In the finite field $mathbb{F}_{101}$ ,where discrete logarithms are $L_2(3)=69$ and $L_2(5)=24$. Compute the...
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Now, I have that $L_2(60)=L_2(4*3*5)=L_2(4)+L_2(3)+L_2(5)=2+69+24=95$.
So from my work $L_2(60)$ is $95$, but the answer on some other website gives $14$.
I just don't see where I went wrong.
cryptography discrete-logarithms
edited Nov 18 at 22:08
Yadati Kiran
1,253417
asked Nov 18 at 19:08
George S
13010
|
show 2 more comments
up vote
0
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Now, I have that $L_2(60)=L_2(4*3*5)=L_2(4)+L_2(3)+L_2(5)=2+69+24=95$.
So from my work $L_2(60)$ is $95$, but the answer on some other website gives $14$.
I just don't see where I went wrong.
cryptography discrete-logarithms
edited Nov 18 at 22:08
Yadati Kiran
1,253417
asked Nov 18 at 19:08
George S
13010
1
This does not look like standard notation. Perhaps some definitions would be a good idea.
– lulu
Nov 18 at 19:10
1
Voting to close the question as it is not clear what you are asking. I am guessing that F101 means $mathbb F_{101}$, the finite field with $101$ elements. But I have no guess as to what $L_2(n)$ might mean. Please edit for clarity.
– lulu
Nov 18 at 19:27
1
Ah, fair enough. So you are saying that $2$ is a primitive root $pmod {101}$ and you are asking to compute the minimal positive $a$ such that $2^aequiv 60pmod {101}$, yes? (Note: please edit your post to reflect this, assuming I have it right. Most people won't read through all the comments to discern your meaning.
– lulu
Nov 18 at 19:34
1
I note, by the way, that $2^{14}equiv 22pmod {101}$. $95$ is absolutely correct and your reasoning is sound.
– lulu
Nov 18 at 19:35
1
I can't imagine where the $14$ comes from. I expect some confusion (or just a good old fashioned error).
– lulu
Nov 18 at 19:37
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Now, I have that $L_2(60)=L_2(4*3*5)=L_2(4)+L_2(3)+L_2(5)=2+69+24=95$.
So from my work $L_2(60)$ is $95$, but the answer on some other website gives $14$.
I just don't see where I went wrong.
cryptography discrete-logarithms
edited Nov 18 at 22:08
Yadati Kiran
1,253417
asked Nov 18 at 19:08
George S
13010
Now, I have that $L_2(60)=L_2(4*3*5)=L_2(4)+L_2(3)+L_2(5)=2+69+24=95$.
So from my work $L_2(60)$ is $95$, but the answer on some other website gives $14$.
I just don't see where I went wrong.
cryptography discrete-logarithms
cryptography discrete-logarithms
edited Nov 18 at 22:08
Yadati Kiran
1,253417
asked Nov 18 at 19:08
George S
13010
edited Nov 18 at 22:08
Yadati Kiran
1,253417
asked Nov 18 at 19:08
George S
13010
edited Nov 18 at 22:08
Yadati Kiran
1,253417
edited Nov 18 at 22:08
Yadati Kiran
1,253417
edited Nov 18 at 22:08
Yadati Kiran
1,253417
1,253417
asked Nov 18 at 19:08
George S
13010
asked Nov 18 at 19:08
George S
13010
asked Nov 18 at 19:08
George S
13010
13010
1
This does not look like standard notation. Perhaps some definitions would be a good idea.
– lulu
Nov 18 at 19:10
1
Voting to close the question as it is not clear what you are asking. I am guessing that F101 means $mathbb F_{101}$, the finite field with $101$ elements. But I have no guess as to what $L_2(n)$ might mean. Please edit for clarity.
– lulu
Nov 18 at 19:27
1
Ah, fair enough. So you are saying that $2$ is a primitive root $pmod {101}$ and you are asking to compute the minimal positive $a$ such that $2^aequiv 60pmod {101}$, yes? (Note: please edit your post to reflect this, assuming I have it right. Most people won't read through all the comments to discern your meaning.
– lulu
Nov 18 at 19:34
1
I note, by the way, that $2^{14}equiv 22pmod {101}$. $95$ is absolutely correct and your reasoning is sound.
– lulu
Nov 18 at 19:35
1
I can't imagine where the $14$ comes from. I expect some confusion (or just a good old fashioned error).
– lulu
Nov 18 at 19:37
|
show 2 more comments
1
This does not look like standard notation. Perhaps some definitions would be a good idea.
