Mixing
Difference of Squares: Always More Than One Solution For Odd Composite #s?
Clash Royale CLAN TAG#URR8PPP
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I've been looking at the pattern in the difference of squares recently. I've seen proofs that for a given prime, there is a unique pair of squares that differ by that prime.
What I'd like to know is: is there guaranteed to be more than one pair of squares with a difference of $m$ where $m$ is an odd, composite integer $> 0$?
The sequence of values $q^2 - p^2$ where $p = q - 1$ will produce a difference of squares of the form $( q - p )( q + p ) = ( 1 )( 2q - 1 ) = m$. There is always a solution for m using this approach, but it's not terribly interesting for factorization purposes. I'd like to find solutions where $p = q - k | k > 1$.
I can perform a search in $O(sqrtm)$ to find one/all such solutions - but I'm not sure that there is always a solution for $m$ (other than $k = 1$). Are there any existing proofs on this matter?
factoring
asked Jul 22 at 21:24
Ryan Pierce Williams
817
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up vote
0
down vote
favorite
I've been looking at the pattern in the difference of squares recently. I've seen proofs that for a given prime, there is a unique pair of squares that differ by that prime.
What I'd like to know is: is there guaranteed to be more than one pair of squares with a difference of $m$ where $m$ is an odd, composite integer $> 0$?
The sequence of values $q^2 - p^2$ where $p = q - 1$ will produce a difference of squares of the form $( q - p )( q + p ) = ( 1 )( 2q - 1 ) = m$. There is always a solution for m using this approach, but it's not terribly interesting for factorization purposes. I'd like to find solutions where $p = q - k | k > 1$.
I can perform a search in $O(sqrtm)$ to find one/all such solutions - but I'm not sure that there is always a solution for $m$ (other than $k = 1$). Are there any existing proofs on this matter?
factoring
asked Jul 22 at 21:24
Ryan Pierce Williams
817
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've been looking at the pattern in the difference of squares recently. I've seen proofs that for a given prime, there is a unique pair of squares that differ by that prime.
What I'd like to know is: is there guaranteed to be more than one pair of squares with a difference of $m$ where $m$ is an odd, composite integer $> 0$?
The sequence of values $q^2 - p^2$ where $p = q - 1$ will produce a difference of squares of the form $( q - p )( q + p ) = ( 1 )( 2q - 1 ) = m$. There is always a solution for m using this approach, but it's not terribly interesting for factorization purposes. I'd like to find solutions where $p = q - k | k > 1$.
I can perform a search in $O(sqrtm)$ to find one/all such solutions - but I'm not sure that there is always a solution for $m$ (other than $k = 1$). Are there any existing proofs on this matter?
factoring
asked Jul 22 at 21:24
Ryan Pierce Williams
817
I've been looking at the pattern in the difference of squares recently. I've seen proofs that for a given prime, there is a unique pair of squares that differ by that prime.
What I'd like to know is: is there guaranteed to be more than one pair of squares with a difference of $m$ where $m$ is an odd, composite integer $> 0$?
The sequence of values $q^2 - p^2$ where $p = q - 1$ will produce a difference of squares of the form $( q - p )( q + p ) = ( 1 )( 2q - 1 ) = m$. There is always a solution for m using this approach, but it's not terribly interesting for factorization purposes. I'd like to find solutions where $p = q - k | k > 1$.
I can perform a search in $O(sqrtm)$ to find one/all such solutions - but I'm not sure that there is always a solution for $m$ (other than $k = 1$). Are there any existing proofs on this matter?
factoring
asked Jul 22 at 21:24
Ryan Pierce Williams
817
edited Jul 22 at 21:38
asked Jul 22 at 21:24
Ryan Pierce Williams
817
asked Jul 22 at 21:24
Ryan Pierce Williams
817
asked Jul 22 at 21:24
Ryan Pierce Williams
817
817
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Write $m^2-n^2 = c$ as $(m+n)(m-n) = c$, for odd composite $c$. Since $c$ is composite, it can be written as a product of two natural numbers in at least two different ways: $c times 1$ and $r times s$, for odd $r > s$ (without loss of generality). So we can have
$$
m = fracc+12, quad n = fracc-12
$$
for the former case, and
$$
m = fracr+s2, quad n = fracr-s2
$$
for the latter case.
answered Jul 22 at 21:29
Brian Tung
25.2k32453
Awesome, thank you very much :) I'll accept as answer once it lets me do that
– Ryan Pierce Williams
Jul 22 at 21:32
@RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
– Brian Tung
Jul 22 at 21:33
Says I gotta wait 8 minutes >.>
– Ryan Pierce Williams
Jul 22 at 21:34
@RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
– Brian Tung
Jul 22 at 21:34
1
Yeah, was looking for that, and found it, eventually, but it would have been easier to find if you had stuck with OPs original letters.
