RSA: Danger of using p to create q
$begingroup$
Assume my prime generation is as follows:
Pick a number $p$ between 1000 and 9999. $p=abcd$.
Make sure $p$ is prime
Construct $q$ such by taking the last 2 digits of $p$ and the first 2 digits of $p$, i.e. $q=cdab$
Make sure $q$ is prime.
Is the resulting $n = p·q$ more easily factorable?
My gut feeling says yes but I can't see why? I thought about Coppersmith but in this case, we don't have any common bit between $p$ and $q$ that are also at the same place. Is there a weakness?
rsa
edited Apr 6 at 11:26
Paŭlo Ebermann
19k562106
asked Apr 5 at 17:00
S. L.S. L.
957
$endgroup$
add a comment |
$begingroup$
Assume my prime generation is as follows:
Pick a number $p$ between 1000 and 9999. $p=abcd$.
Make sure $p$ is prime
Construct $q$ such by taking the last 2 digits of $p$ and the first 2 digits of $p$, i.e. $q=cdab$
Make sure $q$ is prime.
Is the resulting $n = p·q$ more easily factorable?
My gut feeling says yes but I can't see why? I thought about Coppersmith but in this case, we don't have any common bit between $p$ and $q$ that are also at the same place. Is there a weakness?
rsa
edited Apr 6 at 11:26
Paŭlo Ebermann
19k562106
asked Apr 5 at 17:00
S. L.S. L.
957
$endgroup$
2
$begingroup$
Of course, any product of two 4-digit primes is trivially factorable by trial division anyway, since there are only 1061 primes between 1000 and 9999. Add in the digit reversal requirement, and there are only 76(!) possible pairs to consider.
$endgroup$
– Ilmari Karonen
Apr 6 at 1:37
1
$begingroup$
@Nat: My fault, I added the "$= pq$" for context in an edit, and didn't notice the potential ambiguity. I see Paŭlo has already fixed it.
$endgroup$
– Ilmari Karonen
Apr 6 at 12:19
$begingroup$
I'd just like to add that this is inspired by a contest that just ended (12 minutes ago).
$endgroup$
– enedil
Apr 6 at 16:12
add a comment |
$begingroup$
Assume my prime generation is as follows:
Pick a number $p$ between 1000 and 9999. $p=abcd$.
Make sure $p$ is prime
Construct $q$ such by taking the last 2 digits of $p$ and the first 2 digits of $p$, i.e. $q=cdab$
Make sure $q$ is prime.
Is the resulting $n = p·q$ more easily factorable?
My gut feeling says yes but I can't see why? I thought about Coppersmith but in this case, we don't have any common bit between $p$ and $q$ that are also at the same place. Is there a weakness?
rsa
edited Apr 6 at 11:26
Paŭlo Ebermann
19k562106
asked Apr 5 at 17:00
S. L.S. L.
957
$endgroup$
Assume my prime generation is as follows:
Pick a number $p$ between 1000 and 9999. $p=abcd$.
Make sure $p$ is prime
Construct $q$ such by taking the last 2 digits of $p$ and the first 2 digits of $p$, i.e. $q=cdab$
Make sure $q$ is prime.
Is the resulting $n = p·q$ more easily factorable?
My gut feeling says yes but I can't see why? I thought about Coppersmith but in this case, we don't have any common bit between $p$ and $q$ that are also at the same place. Is there a weakness?
rsa
rsa
edited Apr 6 at 11:26
Paŭlo Ebermann
19k562106
asked Apr 5 at 17:00
S. L.S. L.
957
edited Apr 6 at 11:26
Paŭlo Ebermann
19k562106
asked Apr 5 at 17:00
S. L.S. L.
957
edited Apr 6 at 11:26
Paŭlo Ebermann
19k562106
edited Apr 6 at 11:26
Paŭlo Ebermann
19k562106
edited Apr 6 at 11:26
Paŭlo Ebermann
19k562106
19k562106
asked Apr 5 at 17:00
S. L.S. L.
957
asked Apr 5 at 17:00
S. L.S. L.
957
asked Apr 5 at 17:00
S. L.S. L.
957
957
2
$begingroup$
Of course, any product of two 4-digit primes is trivially factorable by trial division anyway, since there are only 1061 primes between 1000 and 9999. Add in the digit reversal requirement, and there are only 76(!) possible pairs to consider.
