Difference of Squares: Always More Than One Solution For Odd Composite #s?












1












$begingroup$


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 $O(sqrt{m})$ 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







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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 $O(sqrt{m})$ 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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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 $O(sqrt{m})$ 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







      share|cite|improve this question











      $endgroup$




      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 $O(sqrt{m})$ 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




      factoring






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jul 22 '18 at 21:38







      Ryan Pierce Williams

















      asked Jul 22 '18 at 21:24









      Ryan Pierce WilliamsRyan Pierce Williams

      1867




      1867






















          1 Answer
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          1












          $begingroup$

          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



          $$
          m = frac{c+1}{2}, quad n = frac{c-1}{2}
          $$



          for the former case, and



          $$
          m = frac{r+s}{2}, quad n = frac{r-s}{2}
          $$



          for the latter case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Awesome, thank you very much :) I'll accept as answer once it lets me do that
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:32










          • $begingroup$
            @RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:33










          • $begingroup$
            Says I gotta wait 8 minutes >.>
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:34










          • $begingroup$
            @RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:34






          • 2




            $begingroup$
            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.
            $endgroup$
            – Thomas Andrews
            Jul 22 '18 at 21:35











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          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



          $$
          m = frac{c+1}{2}, quad n = frac{c-1}{2}
          $$



          for the former case, and



          $$
          m = frac{r+s}{2}, quad n = frac{r-s}{2}
          $$



          for the latter case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Awesome, thank you very much :) I'll accept as answer once it lets me do that
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:32










          • $begingroup$
            @RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:33










          • $begingroup$
            Says I gotta wait 8 minutes >.>
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:34










          • $begingroup$
            @RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:34






          • 2




            $begingroup$
            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.
            $endgroup$
            – Thomas Andrews
            Jul 22 '18 at 21:35
















          1












          $begingroup$

          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



          $$
          m = frac{c+1}{2}, quad n = frac{c-1}{2}
          $$



          for the former case, and



          $$
          m = frac{r+s}{2}, quad n = frac{r-s}{2}
          $$



          for the latter case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Awesome, thank you very much :) I'll accept as answer once it lets me do that
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:32










          • $begingroup$
            @RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:33










          • $begingroup$
            Says I gotta wait 8 minutes >.>
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:34










          • $begingroup$
            @RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:34






          • 2




            $begingroup$
            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.
            $endgroup$
            – Thomas Andrews
            Jul 22 '18 at 21:35














          1












          1








          1





          $begingroup$

          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



          $$
          m = frac{c+1}{2}, quad n = frac{c-1}{2}
          $$



          for the former case, and



          $$
          m = frac{r+s}{2}, quad n = frac{r-s}{2}
          $$



          for the latter case.






          share|cite|improve this answer









          $endgroup$



          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



          $$
          m = frac{c+1}{2}, quad n = frac{c-1}{2}
          $$



          for the former case, and



          $$
          m = frac{r+s}{2}, quad n = frac{r-s}{2}
          $$



          for the latter case.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 22 '18 at 21:29









          Brian TungBrian Tung

          25.9k32555




          25.9k32555












          • $begingroup$
            Awesome, thank you very much :) I'll accept as answer once it lets me do that
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:32










          • $begingroup$
            @RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:33










          • $begingroup$
            Says I gotta wait 8 minutes >.>
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:34










          • $begingroup$
            @RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:34






          • 2




            $begingroup$
            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.
            $endgroup$
            – Thomas Andrews
            Jul 22 '18 at 21:35


















          • $begingroup$
            Awesome, thank you very much :) I'll accept as answer once it lets me do that
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:32










          • $begingroup$
            @RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:33










          • $begingroup$
            Says I gotta wait 8 minutes >.>
            $endgroup$
            – Ryan Pierce Williams
            Jul 22 '18 at 21:34










          • $begingroup$
            @RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
            $endgroup$
            – Brian Tung
            Jul 22 '18 at 21:34






          • 2




            $begingroup$
            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.
            $endgroup$
            – Thomas Andrews
            Jul 22 '18 at 21:35
















          $begingroup$
          Awesome, thank you very much :) I'll accept as answer once it lets me do that
          $endgroup$
          – Ryan Pierce Williams
          Jul 22 '18 at 21:32




          $begingroup$
          Awesome, thank you very much :) I'll accept as answer once it lets me do that
          $endgroup$
          – Ryan Pierce Williams
          Jul 22 '18 at 21:32












          $begingroup$
          @RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
          $endgroup$
          – Brian Tung
          Jul 22 '18 at 21:33




          $begingroup$
          @RyanPierceWilliams: Strange. It should always allow you to accept an answer; what it might not let you do is upvote it. :-/
          $endgroup$
          – Brian Tung
          Jul 22 '18 at 21:33












          $begingroup$
          Says I gotta wait 8 minutes >.>
          $endgroup$
          – Ryan Pierce Williams
          Jul 22 '18 at 21:34




          $begingroup$
          Says I gotta wait 8 minutes >.>
          $endgroup$
          – Ryan Pierce Williams
          Jul 22 '18 at 21:34












          $begingroup$
          @RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
          $endgroup$
          – Brian Tung
          Jul 22 '18 at 21:34




          $begingroup$
          @RyanPierceWilliams: Ahh, I see. Well, I'm in no rush. :-)
          $endgroup$
          – Brian Tung
          Jul 22 '18 at 21:34




          2




          2




          $begingroup$
          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.
          $endgroup$
          – Thomas Andrews
          Jul 22 '18 at 21:35




          $begingroup$
          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.
          $endgroup$
          – Thomas Andrews
          Jul 22 '18 at 21:35


















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