– lulu
Nov 18 at 19:10
1
Voting to close the question as it is not clear what you are asking. I am guessing that F101 means $mathbb F_{101}$, the finite field with $101$ elements. But I have no guess as to what $L_2(n)$ might mean. Please edit for clarity.
– lulu
Nov 18 at 19:27
1
Ah, fair enough. So you are saying that $2$ is a primitive root $pmod {101}$ and you are asking to compute the minimal positive $a$ such that $2^aequiv 60pmod {101}$, yes? (Note: please edit your post to reflect this, assuming I have it right. Most people won't read through all the comments to discern your meaning.
– lulu
Nov 18 at 19:34
1
I note, by the way, that $2^{14}equiv 22pmod {101}$. $95$ is absolutely correct and your reasoning is sound.
– lulu
Nov 18 at 19:35
1
I can't imagine where the $14$ comes from. I expect some confusion (or just a good old fashioned error).
– lulu
Nov 18 at 19:37
1
1
This does not look like standard notation. Perhaps some definitions would be a good idea.
– lulu
Nov 18 at 19:10
This does not look like standard notation. Perhaps some definitions would be a good idea.
– lulu
Nov 18 at 19:10
1
1
Voting to close the question as it is not clear what you are asking. I am guessing that F101 means $mathbb F_{101}$, the finite field with $101$ elements. But I have no guess as to what $L_2(n)$ might mean. Please edit for clarity.
– lulu
Nov 18 at 19:27
Voting to close the question as it is not clear what you are asking. I am guessing that F101 means $mathbb F_{101}$, the finite field with $101$ elements. But I have no guess as to what $L_2(n)$ might mean. Please edit for clarity.
– lulu
Nov 18 at 19:27
1
1
Ah, fair enough. So you are saying that $2$ is a primitive root $pmod {101}$ and you are asking to compute the minimal positive $a$ such that $2^aequiv 60pmod {101}$, yes? (Note: please edit your post to reflect this, assuming I have it right. Most people won't read through all the comments to discern your meaning.
– lulu
Nov 18 at 19:34
Ah, fair enough. So you are saying that $2$ is a primitive root $pmod {101}$ and you are asking to compute the minimal positive $a$ such that $2^aequiv 60pmod {101}$, yes? (Note: please edit your post to reflect this, assuming I have it right. Most people won't read through all the comments to discern your meaning.
– lulu
Nov 18 at 19:34
1
1
I note, by the way, that $2^{14}equiv 22pmod {101}$. $95$ is absolutely correct and your reasoning is sound.
– lulu
Nov 18 at 19:35
I note, by the way, that $2^{14}equiv 22pmod {101}$. $95$ is absolutely correct and your reasoning is sound.
– lulu
Nov 18 at 19:35
1
1
I can't imagine where the $14$ comes from. I expect some confusion (or just a good old fashioned error).
– lulu
Nov 18 at 19:37
I can't imagine where the $14$ comes from. I expect some confusion (or just a good old fashioned error).
– lulu
Nov 18 at 19:37
|
show 2 more comments
1 Answer
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1
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Write $equiv$ for "equal modulo $101$". You are told that $2^{69}equiv 3$ and $2^{24}equiv5$. Furthermore $2^2equiv4$ trivially. From $60=3cdot 5cdot 4$ it then follows that
$$60equiv2^{69}cdot2^{24}cdot2^2equiv2^{95} .$$
This shows that in ${mathbb F}_{101}$ one has $L_2(60)=95$, whereby we have not checked that $2$ is actually a generator of the multiplicative group ${mathbb F}_{101}^*$, vulgo: a primitive root.
answered Nov 18 at 20:15
Christian Blatter
171k7111325
1
I did check and it is a primitive root modulo $101$. I also check $2^{95} equiv 60$ and $2^{14} equiv 22$ so the OP's source was wrong.
– Henno Brandsma
Nov 18 at 22:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Write $equiv$ for "equal modulo $101$". You are told that $2^{69}equiv 3$ and $2^{24}equiv5$. Furthermore $2^2equiv4$ trivially. From $60=3cdot 5cdot 4$ it then follows that
$$60equiv2^{69}cdot2^{24}cdot2^2equiv2^{95} .$$
This shows that in ${mathbb F}_{101}$ one has $L_2(60)=95$, whereby we have not checked that $2$ is actually a generator of the multiplicative group ${mathbb F}_{101}^*$, vulgo: a primitive root.
answered Nov 18 at 20:15
Christian Blatter
171k7111325
1
I did check and it is a primitive root modulo $101$. I also check $2^{95} equiv 60$ and $2^{14} equiv 22$ so the OP's source was wrong.