$endgroup$
– Ilmari Karonen
Apr 6 at 1:37
1
$begingroup$
@Nat: My fault, I added the "$= pq$" for context in an edit, and didn't notice the potential ambiguity. I see Paŭlo has already fixed it.
$endgroup$
– Ilmari Karonen
Apr 6 at 12:19
$begingroup$
I'd just like to add that this is inspired by a contest that just ended (12 minutes ago).
$endgroup$
– enedil
Apr 6 at 16:12
add a comment |
2
$begingroup$
Of course, any product of two 4-digit primes is trivially factorable by trial division anyway, since there are only 1061 primes between 1000 and 9999. Add in the digit reversal requirement, and there are only 76(!) possible pairs to consider.
$endgroup$
– Ilmari Karonen
Apr 6 at 1:37
1
$begingroup$
@Nat: My fault, I added the "$= pq$" for context in an edit, and didn't notice the potential ambiguity. I see Paŭlo has already fixed it.
$endgroup$
– Ilmari Karonen
Apr 6 at 12:19
$begingroup$
I'd just like to add that this is inspired by a contest that just ended (12 minutes ago).
$endgroup$
– enedil
Apr 6 at 16:12
2
2
$begingroup$
Of course, any product of two 4-digit primes is trivially factorable by trial division anyway, since there are only 1061 primes between 1000 and 9999. Add in the digit reversal requirement, and there are only 76(!) possible pairs to consider.
$endgroup$
– Ilmari Karonen
Apr 6 at 1:37
$begingroup$
Of course, any product of two 4-digit primes is trivially factorable by trial division anyway, since there are only 1061 primes between 1000 and 9999. Add in the digit reversal requirement, and there are only 76(!) possible pairs to consider.
$endgroup$
– Ilmari Karonen
Apr 6 at 1:37
1
1
$begingroup$
@Nat: My fault, I added the "$= pq$" for context in an edit, and didn't notice the potential ambiguity. I see Paŭlo has already fixed it.
$endgroup$
– Ilmari Karonen
Apr 6 at 12:19
$begingroup$
@Nat: My fault, I added the "$= pq$" for context in an edit, and didn't notice the potential ambiguity. I see Paŭlo has already fixed it.
$endgroup$
– Ilmari Karonen
Apr 6 at 12:19
$begingroup$
I'd just like to add that this is inspired by a contest that just ended (12 minutes ago).
$endgroup$
– enedil
Apr 6 at 16:12
$begingroup$
I'd just like to add that this is inspired by a contest that just ended (12 minutes ago).
$endgroup$
– enedil
Apr 6 at 16:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You don't need anything fancy like Coppersmith, just simple algebra. The idea is to translate the equations we have involving the digits of $p$ and $q$ in base $B$ ($B = 100$ in your example) into equations involving the digits of $n$ in base $B$, which we know. You have $p = x B + y$ and $q = y B + x$, with $0 lt x, y lt B$. Then $n = x y B^2 + (x^2 + y^2) B + x y$.
The rightmost digit of $n$ in base $B$ is $(x y) bmod B$. Since ${x,y} le B-1$, $(x^2 + y^2) B + x y le 2 (B-1)^2 B + (B-1)^2 lt 2 (B-1)^2 (B+1) = 2 (B-1) (B^2-1) lt 2 B^3$. Hence the $B^3$ digit of $n$ is the $B$ digit of $x y$ plus $z$ where $0 le z lt 2$, i.e. $z in {0, 1}$. So by reading the digits of $n$ in base $B$, we get the digits of $x y$ in base $B$, up to two possibilities, giving just two possibilities for $x y$ itself: $x y in {W_0, W_1}$.
Injecting this knowledge into the equation above gives us $x^2 + y^2 = (n - W_z (B^2 + 1)) / B$. And of course knowing both $x^2 + y^2$ and $x y$ gives $x$ and $y$.
answered Apr 5 at 17:27
GillesGilles
8,44232756
$endgroup$
$begingroup$
Thanks for the explanation! I get most of it but wouldn't $n= xyB^2 + Bx^2 + By^2 + xy$? Do the other equations hold?
$endgroup$
– S. L.
Apr 5 at 19:03
$begingroup$
@S.L. Woops, different equation, but same principle.