– Thomas Andrews
Jul 22 at 21:35
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Write $m^2-n^2 = c$ as $(m+n)(m-n) = c$, for odd composite $c$. Since $c$ is composite, it can be written as a product of two natural numbers in at least two different ways: $c times 1$ and $r times s$, for odd $r > s$ (without loss of generality). So we can have
$$
m = fracc+12, quad n = fracc-12
$$
for the former case, and
$$
m = fracr+s2, quad n = fracr-s2
$$
for the latter case.
answered Jul 22 at 21:29
Brian Tung
25.2k32453
Awesome, thank you very much :) I'll accept as answer once it lets me do that
– Ryan Pierce Williams
Jul 22 at 21:32
@RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
– Brian Tung
Jul 22 at 21:33
Says I gotta wait 8 minutes >.>
– Ryan Pierce Williams
Jul 22 at 21:34
@RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
– Brian Tung
Jul 22 at 21:34
1
Yeah, was looking for that, and found it, eventually, but it would have been easier to find if you had stuck with OPs original letters.
– Thomas Andrews
Jul 22 at 21:35
 |Â
show 4 more comments
up vote
1
down vote
accepted
Write $m^2-n^2 = c$ as $(m+n)(m-n) = c$, for odd composite $c$. Since $c$ is composite, it can be written as a product of two natural numbers in at least two different ways: $c times 1$ and $r times s$, for odd $r > s$ (without loss of generality). So we can have
$$
m = fracc+12, quad n = fracc-12
$$
for the former case, and
$$
m = fracr+s2, quad n = fracr-s2
$$
for the latter case.
answered Jul 22 at 21:29
Brian Tung
25.2k32453
Awesome, thank you very much :) I'll accept as answer once it lets me do that
– Ryan Pierce Williams
Jul 22 at 21:32
@RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
– Brian Tung
Jul 22 at 21:33
Says I gotta wait 8 minutes >.>
– Ryan Pierce Williams
Jul 22 at 21:34
@RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
– Brian Tung
Jul 22 at 21:34
1
Yeah, was looking for that, and found it, eventually, but it would have been easier to find if you had stuck with OPs original letters.
– Thomas Andrews
Jul 22 at 21:35
 |Â
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Write $m^2-n^2 = c$ as $(m+n)(m-n) = c$, for odd composite $c$. Since $c$ is composite, it can be written as a product of two natural numbers in at least two different ways: $c times 1$ and $r times s$, for odd $r > s$ (without loss of generality). So we can have
$$
m = fracc+12, quad n = fracc-12
$$
for the former case, and
$$
m = fracr+s2, quad n = fracr-s2
$$
for the latter case.
answered Jul 22 at 21:29
Brian Tung
25.2k32453
Write $m^2-n^2 = c$ as $(m+n)(m-n) = c$, for odd composite $c$. Since $c$ is composite, it can be written as a product of two natural numbers in at least two different ways: $c times 1$ and $r times s$, for odd $r > s$ (without loss of generality). So we can have
$$
m = fracc+12, quad n = fracc-12
$$
for the former case, and
$$
m = fracr+s2, quad n = fracr-s2
$$
for the latter case.
answered Jul 22 at 21:29
Brian Tung
25.2k32453
answered Jul 22 at 21:29
Brian Tung
25.2k32453
answered Jul 22 at 21:29
Brian Tung
25.2k32453
answered Jul 22 at 21:29
Brian Tung
25.2k32453
25.2k32453
Awesome, thank you very much :) I'll accept as answer once it lets me do that
– Ryan Pierce Williams
Jul 22 at 21:32
@RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
– Brian Tung
Jul 22 at 21:33
Says I gotta wait 8 minutes >.>
– Ryan Pierce Williams
Jul 22 at 21:34
@RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
– Brian Tung
Jul 22 at 21:34
1
Yeah, was looking for that, and found it, eventually, but it would have been easier to find if you had stuck with OPs original letters.
– Thomas Andrews
Jul 22 at 21:35
 |Â
show 4 more comments
Awesome, thank you very much :) I'll accept as answer once it lets me do that
– Ryan Pierce Williams
Jul 22 at 21:32
@RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
– Brian Tung
Jul 22 at 21:33
Says I gotta wait 8 minutes >.>
– Ryan Pierce Williams
Jul 22 at 21:34
@RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
– Brian Tung
Jul 22 at 21:34
1
Yeah, was looking for that, and found it, eventually, but it would have been easier to find if you had stuck with OPs original letters.
– Thomas Andrews
Jul 22 at 21:35
Awesome, thank you very much :) I'll accept as answer once it lets me do that
– Ryan Pierce Williams
Jul 22 at 21:32
Awesome, thank you very much :) I'll accept as answer once it lets me do that
– Ryan Pierce Williams
Jul 22 at 21:32
@RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
– Brian Tung
Jul 22 at 21:33
@RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
– Brian Tung
Jul 22 at 21:33
Says I gotta wait 8 minutes >.>
– Ryan Pierce Williams
Jul 22 at 21:34
Says I gotta wait 8 minutes >.>
– Ryan Pierce Williams
Jul 22 at 21:34
@RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
– Brian Tung
Jul 22 at 21:34
@RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
– Brian Tung
Jul 22 at 21:34
1
1
Yeah, was looking for that, and found it, eventually, but it would have been easier to find if you had stuck with OPs original letters.
– Thomas Andrews
Jul 22 at 21:35
Yeah, was looking for that, and found it, eventually, but it would have been easier to find if you had stuck with OPs original letters.
– Thomas Andrews
Jul 22 at 21:35
 |Â
show 4 more comments
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