RSA: Show how to factor $n=pq$, the product of two primes, given $(p-1)(q-1)$
$begingroup$
As an exercise in my discrete mathematics textbook, for my first-year course, the following question is asked, on the topic of RSA encryption:
Show that we can easily factor $n$ when we know that $n$ is the product of two primes, $p$ and $q$, and we know the value of $(p-1)(q-1)$.
So far my attempted solution has been to expand $(p-1)(q-1)$, to lay a foundation of the known value.
$$(p-1)(q-1)=pq-p-q+1$$
and rewrite
$$(p-1)(q-1)=pq-(p+q)+1$$
The book suggests calling $p+q=s$
and then attempting to use that to find the product $n$, since $q=s-p$, we can write the product as $n=p(s-p)$. Expanding this leads to the quadratic equation
$$0=p^2-ps+n$$
Using the quadratic formula to solve this leads to the following solution:
$$p=frac{spmsqrt{s^2-4n}}{2}$$
As shown by @misterriemann in the comments, the discriminant can not be negative, which leaves out any complex solutions, however i am unsure how to extract $q$ from this, after being given $p$ as a function of $s$.
Any help on this would be appreciated.
discrete-mathematics prime-numbers cryptography
edited Dec 10 '18 at 16:17
Ilmari Karonen
19.6k25183
asked Dec 7 '18 at 18:14
ms_ms_
154
$endgroup$
add a comment |
$begingroup$
As an exercise in my discrete mathematics textbook, for my first-year course, the following question is asked, on the topic of RSA encryption:
Show that we can easily factor $n$ when we know that $n$ is the product of two primes, $p$ and $q$, and we know the value of $(p-1)(q-1)$.
So far my attempted solution has been to expand $(p-1)(q-1)$, to lay a foundation of the known value.
$$(p-1)(q-1)=pq-p-q+1$$
and rewrite
$$(p-1)(q-1)=pq-(p+q)+1$$
The book suggests calling $p+q=s$
and then attempting to use that to find the product $n$, since $q=s-p$, we can write the product as $n=p(s-p)$. Expanding this leads to the quadratic equation
$$0=p^2-ps+n$$
Using the quadratic formula to solve this leads to the following solution:
$$p=frac{spmsqrt{s^2-4n}}{2}$$
As shown by @misterriemann in the comments, the discriminant can not be negative, which leaves out any complex solutions, however i am unsure how to extract $q$ from this, after being given $p$ as a function of $s$.
Any help on this would be appreciated.
discrete-mathematics prime-numbers cryptography
edited Dec 10 '18 at 16:17
Ilmari Karonen
19.6k25183
asked Dec 7 '18 at 18:14
ms_ms_
154
$endgroup$
1
$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n. $$
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:18
$begingroup$
^That's an answer.
$endgroup$
– Randall
Dec 7 '18 at 18:19
$begingroup$
Ah of course, thank you. I will revise my original question, and refer to this.
$endgroup$
– ms_
Dec 7 '18 at 18:22
$begingroup$
"i have no idea how to extract q from this, after being given p as a function of s" .You already know that $q=s-p$
$endgroup$
– Federico
Dec 7 '18 at 18:23
add a comment |
$begingroup$
As an exercise in my discrete mathematics textbook, for my first-year course, the following question is asked, on the topic of RSA encryption:
Show that we can easily factor $n$ when we know that $n$ is the product of two primes, $p$ and $q$, and we know the value of $(p-1)(q-1)$.
So far my attempted solution has been to expand $(p-1)(q-1)$, to lay a foundation of the known value.
$$(p-1)(q-1)=pq-p-q+1$$
and rewrite
$$(p-1)(q-1)=pq-(p+q)+1$$
The book suggests calling $p+q=s$
and then attempting to use that to find the product $n$, since $q=s-p$, we can write the product as $n=p(s-p)$. Expanding this leads to the quadratic equation
$$0=p^2-ps+n$$
Using the quadratic formula to solve this leads to the following solution:
$$p=frac{spmsqrt{s^2-4n}}{2}$$
As shown by @misterriemann in the comments, the discriminant can not be negative, which leaves out any complex solutions, however i am unsure how to extract $q$ from this, after being given $p$ as a function of $s$.
Any help on this would be appreciated.
discrete-mathematics prime-numbers cryptography
edited Dec 10 '18 at 16:17
Ilmari Karonen
19.6k25183
asked Dec 7 '18 at 18:14
ms_ms_
154
$endgroup$
As an exercise in my discrete mathematics textbook, for my first-year course, the following question is asked, on the topic of RSA encryption:
Show that we can easily factor $n$ when we know that $n$ is the product of two primes, $p$ and $q$, and we know the value of $(p-1)(q-1)$.