$endgroup$
– Gilles
Apr 5 at 20:39
add a comment |
$begingroup$
Here's how to recover $x, y$ in a way that's easier than factoring $n$ (I'll use the notation $x, y$ rather than your notation $ab, cd$):
We have $n = xyB^2 + (x^2+y^2)B + xy$
First, compute $n bmod B$, that gives you $xy bmod B$
Then, compute $lfloor (n - B^2(xy bmod B)) / B^3 rfloor$; this gives you $xy / B + epsilon$, where $0 le epsilon le 2$
Pasting those two together will give you a total of three possibilities of $xy$.
Then, for each possibility, compute $(n - xyB^2 - xy) / B + 2xy$ and $(n - xyB^2 - xy) / B - 2xy$; if the guess of $epsilon$ is correct, these will be $(x+y)^2$ and $(x-y)^2$; take squareroots, and extract $x, y$ directly.
(Thanks for Giles for pointing out this last part)
answered Apr 5 at 20:16
ponchoponcho
94k2148247
$endgroup$
$begingroup$
Yeah, right, the $B^3$ digit of $n$ gives the other digit of $x y$. And there's no need to factor anything: once you know $x y$, you know $x^2 + y^2$.
$endgroup$
– Gilles
Apr 5 at 20:39
$begingroup$
@Gilles: yup, you're right; I'll update the answer
$endgroup$
– poncho
Apr 5 at 21:01
$begingroup$
I don't get this part: Then, compute $⌊(n−B^2(xymod B))/B^3⌋$ this gives you $xy/B+ϵ$, where $0≤ϵ≤2$. I have $xymod B$ but not $xy$?
$endgroup$
– S. L.
Apr 5 at 21:10
1
$begingroup$
$(n - B^2(xy bmod B)) / B^3 = lfloor(xy/B) rfloor + x^2 / B^2 + y^2 / B^2 + xy / B^3$; we know that $x^2 / B^2, y^2 / B^2, xy / B^3$ are all less than 1 (and $ge 0$), and so the sum must be in the interval $[0, 3)$, that is, two or less once you round down...
$endgroup$
– poncho
Apr 5 at 22:18
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You don't need anything fancy like Coppersmith, just simple algebra. The idea is to translate the equations we have involving the digits of $p$ and $q$ in base $B$ ($B = 100$ in your example) into equations involving the digits of $n$ in base $B$, which we know. You have $p = x B + y$ and $q = y B + x$, with $0 lt x, y lt B$. Then $n = x y B^2 + (x^2 + y^2) B + x y$.
The rightmost digit of $n$ in base $B$ is $(x y) bmod B$. Since ${x,y} le B-1$, $(x^2 + y^2) B + x y le 2 (B-1)^2 B + (B-1)^2 lt 2 (B-1)^2 (B+1) = 2 (B-1) (B^2-1) lt 2 B^3$. Hence the $B^3$ digit of $n$ is the $B$ digit of $x y$ plus $z$ where $0 le z lt 2$, i.e. $z in {0, 1}$. So by reading the digits of $n$ in base $B$, we get the digits of $x y$ in base $B$, up to two possibilities, giving just two possibilities for $x y$ itself: $x y in {W_0, W_1}$.
Injecting this knowledge into the equation above gives us $x^2 + y^2 = (n - W_z (B^2 + 1)) / B$. And of course knowing both $x^2 + y^2$ and $x y$ gives $x$ and $y$.
answered Apr 5 at 17:27
GillesGilles
8,44232756
$endgroup$
$begingroup$
Thanks for the explanation! I get most of it but wouldn't $n= xyB^2 + Bx^2 + By^2 + xy$? Do the other equations hold?
$endgroup$
– S. L.
Apr 5 at 19:03
$begingroup$
@S.L. Woops, different equation, but same principle.
$endgroup$
– Gilles
Apr 5 at 20:39
add a comment |
$begingroup$
You don't need anything fancy like Coppersmith, just simple algebra. The idea is to translate the equations we have involving the digits of $p$ and $q$ in base $B$ ($B = 100$ in your example) into equations involving the digits of $n$ in base $B$, which we know. You have $p = x B + y$ and $q = y B + x$, with $0 lt x, y lt B$. Then $n = x y B^2 + (x^2 + y^2) B + x y$.
