Prime sequence.












4














Suppose we have $ p, q$ where $ p$ and $q$ are twin primes between 6,10000. Now the following sequence $pq + 2$ generates 22 primes . These primes are all in the congruence class $ [1]_{100}$ I've been asked to prove why this is the case. I've spent a good time thinking about it and not got any ideas how to even start. Wondering if someone could give me just a direction at least to follow. Was considering the notion of parity as pq is always going to be odd.



Best regards






prime-numbers modular-arithmetic







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  • 1




    Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
    – lulu
    Dec 10 at 16:10












  • Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
    – Dead_Ling0
    Dec 10 at 16:12










  • $p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
    – reuns
    Dec 10 at 16:17










  • It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
    – Dead_Ling0
    Dec 10 at 16:19
















4














Suppose we have $ p, q$ where $ p$ and $q$ are twin primes between 6,10000. Now the following sequence $pq + 2$ generates 22 primes . These primes are all in the congruence class $ [1]_{100}$ I've been asked to prove why this is the case. I've spent a good time thinking about it and not got any ideas how to even start. Wondering if someone could give me just a direction at least to follow. Was considering the notion of parity as pq is always going to be odd.



Best regards






prime-numbers modular-arithmetic







share|cite|improve this question




















  • 1




    Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
    – lulu
    Dec 10 at 16:10












  • Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
    – Dead_Ling0
    Dec 10 at 16:12










  • $p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
    – reuns
    Dec 10 at 16:17










  • It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
    – Dead_Ling0
    Dec 10 at 16:19














4












4








4







Suppose we have $ p, q$ where $ p$ and $q$ are twin primes between 6,10000. Now the following sequence $pq + 2$ generates 22 primes . These primes are all in the congruence class $ [1]_{100}$ I've been asked to prove why this is the case. I've spent a good time thinking about it and not got any ideas how to even start. Wondering if someone could give me just a direction at least to follow. Was considering the notion of parity as pq is always going to be odd.



Best regards






prime-numbers modular-arithmetic







share|cite|improve this question















Suppose we have $ p, q$ where $ p$ and $q$ are twin primes between 6,10000. Now the following sequence $pq + 2$ generates 22 primes . These primes are all in the congruence class $ [1]_{100}$ I've been asked to prove why this is the case. I've spent a good time thinking about it and not got any ideas how to even start. Wondering if someone could give me just a direction at least to follow. Was considering the notion of parity as pq is always going to be odd.



Best regards






prime-numbers modular-arithmetic




prime-numbers modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 at 17:49

























asked Dec 10 at 16:06









Dead_Ling0

406




406








  • 1




    Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
    – lulu
    Dec 10 at 16:10












  • Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
    – Dead_Ling0
    Dec 10 at 16:12










  • $p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
    – reuns
    Dec 10 at 16:17










  • It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
    – Dead_Ling0
    Dec 10 at 16:19














  • 1




    Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
    – lulu
    Dec 10 at 16:10












  • Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
    – Dead_Ling0
    Dec 10 at 16:12










  • $p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
    – reuns
    Dec 10 at 16:17










  • It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
    – Dead_Ling0
    Dec 10 at 16:19








1




1




Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
– lulu
Dec 10 at 16:10






Not following. $(3,5)$ are twin primes and $17=3times 5+2$ is a prime. But $17not equiv 1pmod {100}$. Similarly with $37=5times 7+2$. What do you mean?
– lulu
Dec 10 at 16:10














Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
– Dead_Ling0
Dec 10 at 16:12




Sorry just realised I didn't write the question particularly well, p, q should be twin primes between 6, and 10000.
– Dead_Ling0
Dec 10 at 16:12












$p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
– reuns
Dec 10 at 16:17




$p,p+2,p(p+2)+2$ are primes implies $pequiv -1 bmod 5$
– reuns
Dec 10 at 16:17












It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
– Dead_Ling0
Dec 10 at 16:19




It doesn't have to be prime. Like I said in my question only 22 primes are generated this way the first is when we have (149,151).
– Dead_Ling0
Dec 10 at 16:19










4 Answers
4






active

oldest

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1














Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then



$$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$




Therefore $p(p+2)+2 equiv 1 pmod {100}$







share|cite|improve this answer































    2














    Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).






    share|cite|improve this answer























    • Will give that a look now thanks for posting :)
      – Dead_Ling0
      Dec 10 at 16:28



















    2














    The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.






    share|cite|improve this answer





























      0














      Rather than considering parity, you might want to consider the situation modulo $6$ to start with.



      If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$



      We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$






      share|cite|improve this answer























      • How does mod 6 help any more than mod 2 with finding results mod 100?
        – Arthur
        Dec 10 at 16:25












      • @Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
        – Mark Bennet
        Dec 10 at 19:48










      • The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
        – Arthur
        Dec 10 at 19:57












      • @Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
        – Mark Bennet
        Dec 10 at 20:09











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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then



      $$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$




      Therefore $p(p+2)+2 equiv 1 pmod {100}$







      share|cite|improve this answer




























        1














        Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then



        $$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$




        Therefore $p(p+2)+2 equiv 1 pmod {100}$







        share|cite|improve this answer


























          1












          1








          1






          Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then



          $$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$




          Therefore $p(p+2)+2 equiv 1 pmod {100}$







          share|cite|improve this answer














          Look at the possible remainders $bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $pequiv 9pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then



          $$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$




          Therefore $p(p+2)+2 equiv 1 pmod {100}$








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 at 17:02

























          answered Dec 10 at 16:55









          gammatester

          16.7k21632




          16.7k21632























              2














              Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).