– Henno Brandsma
Nov 18 at 22:43
add a comment |
up vote
1
down vote
Write $equiv$ for "equal modulo $101$". You are told that $2^{69}equiv 3$ and $2^{24}equiv5$. Furthermore $2^2equiv4$ trivially. From $60=3cdot 5cdot 4$ it then follows that
$$60equiv2^{69}cdot2^{24}cdot2^2equiv2^{95} .$$
This shows that in ${mathbb F}_{101}$ one has $L_2(60)=95$, whereby we have not checked that $2$ is actually a generator of the multiplicative group ${mathbb F}_{101}^*$, vulgo: a primitive root.
answered Nov 18 at 20:15
Christian Blatter
171k7111325
1
I did check and it is a primitive root modulo $101$. I also check $2^{95} equiv 60$ and $2^{14} equiv 22$ so the OP's source was wrong.
– Henno Brandsma
Nov 18 at 22:43
add a comment |
up vote
1
down vote
up vote
1
down vote
Write $equiv$ for "equal modulo $101$". You are told that $2^{69}equiv 3$ and $2^{24}equiv5$. Furthermore $2^2equiv4$ trivially. From $60=3cdot 5cdot 4$ it then follows that
$$60equiv2^{69}cdot2^{24}cdot2^2equiv2^{95} .$$
This shows that in ${mathbb F}_{101}$ one has $L_2(60)=95$, whereby we have not checked that $2$ is actually a generator of the multiplicative group ${mathbb F}_{101}^*$, vulgo: a primitive root.
answered Nov 18 at 20:15
Christian Blatter
171k7111325
Write $equiv$ for "equal modulo $101$". You are told that $2^{69}equiv 3$ and $2^{24}equiv5$. Furthermore $2^2equiv4$ trivially. From $60=3cdot 5cdot 4$ it then follows that
$$60equiv2^{69}cdot2^{24}cdot2^2equiv2^{95} .$$
This shows that in ${mathbb F}_{101}$ one has $L_2(60)=95$, whereby we have not checked that $2$ is actually a generator of the multiplicative group ${mathbb F}_{101}^*$, vulgo: a primitive root.
answered Nov 18 at 20:15
Christian Blatter
171k7111325
answered Nov 18 at 20:15
Christian Blatter
171k7111325
answered Nov 18 at 20:15
Christian Blatter
171k7111325
answered Nov 18 at 20:15
Christian Blatter
171k7111325
171k7111325
1
I did check and it is a primitive root modulo $101$. I also check $2^{95} equiv 60$ and $2^{14} equiv 22$ so the OP's source was wrong.
– Henno Brandsma
Nov 18 at 22:43
add a comment |
1
I did check and it is a primitive root modulo $101$. I also check $2^{95} equiv 60$ and $2^{14} equiv 22$ so the OP's source was wrong.
– Henno Brandsma
Nov 18 at 22:43
1
1
I did check and it is a primitive root modulo $101$. I also check $2^{95} equiv 60$ and $2^{14} equiv 22$ so the OP's source was wrong.
– Henno Brandsma
Nov 18 at 22:43
I did check and it is a primitive root modulo $101$. I also check $2^{95} equiv 60$ and $2^{14} equiv 22$ so the OP's source was wrong.
– Henno Brandsma
Nov 18 at 22:43
add a comment |
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1
This does not look like standard notation. Perhaps some definitions would be a good idea.
– lulu
Nov 18 at 19:10
1
Voting to close the question as it is not clear what you are asking. I am guessing that F101 means $mathbb F_{101}$, the finite field with $101$ elements. But I have no guess as to what $L_2(n)$ might mean. Please edit for clarity.
– lulu
Nov 18 at 19:27
1
Ah, fair enough. So you are saying that $2$ is a primitive root $pmod {101}$ and you are asking to compute the minimal positive $a$ such that $2^aequiv 60pmod {101}$, yes? (Note: please edit your post to reflect this, assuming I have it right. Most people won't read through all the comments to discern your meaning.
– lulu
Nov 18 at 19:34
1
I note, by the way, that $2^{14}equiv 22pmod {101}$. $95$ is absolutely correct and your reasoning is sound.
– lulu
Nov 18 at 19:35
1
I can't imagine where the $14$ comes from. I expect some confusion (or just a good old fashioned error).
– lulu
Nov 18 at 19:37