The rightmost digit of $n$ in base $B$ is $(x y) bmod B$. Since ${x,y} le B-1$, $(x^2 + y^2) B + x y le 2 (B-1)^2 B + (B-1)^2 lt 2 (B-1)^2 (B+1) = 2 (B-1) (B^2-1) lt 2 B^3$. Hence the $B^3$ digit of $n$ is the $B$ digit of $x y$ plus $z$ where $0 le z lt 2$, i.e. $z in {0, 1}$. So by reading the digits of $n$ in base $B$, we get the digits of $x y$ in base $B$, up to two possibilities, giving just two possibilities for $x y$ itself: $x y in {W_0, W_1}$.
Injecting this knowledge into the equation above gives us $x^2 + y^2 = (n - W_z (B^2 + 1)) / B$. And of course knowing both $x^2 + y^2$ and $x y$ gives $x$ and $y$.
answered Apr 5 at 17:27
GillesGilles
8,44232756
$endgroup$
$begingroup$
Thanks for the explanation! I get most of it but wouldn't $n= xyB^2 + Bx^2 + By^2 + xy$? Do the other equations hold?
$endgroup$
– S. L.
Apr 5 at 19:03
$begingroup$
@S.L. Woops, different equation, but same principle.
$endgroup$
– Gilles
Apr 5 at 20:39
add a comment |
$begingroup$
You don't need anything fancy like Coppersmith, just simple algebra. The idea is to translate the equations we have involving the digits of $p$ and $q$ in base $B$ ($B = 100$ in your example) into equations involving the digits of $n$ in base $B$, which we know. You have $p = x B + y$ and $q = y B + x$, with $0 lt x, y lt B$. Then $n = x y B^2 + (x^2 + y^2) B + x y$.
The rightmost digit of $n$ in base $B$ is $(x y) bmod B$. Since ${x,y} le B-1$, $(x^2 + y^2) B + x y le 2 (B-1)^2 B + (B-1)^2 lt 2 (B-1)^2 (B+1) = 2 (B-1) (B^2-1) lt 2 B^3$. Hence the $B^3$ digit of $n$ is the $B$ digit of $x y$ plus $z$ where $0 le z lt 2$, i.e. $z in {0, 1}$. So by reading the digits of $n$ in base $B$, we get the digits of $x y$ in base $B$, up to two possibilities, giving just two possibilities for $x y$ itself: $x y in {W_0, W_1}$.
Injecting this knowledge into the equation above gives us $x^2 + y^2 = (n - W_z (B^2 + 1)) / B$. And of course knowing both $x^2 + y^2$ and $x y$ gives $x$ and $y$.
answered Apr 5 at 17:27
GillesGilles
8,44232756
$endgroup$
You don't need anything fancy like Coppersmith, just simple algebra. The idea is to translate the equations we have involving the digits of $p$ and $q$ in base $B$ ($B = 100$ in your example) into equations involving the digits of $n$ in base $B$, which we know. You have $p = x B + y$ and $q = y B + x$, with $0 lt x, y lt B$. Then $n = x y B^2 + (x^2 + y^2) B + x y$.
The rightmost digit of $n$ in base $B$ is $(x y) bmod B$. Since ${x,y} le B-1$, $(x^2 + y^2) B + x y le 2 (B-1)^2 B + (B-1)^2 lt 2 (B-1)^2 (B+1) = 2 (B-1) (B^2-1) lt 2 B^3$. Hence the $B^3$ digit of $n$ is the $B$ digit of $x y$ plus $z$ where $0 le z lt 2$, i.e. $z in {0, 1}$. So by reading the digits of $n$ in base $B$, we get the digits of $x y$ in base $B$, up to two possibilities, giving just two possibilities for $x y$ itself: $x y in {W_0, W_1}$.
Injecting this knowledge into the equation above gives us $x^2 + y^2 = (n - W_z (B^2 + 1)) / B$. And of course knowing both $x^2 + y^2$ and $x y$ gives $x$ and $y$.
answered Apr 5 at 17:27
GillesGilles
8,44232756
edited Apr 5 at 20:38
answered Apr 5 at 17:27
GillesGilles
8,44232756
answered Apr 5 at 17:27
GillesGilles
8,44232756
answered Apr 5 at 17:27
GillesGilles
8,44232756
8,44232756
$begingroup$
Thanks for the explanation! I get most of it but wouldn't $n= xyB^2 + Bx^2 + By^2 + xy$? Do the other equations hold?