So far my attempted solution has been to expand $(p-1)(q-1)$, to lay a foundation of the known value.
$$(p-1)(q-1)=pq-p-q+1$$
and rewrite
$$(p-1)(q-1)=pq-(p+q)+1$$
The book suggests calling $p+q=s$
and then attempting to use that to find the product $n$, since $q=s-p$, we can write the product as $n=p(s-p)$. Expanding this leads to the quadratic equation
$$0=p^2-ps+n$$
Using the quadratic formula to solve this leads to the following solution:
$$p=frac{spmsqrt{s^2-4n}}{2}$$
As shown by @misterriemann in the comments, the discriminant can not be negative, which leaves out any complex solutions, however i am unsure how to extract $q$ from this, after being given $p$ as a function of $s$.
Any help on this would be appreciated.
discrete-mathematics prime-numbers cryptography
discrete-mathematics prime-numbers cryptography
edited Dec 10 '18 at 16:17
Ilmari Karonen
19.6k25183
asked Dec 7 '18 at 18:14
ms_ms_
154
edited Dec 10 '18 at 16:17
Ilmari Karonen
19.6k25183
asked Dec 7 '18 at 18:14
ms_ms_
154
edited Dec 10 '18 at 16:17
Ilmari Karonen
19.6k25183
edited Dec 10 '18 at 16:17
Ilmari Karonen
19.6k25183
edited Dec 10 '18 at 16:17
Ilmari Karonen
19.6k25183
19.6k25183
asked Dec 7 '18 at 18:14
ms_ms_
154
asked Dec 7 '18 at 18:14
ms_ms_
154
asked Dec 7 '18 at 18:14
ms_ms_
154
154
1
$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n. $$
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:18
$begingroup$
^That's an answer.
$endgroup$
– Randall
Dec 7 '18 at 18:19
$begingroup$
Ah of course, thank you. I will revise my original question, and refer to this.
$endgroup$
– ms_
Dec 7 '18 at 18:22
$begingroup$
"i have no idea how to extract q from this, after being given p as a function of s" .You already know that $q=s-p$
$endgroup$
– Federico
Dec 7 '18 at 18:23
add a comment |
1
$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n. $$
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:18
$begingroup$
^That's an answer.
$endgroup$
– Randall
Dec 7 '18 at 18:19
$begingroup$
Ah of course, thank you. I will revise my original question, and refer to this.
$endgroup$
– ms_
Dec 7 '18 at 18:22
$begingroup$
"i have no idea how to extract q from this, after being given p as a function of s" .You already know that $q=s-p$
$endgroup$
– Federico
Dec 7 '18 at 18:23
1
1
$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n. $$
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:18
$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n. $$
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:18
$begingroup$
^That's an answer.
$endgroup$
– Randall
Dec 7 '18 at 18:19
$begingroup$
^That's an answer.
$endgroup$
– Randall
Dec 7 '18 at 18:19
$begingroup$
Ah of course, thank you. I will revise my original question, and refer to this.
$endgroup$
– ms_
Dec 7 '18 at 18:22
$begingroup$
Ah of course, thank you. I will revise my original question, and refer to this.
$endgroup$
– ms_
Dec 7 '18 at 18:22
$begingroup$
"i have no idea how to extract q from this, after being given p as a function of s" .You already know that $q=s-p$
$endgroup$
– Federico
Dec 7 '18 at 18:23
$begingroup$
"i have no idea how to extract q from this, after being given p as a function of s" .You already know that $q=s-p$
$endgroup$
– Federico
Dec 7 '18 at 18:23
add a comment |
1 Answer
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$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n quad Rightarrow quad s^2-4n geq 0.$$
Furthermore, since you know the value of $(p-1)(q-1),$ (as well as $n=pq)$, you also know $s$, since
$$ s = p+q = pq-(p-1)(q-1)+1. $$
Then you can find $p$ from your formula, which immediately gives you $q$ as well, since
$$ p+q=s quad iff quad q=s-p, $$
or if you prefer to do division,
$$ n=pq quad iff quad q = frac{n}{p}. $$
answered Dec 7 '18 at 18:24
MisterRiemannMisterRiemann
5,8451624
$endgroup$
1
$begingroup$
Your third equation made me realize that i totally missed a solution for s, given in terms of "known" values. Thank you for this answer.
$endgroup$
– ms_
Dec 7 '18 at 18:34
1
$begingroup$
You're very welcome!
