How to solve such a quadratic congruence equation?
$begingroup$
I have the following equation: $y^2 equiv r^2 pmod n $
I know the values of y and n, I just need to find the values of r.
Assuming that $y = 12654$ and $n = 79061$, my working is as follows:
$ 12654^2$ mod $79061 = r^2$ mod $79061$
$25191 = r^2$ mod $79061$
The prime factorization of 79061 is $173*457$
Hence,
$r^2 = 25191$ mod $173$ $=>$ $106$ mod $173$
$r^2 = 25191$ mod $457$ $=>$ $56$ mod $457$
So now I have two equations,
$r^2 = 106$ mod $173$ and $r^2 = 56$ mod $457$
I am stuck here, I would appreciate if someone can help me move forward.
I've stumbled upon other similar questions where the answers show that they get rid of the squared but I cannot understand how they do it.
elementary-number-theory modular-arithmetic chinese-remainder-theorem
asked Dec 22 '18 at 11:22
MulishiaMulishia
62
$endgroup$
add a comment |
$begingroup$
I have the following equation: $y^2 equiv r^2 pmod n $
I know the values of y and n, I just need to find the values of r.
Assuming that $y = 12654$ and $n = 79061$, my working is as follows:
$ 12654^2$ mod $79061 = r^2$ mod $79061$
$25191 = r^2$ mod $79061$
The prime factorization of 79061 is $173*457$
Hence,
$r^2 = 25191$ mod $173$ $=>$ $106$ mod $173$
$r^2 = 25191$ mod $457$ $=>$ $56$ mod $457$
So now I have two equations,
$r^2 = 106$ mod $173$ and $r^2 = 56$ mod $457$
I am stuck here, I would appreciate if someone can help me move forward.
I've stumbled upon other similar questions where the answers show that they get rid of the squared but I cannot understand how they do it.
elementary-number-theory modular-arithmetic chinese-remainder-theorem
asked Dec 22 '18 at 11:22
MulishiaMulishia
62
$endgroup$
add a comment |
$begingroup$
I have the following equation: $y^2 equiv r^2 pmod n $
I know the values of y and n, I just need to find the values of r.
Assuming that $y = 12654$ and $n = 79061$, my working is as follows:
$ 12654^2$ mod $79061 = r^2$ mod $79061$
$25191 = r^2$ mod $79061$
The prime factorization of 79061 is $173*457$
Hence,
$r^2 = 25191$ mod $173$ $=>$ $106$ mod $173$
$r^2 = 25191$ mod $457$ $=>$ $56$ mod $457$
So now I have two equations,
$r^2 = 106$ mod $173$ and $r^2 = 56$ mod $457$
I am stuck here, I would appreciate if someone can help me move forward.
I've stumbled upon other similar questions where the answers show that they get rid of the squared but I cannot understand how they do it.
elementary-number-theory modular-arithmetic chinese-remainder-theorem
asked Dec 22 '18 at 11:22
MulishiaMulishia
62
$endgroup$
I have the following equation: $y^2 equiv r^2 pmod n $
I know the values of y and n, I just need to find the values of r.
Assuming that $y = 12654$ and $n = 79061$, my working is as follows:
$ 12654^2$ mod $79061 = r^2$ mod $79061$
$25191 = r^2$ mod $79061$
The prime factorization of 79061 is $173*457$
Hence,
$r^2 = 25191$ mod $173$ $=>$ $106$ mod $173$
$r^2 = 25191$ mod $457$ $=>$ $56$ mod $457$
So now I have two equations,
$r^2 = 106$ mod $173$ and $r^2 = 56$ mod $457$
I am stuck here, I would appreciate if someone can help me move forward.
I've stumbled upon other similar questions where the answers show that they get rid of the squared but I cannot understand how they do it.
elementary-number-theory modular-arithmetic chinese-remainder-theorem
elementary-number-theory modular-arithmetic chinese-remainder-theorem
asked Dec 22 '18 at 11:22
MulishiaMulishia
62
asked Dec 22 '18 at 11:22
MulishiaMulishia
62
asked Dec 22 '18 at 11:22
MulishiaMulishia
62
asked Dec 22 '18 at 11:22
MulishiaMulishia
62
asked Dec 22 '18 at 11:22
MulishiaMulishia
62
62
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).
Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.
So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
$xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.
answered Dec 22 '18 at 12:38
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
$begingroup$
Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
$endgroup$
– Mulishia
Dec 22 '18 at 13:11
$begingroup$
@Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:50
$begingroup$
@Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:54
$begingroup$
@Mulishia You can continue with this answer where I completely solve an analogous example.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 20:15
add a comment |
$begingroup$
Similar to 'random' comment we have:
$(y-r)(y+r)≡0mod n$
$n=79061=173times457$
Following cases can be considered:
a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$
b: $y+r=173$ ⇒ $r=173-12654=-12421$
And we have:
$ 12654^2-(±12421)^2=55times 79061$
c: $y-r=457$⇒ $r=12654+457=13111$
d: $y+r=457$⇒ $r=-13111$
And we have:
$12654^2-13111^2=148.93...times79061$
So $r= ±12421$ is acceptable.
answered Dec 22 '18 at 13:20
siroussirous
1,6871514
$endgroup$
add a comment |
$begingroup$
The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.
answered Dec 22 '18 at 12:29
randomrandom
57626
$endgroup$
add a comment |
$begingroup$
$$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
$$ r=91542 - 79061 m $$
answered Dec 22 '18 at 18:17
S. I.S. I.
112
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).
Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.
So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
$xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.
answered Dec 22 '18 at 12:38
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
$begingroup$
Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
$endgroup$
– Mulishia
Dec 22 '18 at 13:11
$begingroup$
@Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:50
$begingroup$
@Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:54
$begingroup$
@Mulishia You can continue with this answer where I completely solve an analogous example.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 20:15
add a comment |
$begingroup$
You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).
Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.
So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
$xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.
answered Dec 22 '18 at 12:38
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
$begingroup$
Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
$endgroup$
– Mulishia
Dec 22 '18 at 13:11
$begingroup$
@Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:50
$begingroup$
@Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:54
$begingroup$
@Mulishia You can continue with this answer where I completely solve an analogous example.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 20:15
add a comment |
$begingroup$
You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).
Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.
So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
$xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.
answered Dec 22 '18 at 12:38
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).
Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.
So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
$xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.
answered Dec 22 '18 at 12:38
Henno BrandsmaHenno Brandsma
111k348119
edited Dec 22 '18 at 12:46
answered Dec 22 '18 at 12:38
Henno BrandsmaHenno Brandsma
111k348119
answered Dec 22 '18 at 12:38
Henno BrandsmaHenno Brandsma
111k348119
answered Dec 22 '18 at 12:38
Henno BrandsmaHenno Brandsma
111k348119
111k348119
$begingroup$
Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
$endgroup$
– Mulishia
Dec 22 '18 at 13:11
$begingroup$
@Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:50
$begingroup$
@Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:54
$begingroup$
@Mulishia You can continue with this answer where I completely solve an analogous example.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 20:15
add a comment |
$begingroup$
Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
$endgroup$
– Mulishia
Dec 22 '18 at 13:11
$begingroup$
@Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:50
$begingroup$
@Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:54
$begingroup$
@Mulishia You can continue with this answer where I completely solve an analogous example.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 20:15
$begingroup$
Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
$endgroup$
– Mulishia
Dec 22 '18 at 13:11
$begingroup$
Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
$endgroup$
– Mulishia
Dec 22 '18 at 13:11
$begingroup$
@Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:50
$begingroup$
@Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:50
$begingroup$
@Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:54
$begingroup$
@Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 22:54
$begingroup$
@Mulishia You can continue with this answer where I completely solve an analogous example.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 20:15
$begingroup$
@Mulishia You can continue with this answer where I completely solve an analogous example.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 20:15
add a comment |
$begingroup$
Similar to 'random' comment we have:
$(y-r)(y+r)≡0mod n$
$n=79061=173times457$
Following cases can be considered:
a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$
b: $y+r=173$ ⇒ $r=173-12654=-12421$
And we have:
$ 12654^2-(±12421)^2=55times 79061$
c: $y-r=457$⇒ $r=12654+457=13111$
d: $y+r=457$⇒ $r=-13111$
And we have:
$12654^2-13111^2=148.93...times79061$
So $r= ±12421$ is acceptable.