$endgroup$
– S. L.
Apr 5 at 19:03
$begingroup$
@S.L. Woops, different equation, but same principle.
$endgroup$
– Gilles
Apr 5 at 20:39
add a comment |
$begingroup$
Thanks for the explanation! I get most of it but wouldn't $n= xyB^2 + Bx^2 + By^2 + xy$? Do the other equations hold?
$endgroup$
– S. L.
Apr 5 at 19:03
$begingroup$
@S.L. Woops, different equation, but same principle.
$endgroup$
– Gilles
Apr 5 at 20:39
$begingroup$
Thanks for the explanation! I get most of it but wouldn't $n= xyB^2 + Bx^2 + By^2 + xy$? Do the other equations hold?
$endgroup$
– S. L.
Apr 5 at 19:03
$begingroup$
Thanks for the explanation! I get most of it but wouldn't $n= xyB^2 + Bx^2 + By^2 + xy$? Do the other equations hold?
$endgroup$
– S. L.
Apr 5 at 19:03
$begingroup$
@S.L. Woops, different equation, but same principle.
$endgroup$
– Gilles
Apr 5 at 20:39
$begingroup$
@S.L. Woops, different equation, but same principle.
$endgroup$
– Gilles
Apr 5 at 20:39
add a comment |
$begingroup$
Here's how to recover $x, y$ in a way that's easier than factoring $n$ (I'll use the notation $x, y$ rather than your notation $ab, cd$):
We have $n = xyB^2 + (x^2+y^2)B + xy$
First, compute $n bmod B$, that gives you $xy bmod B$
Then, compute $lfloor (n - B^2(xy bmod B)) / B^3 rfloor$; this gives you $xy / B + epsilon$, where $0 le epsilon le 2$
Pasting those two together will give you a total of three possibilities of $xy$.
Then, for each possibility, compute $(n - xyB^2 - xy) / B + 2xy$ and $(n - xyB^2 - xy) / B - 2xy$; if the guess of $epsilon$ is correct, these will be $(x+y)^2$ and $(x-y)^2$; take squareroots, and extract $x, y$ directly.
(Thanks for Giles for pointing out this last part)
answered Apr 5 at 20:16
ponchoponcho
94k2148247
$endgroup$
$begingroup$
Yeah, right, the $B^3$ digit of $n$ gives the other digit of $x y$. And there's no need to factor anything: once you know $x y$, you know $x^2 + y^2$.
$endgroup$
– Gilles
Apr 5 at 20:39
$begingroup$
@Gilles: yup, you're right; I'll update the answer
$endgroup$
– poncho
Apr 5 at 21:01
$begingroup$
I don't get this part: Then, compute $⌊(n−B^2(xymod B))/B^3⌋$ this gives you $xy/B+ϵ$, where $0≤ϵ≤2$. I have $xymod B$ but not $xy$?
$endgroup$
– S. L.
Apr 5 at 21:10
1
$begingroup$
$(n - B^2(xy bmod B)) / B^3 = lfloor(xy/B) rfloor + x^2 / B^2 + y^2 / B^2 + xy / B^3$; we know that $x^2 / B^2, y^2 / B^2, xy / B^3$ are all less than 1 (and $ge 0$), and so the sum must be in the interval $[0, 3)$, that is, two or less once you round down...
$endgroup$
– poncho
Apr 5 at 22:18
add a comment |
$begingroup$
Here's how to recover $x, y$ in a way that's easier than factoring $n$ (I'll use the notation $x, y$ rather than your notation $ab, cd$):
We have $n = xyB^2 + (x^2+y^2)B + xy$
First, compute $n bmod B$, that gives you $xy bmod B$
Then, compute $lfloor (n - B^2(xy bmod B)) / B^3 rfloor$; this gives you $xy / B + epsilon$, where $0 le epsilon le 2$
Pasting those two together will give you a total of three possibilities of $xy$.
Then, for each possibility, compute $(n - xyB^2 - xy) / B + 2xy$ and $(n - xyB^2 - xy) / B - 2xy$; if the guess of $epsilon$ is correct, these will be $(x+y)^2$ and $(x-y)^2$; take squareroots, and extract $x, y$ directly.