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:35
add a comment |
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$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n quad Rightarrow quad s^2-4n geq 0.$$
Furthermore, since you know the value of $(p-1)(q-1),$ (as well as $n=pq)$, you also know $s$, since
$$ s = p+q = pq-(p-1)(q-1)+1. $$
Then you can find $p$ from your formula, which immediately gives you $q$ as well, since
$$ p+q=s quad iff quad q=s-p, $$
or if you prefer to do division,
$$ n=pq quad iff quad q = frac{n}{p}. $$
answered Dec 7 '18 at 18:24
MisterRiemannMisterRiemann
5,8451624
$endgroup$
1
$begingroup$
Your third equation made me realize that i totally missed a solution for s, given in terms of "known" values. Thank you for this answer.
$endgroup$
– ms_
Dec 7 '18 at 18:34
1
$begingroup$
You're very welcome!
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:35
add a comment |
$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n quad Rightarrow quad s^2-4n geq 0.$$
Furthermore, since you know the value of $(p-1)(q-1),$ (as well as $n=pq)$, you also know $s$, since
$$ s = p+q = pq-(p-1)(q-1)+1. $$
Then you can find $p$ from your formula, which immediately gives you $q$ as well, since
$$ p+q=s quad iff quad q=s-p, $$
or if you prefer to do division,
$$ n=pq quad iff quad q = frac{n}{p}. $$
answered Dec 7 '18 at 18:24
MisterRiemannMisterRiemann
5,8451624
$endgroup$
1
$begingroup$
Your third equation made me realize that i totally missed a solution for s, given in terms of "known" values. Thank you for this answer.
$endgroup$
– ms_
Dec 7 '18 at 18:34
1
$begingroup$
You're very welcome!
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:35
add a comment |
$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n quad Rightarrow quad s^2-4n geq 0.$$
Furthermore, since you know the value of $(p-1)(q-1),$ (as well as $n=pq)$, you also know $s$, since
$$ s = p+q = pq-(p-1)(q-1)+1. $$
Then you can find $p$ from your formula, which immediately gives you $q$ as well, since
$$ p+q=s quad iff quad q=s-p, $$
or if you prefer to do division,
$$ n=pq quad iff quad q = frac{n}{p}. $$
answered Dec 7 '18 at 18:24
MisterRiemannMisterRiemann
5,8451624
$endgroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n quad Rightarrow quad s^2-4n geq 0.$$
Furthermore, since you know the value of $(p-1)(q-1),$ (as well as $n=pq)$, you also know $s$, since
$$ s = p+q = pq-(p-1)(q-1)+1. $$
Then you can find $p$ from your formula, which immediately gives you $q$ as well, since
$$ p+q=s quad iff quad q=s-p, $$
or if you prefer to do division,
$$ n=pq quad iff quad q = frac{n}{p}. $$
answered Dec 7 '18 at 18:24
MisterRiemannMisterRiemann
5,8451624
answered Dec 7 '18 at 18:24
MisterRiemannMisterRiemann
5,8451624
answered Dec 7 '18 at 18:24
MisterRiemannMisterRiemann
5,8451624
answered Dec 7 '18 at 18:24
MisterRiemannMisterRiemann
5,8451624
5,8451624
1
$begingroup$
Your third equation made me realize that i totally missed a solution for s, given in terms of "known" values. Thank you for this answer.
$endgroup$
– ms_
Dec 7 '18 at 18:34
1
$begingroup$
You're very welcome!
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:35
add a comment |
1
$begingroup$
Your third equation made me realize that i totally missed a solution for s, given in terms of "known" values. Thank you for this answer.
$endgroup$
– ms_
Dec 7 '18 at 18:34
1
$begingroup$
You're very welcome!
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:35
1
1
$begingroup$
Your third equation made me realize that i totally missed a solution for s, given in terms of "known" values. Thank you for this answer.
$endgroup$
– ms_
Dec 7 '18 at 18:34
$begingroup$
Your third equation made me realize that i totally missed a solution for s, given in terms of "known" values. Thank you for this answer.
$endgroup$
– ms_
Dec 7 '18 at 18:34
1
1
$begingroup$
You're very welcome!
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:35
$begingroup$
You're very welcome!
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:35
add a comment |
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$begingroup$
The discriminant cannot be negative, since by AM-GM, $$ frac{p+q}{2} geq sqrt{pq}, $$ and so $$ s^2 = (p+q)^2 geq 4pq = 4n. $$
$endgroup$
– MisterRiemann
Dec 7 '18 at 18:18
$begingroup$
^That's an answer.
$endgroup$
– Randall
Dec 7 '18 at 18:19
$begingroup$
Ah of course, thank you. I will revise my original question, and refer to this.
$endgroup$
– ms_
Dec 7 '18 at 18:22
$begingroup$
"i have no idea how to extract q from this, after being given p as a function of s" .You already know that $q=s-p$
$endgroup$
– Federico
Dec 7 '18 at 18:23