answered Dec 22 '18 at 13:20
siroussirous
1,6871514
$endgroup$
add a comment |
$begingroup$
Similar to 'random' comment we have:
$(y-r)(y+r)≡0mod n$
$n=79061=173times457$
Following cases can be considered:
a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$
b: $y+r=173$ ⇒ $r=173-12654=-12421$
And we have:
$ 12654^2-(±12421)^2=55times 79061$
c: $y-r=457$⇒ $r=12654+457=13111$
d: $y+r=457$⇒ $r=-13111$
And we have:
$12654^2-13111^2=148.93...times79061$
So $r= ±12421$ is acceptable.
answered Dec 22 '18 at 13:20
siroussirous
1,6871514
$endgroup$
add a comment |
$begingroup$
Similar to 'random' comment we have:
$(y-r)(y+r)≡0mod n$
$n=79061=173times457$
Following cases can be considered:
a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$
b: $y+r=173$ ⇒ $r=173-12654=-12421$
And we have:
$ 12654^2-(±12421)^2=55times 79061$
c: $y-r=457$⇒ $r=12654+457=13111$
d: $y+r=457$⇒ $r=-13111$
And we have:
$12654^2-13111^2=148.93...times79061$
So $r= ±12421$ is acceptable.
answered Dec 22 '18 at 13:20
siroussirous
1,6871514
$endgroup$
Similar to 'random' comment we have:
$(y-r)(y+r)≡0mod n$
$n=79061=173times457$
Following cases can be considered:
a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$
b: $y+r=173$ ⇒ $r=173-12654=-12421$
And we have:
$ 12654^2-(±12421)^2=55times 79061$
c: $y-r=457$⇒ $r=12654+457=13111$
d: $y+r=457$⇒ $r=-13111$
And we have:
$12654^2-13111^2=148.93...times79061$
So $r= ±12421$ is acceptable.
answered Dec 22 '18 at 13:20
siroussirous
1,6871514
answered Dec 22 '18 at 13:20
siroussirous
1,6871514
answered Dec 22 '18 at 13:20
siroussirous
1,6871514
answered Dec 22 '18 at 13:20
siroussirous
1,6871514
1,6871514
add a comment |
add a comment |
$begingroup$
The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.
answered Dec 22 '18 at 12:29
randomrandom
57626
$endgroup$
add a comment |
$begingroup$
The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.
answered Dec 22 '18 at 12:29
randomrandom
57626
$endgroup$
add a comment |
$begingroup$
The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.
answered Dec 22 '18 at 12:29
randomrandom
57626
$endgroup$
The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.
answered Dec 22 '18 at 12:29
randomrandom
57626
answered Dec 22 '18 at 12:29
randomrandom
57626
answered Dec 22 '18 at 12:29
randomrandom
57626
answered Dec 22 '18 at 12:29
randomrandom
57626
57626
add a comment |
add a comment |
$begingroup$
$$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
$$ r=91542 - 79061 m $$
answered Dec 22 '18 at 18:17
S. I.S. I.
112
$endgroup$
add a comment |
$begingroup$
$$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
$$ r=91542 - 79061 m $$
answered Dec 22 '18 at 18:17
S. I.S. I.
112
$endgroup$
add a comment |
$begingroup$
$$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
$$ r=91542 - 79061 m $$
answered Dec 22 '18 at 18:17
S. I.S. I.
112
$endgroup$
$$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
$$ r=91542 - 79061 m $$
answered Dec 22 '18 at 18:17
S. I.S. I.
112
answered Dec 22 '18 at 18:17
S. I.S. I.
112
answered Dec 22 '18 at 18:17
S. I.S. I.
112
answered Dec 22 '18 at 18:17
S. I.S. I.
112
112
add a comment |
add a comment |
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