(Thanks for Giles for pointing out this last part)
answered Apr 5 at 20:16
ponchoponcho
94k2148247
$endgroup$
$begingroup$
Yeah, right, the $B^3$ digit of $n$ gives the other digit of $x y$. And there's no need to factor anything: once you know $x y$, you know $x^2 + y^2$.
$endgroup$
– Gilles
Apr 5 at 20:39
$begingroup$
@Gilles: yup, you're right; I'll update the answer
$endgroup$
– poncho
Apr 5 at 21:01
$begingroup$
I don't get this part: Then, compute $⌊(n−B^2(xymod B))/B^3⌋$ this gives you $xy/B+ϵ$, where $0≤ϵ≤2$. I have $xymod B$ but not $xy$?
$endgroup$
– S. L.
Apr 5 at 21:10
1
$begingroup$
$(n - B^2(xy bmod B)) / B^3 = lfloor(xy/B) rfloor + x^2 / B^2 + y^2 / B^2 + xy / B^3$; we know that $x^2 / B^2, y^2 / B^2, xy / B^3$ are all less than 1 (and $ge 0$), and so the sum must be in the interval $[0, 3)$, that is, two or less once you round down...
$endgroup$
– poncho
Apr 5 at 22:18
add a comment |
$begingroup$
Here's how to recover $x, y$ in a way that's easier than factoring $n$ (I'll use the notation $x, y$ rather than your notation $ab, cd$):
We have $n = xyB^2 + (x^2+y^2)B + xy$
First, compute $n bmod B$, that gives you $xy bmod B$
Then, compute $lfloor (n - B^2(xy bmod B)) / B^3 rfloor$; this gives you $xy / B + epsilon$, where $0 le epsilon le 2$
Pasting those two together will give you a total of three possibilities of $xy$.
Then, for each possibility, compute $(n - xyB^2 - xy) / B + 2xy$ and $(n - xyB^2 - xy) / B - 2xy$; if the guess of $epsilon$ is correct, these will be $(x+y)^2$ and $(x-y)^2$; take squareroots, and extract $x, y$ directly.
(Thanks for Giles for pointing out this last part)
answered Apr 5 at 20:16
ponchoponcho
94k2148247
$endgroup$
Here's how to recover $x, y$ in a way that's easier than factoring $n$ (I'll use the notation $x, y$ rather than your notation $ab, cd$):
We have $n = xyB^2 + (x^2+y^2)B + xy$
First, compute $n bmod B$, that gives you $xy bmod B$
Then, compute $lfloor (n - B^2(xy bmod B)) / B^3 rfloor$; this gives you $xy / B + epsilon$, where $0 le epsilon le 2$
Pasting those two together will give you a total of three possibilities of $xy$.
Then, for each possibility, compute $(n - xyB^2 - xy) / B + 2xy$ and $(n - xyB^2 - xy) / B - 2xy$; if the guess of $epsilon$ is correct, these will be $(x+y)^2$ and $(x-y)^2$; take squareroots, and extract $x, y$ directly.
(Thanks for Giles for pointing out this last part)
answered Apr 5 at 20:16
ponchoponcho
94k2148247
edited Apr 5 at 21:07
answered Apr 5 at 20:16
ponchoponcho
94k2148247
answered Apr 5 at 20:16
ponchoponcho
94k2148247
answered Apr 5 at 20:16
ponchoponcho
94k2148247
94k2148247
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Yeah, right, the $B^3$ digit of $n$ gives the other digit of $x y$. And there's no need to factor anything: once you know $x y$, you know $x^2 + y^2$.
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– Gilles
Apr 5 at 20:39
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@Gilles: yup, you're right; I'll update the answer
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– poncho
Apr 5 at 21:01
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I don't get this part: Then, compute $⌊(n−B^2(xymod B))/B^3⌋$ this gives you $xy/B+ϵ$, where $0≤ϵ≤2$. I have $xymod B$ but not $xy$?
$endgroup$
– S. L.
Apr 5 at 21:10
1
$begingroup$
$(n - B^2(xy bmod B)) / B^3 = lfloor(xy/B) rfloor + x^2 / B^2 + y^2 / B^2 + xy / B^3$; we know that $x^2 / B^2, y^2 / B^2, xy / B^3$ are all less than 1 (and $ge 0$), and so the sum must be in the interval $[0, 3)$, that is, two or less once you round down...
