Generating Prime Numbers From Composite Numbers











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If $a$ and $b$ are non-perfect-square composite numbers, and $gcd(a,b)=1$,
then at least one element of {$ab-1,ab+1$} is a prime number.




For example, if we let $a=35$, and $b=18$; clearly the are free-square composite numbers, and gcd$(35,18)=1$



$ab-1=35times18-1=629$ which is a composite number, and



$ab+1=35times18+1=631$ which is a prime number.



Is the statement true?






elementary-number-theory prime-numbers







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  • 3




    $18 = 3^2cdot 2$ is not square-free.
    – Arthur
    Nov 22 at 14:16






  • 1




    If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
    – Arthur
    Nov 22 at 14:38

















up vote
1
down vote

favorite
1













If $a$ and $b$ are non-perfect-square composite numbers, and $gcd(a,b)=1$,
then at least one element of {$ab-1,ab+1$} is a prime number.




For example, if we let $a=35$, and $b=18$; clearly the are free-square composite numbers, and gcd$(35,18)=1$



$ab-1=35times18-1=629$ which is a composite number, and



$ab+1=35times18+1=631$ which is a prime number.



Is the statement true?






elementary-number-theory prime-numbers







share|cite|improve this question




















  • 3




    $18 = 3^2cdot 2$ is not square-free.
    – Arthur
    Nov 22 at 14:16






  • 1




    If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
    – Arthur
    Nov 22 at 14:38















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






If $a$ and $b$ are non-perfect-square composite numbers, and $gcd(a,b)=1$,
then at least one element of {$ab-1,ab+1$} is a prime number.




For example, if we let $a=35$, and $b=18$; clearly the are free-square composite numbers, and gcd$(35,18)=1$



$ab-1=35times18-1=629$ which is a composite number, and



$ab+1=35times18+1=631$ which is a prime number.



Is the statement true?






elementary-number-theory prime-numbers







share|cite|improve this question
















If $a$ and $b$ are non-perfect-square composite numbers, and $gcd(a,b)=1$,
then at least one element of {$ab-1,ab+1$} is a prime number.




For example, if we let $a=35$, and $b=18$; clearly the are free-square composite numbers, and gcd$(35,18)=1$



$ab-1=35times18-1=629$ which is a composite number, and



$ab+1=35times18+1=631$ which is a prime number.



Is the statement true?






elementary-number-theory prime-numbers




elementary-number-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 10:02

























asked Nov 22 at 14:14









Hussain-Alqatari

2977




2977








  • 3




    $18 = 3^2cdot 2$ is not square-free.
    – Arthur
    Nov 22 at 14:16






  • 1




    If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
    – Arthur
    Nov 22 at 14:38
















  • 3




    $18 = 3^2cdot 2$ is not square-free.
    – Arthur
    Nov 22 at 14:16






  • 1




    If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
    – Arthur
    Nov 22 at 14:38










3




3




$18 = 3^2cdot 2$ is not square-free.
– Arthur
Nov 22 at 14:16




$18 = 3^2cdot 2$ is not square-free.
– Arthur
Nov 22 at 14:16




1




1




If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
– Arthur
Nov 22 at 14:38






If you multiply two coprime numbers with lots of small prime factors in them, you're going to get a medium-sized (3-4 digit) number with lots of small prime factors. If you add or subtract $1$ to / from that, you get a new number of the same size with no small prime factors. The odds of that number itself being prime is thus quite high. It's not a guarantee in any way, but it explains why it took me a minute or two of searching to find a counterexample.
– Arthur
Nov 22 at 14:38












2 Answers
2






active

oldest

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up vote
2
down vote



accepted










You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.



[edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:



714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262






share|cite|improve this answer






























    up vote
    3
    down vote













    If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.



    Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
    $$
    ab + 1 = 715 = 5cdot 11cdot 13\
    ab - 1 = 713 = 23cdot 31
    $$






    share|cite|improve this answer



















    • 1




      $a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
      – Especially Lime
      Nov 22 at 14:23










    • @EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
      – Arthur
      Nov 22 at 14:30













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    2 Answers
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    2 Answers
    2






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    up vote
    2
    down vote



    accepted










    You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.



    [edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:



    714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted










      You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.



      [edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:



      714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.



        [edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:



        714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262






        share|cite|improve this answer














        You certainly need at least one of $a,b$ to be even to have any hope that one of these is always prime. However, even then it's not true. For example, take $a=38,b=143$, so $ab=5434$. neither $5433$ nor $5435$ is prime.



        [edit] As Arthur points out my example is rather larger than necessary. $ab$ must be a product of at least four distinct primes. If it is a product of more than four primes, it must be at least $2310$ (which isn't a counterexample). So we can look at the values in this sequence below $2310$ and check whether the neighbouring numbers are composite to find the first (i.e. smallest $ab$) few counterexamples. They are:



        714, 870, 1155, 1190, 1254, 1330, 1365, 1518, 1590, 1770, 1785, 1794, 1806, 1938, 1995, 2046, 2145, 2170, 2190, 2210, 2226, 2262







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 23 at 9:35

























        answered Nov 22 at 14:29









        Especially Lime

        21.3k22656




        21.3k22656






















            up vote
            3
            down vote













            If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.



            Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
            $$
            ab + 1 = 715 = 5cdot 11cdot 13\
            ab - 1 = 713 = 23cdot 31
            $$






            share|cite|improve this answer



















            • 1




              $a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
              – Especially Lime
              Nov 22 at 14:23










            • @EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
              – Arthur
              Nov 22 at 14:30

















            up vote
            3
            down vote













            If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.



            Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
            $$
            ab + 1 = 715 = 5cdot 11cdot 13\
            ab - 1 = 713 = 23cdot 31
            $$






            share|cite|improve this answer



















            • 1




              $a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
              – Especially Lime
              Nov 22 at 14:23










            • @EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
              – Arthur
              Nov 22 at 14:30















            up vote
            3
            down vote










            up vote
            3
            down vote









            If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.



            Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
            $$
            ab + 1 = 715 = 5cdot 11cdot 13\
            ab - 1 = 713 = 23cdot 31
            $$






            share|cite|improve this answer














            If both $a$ and $b$ are odd, then $abpm 1$ are both even and therefore not prime.



            Even if one of them are even, there are counterexamples. For instance, $a = 34, b = 21$ gives
            $$
            ab + 1 = 715 = 5cdot 11cdot 13\
            ab - 1 = 713 = 23cdot 31
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 22 at 14:29

























            answered Nov 22 at 14:19









            Arthur

            110k7104186




            110k7104186








            • 1




              $a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
              – Especially Lime
              Nov 22 at 14:23










            • @EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
              – Arthur
              Nov 22 at 14:30
















            • 1




              $a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
              – Especially Lime
              Nov 22 at 14:23










            • @EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
              – Arthur
              Nov 22 at 14:30










            1




            1




            $a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
            – Especially Lime
            Nov 22 at 14:23




            $a$ and $b$ are required to be composite, so $ab$ is not a general square-free number. Any square-free number with at least four prime factors would work though.
            – Especially Lime
            Nov 22 at 14:23












            @EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
            – Arthur
            Nov 22 at 14:30






            @EspeciallyLime I found a (hopefully correct this time) example. (Wonder if it's the smallest...)
            – Arthur
            Nov 22 at 14:30




















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