              share|cite|improve this answer























              • Will give that a look now thanks for posting :)
                – Dead_Ling0
                Dec 10 at 16:28
















              2














              Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).






              share|cite|improve this answer























              • Will give that a look now thanks for posting :)
                – Dead_Ling0
                Dec 10 at 16:28














              2












              2








              2






              Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).






              share|cite|improve this answer














              Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 10 at 16:25

























              answered Dec 10 at 16:15









              Arthur

              110k7105186




              110k7105186












              • Will give that a look now thanks for posting :)
                – Dead_Ling0
                Dec 10 at 16:28


















              • Will give that a look now thanks for posting :)
                – Dead_Ling0
                Dec 10 at 16:28
















              Will give that a look now thanks for posting :)
              – Dead_Ling0
              Dec 10 at 16:28




              Will give that a look now thanks for posting :)
              – Dead_Ling0
              Dec 10 at 16:28











              2














              The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.






              share|cite|improve this answer


























                2














                The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.






                share|cite|improve this answer
























                  2












                  2








                  2






                  The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.






                  share|cite|improve this answer












                  The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $pequiv -1 mod 5$. But consider further that if $pequiv -1 mod 5$, then $p+2equiv 1 mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2equiv 1 mod 10Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2equiv 1 mod 100$ as you observe.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 at 16:59









                  Keith Backman

                  9941612




                  9941612























                      0














                      Rather than considering parity, you might want to consider the situation modulo $6$ to start with.



                      If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$



                      We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$






                      share|cite|improve this answer























                      • How does mod 6 help any more than mod 2 with finding results mod 100?
                        – Arthur
                        Dec 10 at 16:25












                      • @Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
                        – Mark Bennet
                        Dec 10 at 19:48










                      • The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
                        – Arthur
                        Dec 10 at 19:57












                      • @Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
                        – Mark Bennet
                        Dec 10 at 20:09
















                      0














                      Rather than considering parity, you might want to consider the situation modulo $6$ to start with.



                      If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$



                      We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$






                      share|cite|improve this answer























                      • How does mod 6 help any more than mod 2 with finding results mod 100?
                        – Arthur
                        Dec 10 at 16:25












                      • @Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
                        – Mark Bennet
                        Dec 10 at 19:48










                      • The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
                        – Arthur
                        Dec 10 at 19:57












                      • @Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
                        – Mark Bennet
                        Dec 10 at 20:09














                      0












                      0








                      0






                      Rather than considering parity, you might want to consider the situation modulo $6$ to start with.



                      If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$



                      We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$






                      share|cite|improve this answer














                      Rather than considering parity, you might want to consider the situation modulo $6$ to start with.



                      If our three primes are $p, p+2, p^2+2p+2$ we have $pequiv -1 bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $pequiv -1 bmod 5$ to avoid divisibility by $5$



                      We therefore have $pequiv -1 bmod 30$ and $p(p+2)+2equiv 1 bmod 900$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 10 at 20:16

























                      answered Dec 10 at 16:12









                      Mark Bennet

                      80.5k981179




                      80.5k981179












                      • How does mod 6 help any more than mod 2 with finding results mod 100?
                        – Arthur
                        Dec 10 at 16:25












                      • @Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
                        – Mark Bennet
                        Dec 10 at 19:48










                      • The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
                        – Arthur
                        Dec 10 at 19:57












                      • @Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
                        – Mark Bennet
                        Dec 10 at 20:09


















                      • How does mod 6 help any more than mod 2 with finding results mod 100?
                        – Arthur
                        Dec 10 at 16:25












                      • @Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
                        – Mark Bennet
                        Dec 10 at 19:48










                      • The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
                        – Arthur
                        Dec 10 at 19:57












                      • @Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
                        – Mark Bennet
                        Dec 10 at 20:09
















                      How does mod 6 help any more than mod 2 with finding results mod 100?
                      – Arthur
                      Dec 10 at 16:25






                      How does mod 6 help any more than mod 2 with finding results mod 100?
                      – Arthur
                      Dec 10 at 16:25














                      @Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
                      – Mark Bennet
                      Dec 10 at 19:48




                      @Arthur Your two primes are $6n+1$ and $6n-1$ and $pq+2=36n^2+1$. That deals with the issue modulo $4$. Then (since the two are prime) we can't have $nequiv pm 1 bmod 5$ (we'd have a factor $5$) etc. I was suggesting it as a way into the problem which got further than "parity".
                      – Mark Bennet
                      Dec 10 at 19:48












                      The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
                      – Arthur
                      Dec 10 at 19:57






                      The two primes are $2npm 1$, and $pq+2=4n^2+1$. That deals with the issue modulo $4$ just as well as your suggestion, and you can draw exactly the same conclusions from there. It doesn't get further than "parity" in any way. Mod 6 is just mod 2 and mod 3 combined, and the mod 3 doesn't help us in any way here.
                      – Arthur
                      Dec 10 at 19:57














                      @Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
                      – Mark Bennet
                      Dec 10 at 20:09




                      @Arthur You are right, of course - I was posting before going off to a meeting, and in too much haste. However, looking modulo $6$ and combining with modulo $5$ gives an interestingly stronger result, which I have noted in an amended answer. My instinct with twin primes is to start with $6npm 1$ and deal with the case of $3$ separately.
                      – Mark Bennet
                      Dec 10 at 20:09


















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