique modulo 22) such that $5^nequiv20pmod23)$. Although @EthanBolker is right about this being hard to compute in general, this particular problem could be solved by just trying one $n$ after another.
$endgroup$
– Andreas Blass
Dec 8 '18 at 2:44
$begingroup$
@AndreasBlass Excellent job of decryption!
$endgroup$
– saulspatz
Dec 8 '18 at 2:57
$begingroup$
@saulspatz Thanks, but I see (now that it's too late to edit my comment) that I didn't do a very good job of including necessary braces in TeX. That congruence should be $5^nequiv20pmod{23}$.
$endgroup$
– Andreas Blass
Dec 8 '18 at 3:01
$begingroup$
@AndreasBlass is 5^n ≡ 20 (mod23) the answer? I'm sorry about the format, I'm new to the website and still struggle to do it properly. The Z I was mentioning is exactly the one you showed. Is there any helpful resources besides Wikipedia that can help me understand what is going on? At least you helped me clarify that Z is basically the mod, but I'm still very confused about how I can compute this by hand.
$endgroup$
– Curious
Dec 8 '18 at 3:06
|
show 3 more comments
2
$begingroup$
I think you are asking about the discrete logarithm: en.wikipedia.org/wiki/Discrete_logarithm In general, it is hard to compute, which is why it's important in cryptography.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 2:20
4
$begingroup$
My best guess is that Z*base23 was intended to mean $(mathbb Z/23)^*$, i.e., the multiplicative group of non-zero elements of the 23-element field. So the log in the question would mean the $n$ (unique modulo 22) such that $5^nequiv20pmod23)$. Although @EthanBolker is right about this being hard to compute in general, this particular problem could be solved by just trying one $n$ after another.
$endgroup$
– Andreas Blass
Dec 8 '18 at 2:44
$begingroup$
@AndreasBlass Excellent job of decryption!
$endgroup$
– saulspatz
Dec 8 '18 at 2:57
$begingroup$
@saulspatz Thanks, but I see (now that it's too late to edit my comment) that I didn't do a very good job of including necessary braces in TeX. That congruence should be $5^nequiv20pmod{23}$.
$endgroup$
– Andreas Blass
Dec 8 '18 at 3:01
$begingroup$
@AndreasBlass is 5^n ≡ 20 (mod23) the answer? I'm sorry about the format, I'm new to the website and still struggle to do it properly. The Z I was mentioning is exactly the one you showed. Is there any helpful resources besides Wikipedia that can help me understand what is going on? At least you helped me clarify that Z is basically the mod, but I'm still very confused about how I can compute this by hand.
$endgroup$
– Curious
Dec 8 '18 at 3:06
2
2
$begingroup$
I think you are asking about the discrete logarithm: en.wikipedia.org/wiki/Discrete_logarithm In general, it is hard to compute, which is why it's important in cryptography.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 2:20
$begingroup$
I think you are asking about the discrete logarithm: en.wikipedia.org/wiki/Discrete_logarithm In general, it is hard to compute, which is why it's important in cryptography.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 2:20
4
4
$begingroup$
My best guess is that Z*base23 was intended to mean $(mathbb Z/23)^*$, i.e., the multiplicative group of non-zero elements of the 23-element field. So the log in the question would mean the $n$ (unique modulo 22) such that $5^nequiv20pmod23)$. Although @EthanBolker is right about this being hard to compute in general, this particular problem could be solved by just trying one $n$ after another.
$endgroup$
– Andreas Blass
Dec 8 '18 at 2:44
$begingroup$
My best guess is that Z*base23 was intended to mean $(mathbb Z/23)^*$, i.e., the multiplicative group of non-zero elements of the 23-element field. So the log in the question would mean the $n$ (unique modulo 22) such that $5^nequiv20pmod23)$. Although @EthanBolker is right about this being hard to compute in general, this particular problem could be solved by just trying one $n$ after another.
$endgroup$
– Andreas Blass
Dec 8 '18 at 2:44
$begingroup$
@AndreasBlass Excellent job of decryption!
$endgroup$
– saulspatz
Dec 8 '18 at 2:57
$begingroup$
@AndreasBlass Excellent job of decryption!
$endgroup$
– saulspatz
Dec 8 '18 at 2:57
$begingroup$
@saulspatz Thanks, but I see (now that it's too late to edit my comment) that I didn't do a very good job of including necessary braces in TeX. That congruence should be $5^nequiv20pmod{23}$.
$endgroup$
– Andreas Blass
Dec 8 '18 at 3:01
$begingroup$
@saulspatz Thanks, but I see (now that it's too late to edit my comment) that I didn't do a very good job of including necessary braces in TeX. That congruence should be $5^nequiv20pmod{23}$.
$endgroup$
– Andreas Blass
Dec 8 '18 at 3:01
$begingroup$
@AndreasBlass is 5^n ≡ 20 (mod23) the answer? I'm sorry about the format, I'm new to the website and still struggle to do it properly. The Z I was mentioning is exactly the one you showed. Is there any helpful resources besides Wikipedia that can help me understand what is going on? At least you helped me clarify that Z is basically the mod, but I'm still very confused about how I can compute this by hand.
$endgroup$
– Curious
Dec 8 '18 at 3:06
$begingroup$
@AndreasBlass is 5^n ≡ 20 (mod23) the answer? I'm sorry about the format, I'm new to the website and still struggle to do it properly. The Z I was mentioning is exactly the one you showed. Is there any helpful resources besides Wikipedia that can help me understand what is going on? At least you helped me clarify that Z is basically the mod, but I'm still very confused about how I can compute this by hand.
$endgroup$
– Curious
Dec 8 '18 at 3:06
|
show 3 more comments
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2
$begingroup$
I think you are asking about the discrete logarithm: en.wikipedia.org/wiki/Discrete_logarithm In general, it is hard to compute, which is why it's important in cryptography.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 2:20
4
$begingroup$
My best guess is that Z*base23 was intended to mean $(mathbb Z/23)^*$, i.e., the multiplicative group of non-zero elements of the 23-element field. So the log in the question would mean the $n$ (unique modulo 22) such that $5^nequiv20pmod23)$. Although @EthanBolker is right about this being hard to compute in general, this particular problem could be solved by just trying one $n$ after another.
$endgroup$
– Andreas Blass
Dec 8 '18 at 2:44
$begingroup$
@AndreasBlass Excellent job of decryption!
$endgroup$
– saulspatz
Dec 8 '18 at 2:57
$begingroup$
@saulspatz Thanks, but I see (now that it's too late to edit my comment) that I didn't do a very good job of including necessary braces in TeX. That congruence should be $5^nequiv20pmod{23}$.
$endgroup$
– Andreas Blass
Dec 8 '18 at 3:01
$begingroup$
@AndreasBlass is 5^n ≡ 20 (mod23) the answer? I'm sorry about the format, I'm new to the website and still struggle to do it properly. The Z I was mentioning is exactly the one you showed. Is there any helpful resources besides Wikipedia that can help me understand what is going on? At least you helped me clarify that Z is basically the mod, but I'm still very confused about how I can compute this by hand.
$endgroup$
– Curious
Dec 8 '18 at 3:06