$endgroup$
– poncho
Apr 5 at 22:18
add a comment |
$begingroup$
Yeah, right, the $B^3$ digit of $n$ gives the other digit of $x y$. And there's no need to factor anything: once you know $x y$, you know $x^2 + y^2$.
$endgroup$
– Gilles
Apr 5 at 20:39
$begingroup$
@Gilles: yup, you're right; I'll update the answer
$endgroup$
– poncho
Apr 5 at 21:01
$begingroup$
I don't get this part: Then, compute $⌊(n−B^2(xymod B))/B^3⌋$ this gives you $xy/B+ϵ$, where $0≤ϵ≤2$. I have $xymod B$ but not $xy$?
$endgroup$
– S. L.
Apr 5 at 21:10
1
$begingroup$
$(n - B^2(xy bmod B)) / B^3 = lfloor(xy/B) rfloor + x^2 / B^2 + y^2 / B^2 + xy / B^3$; we know that $x^2 / B^2, y^2 / B^2, xy / B^3$ are all less than 1 (and $ge 0$), and so the sum must be in the interval $[0, 3)$, that is, two or less once you round down...
$endgroup$
– poncho
Apr 5 at 22:18
$begingroup$
Yeah, right, the $B^3$ digit of $n$ gives the other digit of $x y$. And there's no need to factor anything: once you know $x y$, you know $x^2 + y^2$.
$endgroup$
– Gilles
Apr 5 at 20:39
$begingroup$
Yeah, right, the $B^3$ digit of $n$ gives the other digit of $x y$. And there's no need to factor anything: once you know $x y$, you know $x^2 + y^2$.
$endgroup$
– Gilles
Apr 5 at 20:39
$begingroup$
@Gilles: yup, you're right; I'll update the answer
$endgroup$
– poncho
Apr 5 at 21:01
$begingroup$
@Gilles: yup, you're right; I'll update the answer
$endgroup$
– poncho
Apr 5 at 21:01
$begingroup$
I don't get this part: Then, compute $⌊(n−B^2(xymod B))/B^3⌋$ this gives you $xy/B+ϵ$, where $0≤ϵ≤2$. I have $xymod B$ but not $xy$?
$endgroup$
– S. L.
Apr 5 at 21:10
$begingroup$
I don't get this part: Then, compute $⌊(n−B^2(xymod B))/B^3⌋$ this gives you $xy/B+ϵ$, where $0≤ϵ≤2$. I have $xymod B$ but not $xy$?
$endgroup$
– S. L.
Apr 5 at 21:10
1
1
$begingroup$
$(n - B^2(xy bmod B)) / B^3 = lfloor(xy/B) rfloor + x^2 / B^2 + y^2 / B^2 + xy / B^3$; we know that $x^2 / B^2, y^2 / B^2, xy / B^3$ are all less than 1 (and $ge 0$), and so the sum must be in the interval $[0, 3)$, that is, two or less once you round down...
$endgroup$
– poncho
Apr 5 at 22:18
$begingroup$
$(n - B^2(xy bmod B)) / B^3 = lfloor(xy/B) rfloor + x^2 / B^2 + y^2 / B^2 + xy / B^3$; we know that $x^2 / B^2, y^2 / B^2, xy / B^3$ are all less than 1 (and $ge 0$), and so the sum must be in the interval $[0, 3)$, that is, two or less once you round down...
$endgroup$
– poncho
Apr 5 at 22:18
add a comment |
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Of course, any product of two 4-digit primes is trivially factorable by trial division anyway, since there are only 1061 primes between 1000 and 9999. Add in the digit reversal requirement, and there are only 76(!) possible pairs to consider.
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– Ilmari Karonen
Apr 6 at 1:37
1
$begingroup$
@Nat: My fault, I added the "$= pq$" for context in an edit, and didn't notice the potential ambiguity. I see Paŭlo has already fixed it.
$endgroup$
– Ilmari Karonen
Apr 6 at 12:19
$begingroup$
I'd just like to add that this is inspired by a contest that just ended (12 minutes ago).
$endgroup$
– enedil
Apr 6 at 16:12