The general proposition of Fermat












13












$begingroup$


In his letter to Frenicle, dated 18th October, 1640, Fermat states the following (Point 8, translated) :




If you subtract $2$ from a square, the remaining value cannot be divided by a prime which is greater than a square by $2$



For example, take for a square $1,000,000$, from which, subtracted by two, remains $999,998$. I say that the given remainder cannot be divided by $11$ or by $83$, by $227$, and so on.



You can prove the same rule for odd squares and, if I wanted, I would give you the lovely and general rule; but I'm content with having only indicated it to you.




In other words, numbers of the form $x^2 -2$ are not divisible by primes of the form $a^2 + 2$, where $x$ and $a$ are integers.



Questions :



$1)$ What is the general rule Fermat is talking about?



$2)$ Are there any modern references to this problem?



$3)$ How would Fermat have proved it?






elementary-number-theory reference-request prime-numbers math-history







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I believe he considered 3 cases : 1) $a^2+2$ is prime, 2) $a^2+2$ is odd, 3) $a^2+2$ is any integer (general case)
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 4:19








  • 1




    $begingroup$
    $x^2-2 = m*(a^2+2)$ has no solutions over positive integers, general case/rule
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 4:21












  • $begingroup$
    Well, $a^2+2$ primes are always $2$ or $4k+3$ primes (the converse is of course not true). Not sure how that helps though
    $endgroup$
    – YiFan
    Dec 7 '18 at 6:04






  • 1




    $begingroup$
    Could the general rule be that numbers of the form $x^2-n$ are not divisible by primes of the form $a^2+n$?
    $endgroup$
    – tyobrien
    Dec 7 '18 at 6:20
















13












$begingroup$


In his letter to Frenicle, dated 18th October, 1640, Fermat states the following (Point 8, translated) :




If you subtract $2$ from a square, the remaining value cannot be divided by a prime which is greater than a square by $2$



For example, take for a square $1,000,000$, from which, subtracted by two, remains $999,998$. I say that the given remainder cannot be divided by $11$ or by $83$, by $227$, and so on.



You can prove the same rule for odd squares and, if I wanted, I would give you the lovely and general rule; but I'm content with having only indicated it to you.




In other words, numbers of the form $x^2 -2$ are not divisible by primes of the form $a^2 + 2$, where $x$ and $a$ are integers.



Questions :



$1)$ What is the general rule Fermat is talking about?



$2)$ Are there any modern references to this problem?



$3)$ How would Fermat have proved it?






elementary-number-theory reference-request prime-numbers math-history







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I believe he considered 3 cases : 1) $a^2+2$ is prime, 2) $a^2+2$ is odd, 3) $a^2+2$ is any integer (general case)
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 4:19








  • 1




    $begingroup$
    $x^2-2 = m*(a^2+2)$ has no solutions over positive integers, general case/rule
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 4:21












  • $begingroup$
    Well, $a^2+2$ primes are always $2$ or $4k+3$ primes (the converse is of course not true). Not sure how that helps though
    $endgroup$
    – YiFan
    Dec 7 '18 at 6:04






  • 1




    $begingroup$
    Could the general rule be that numbers of the form $x^2-n$ are not divisible by primes of the form $a^2+n$?
    $endgroup$
    – tyobrien
    Dec 7 '18 at 6:20














13












13








13


2



$begingroup$


In his letter to Frenicle, dated 18th October, 1640, Fermat states the following (Point 8, translated) :




If you subtract $2$ from a square, the remaining value cannot be divided by a prime which is greater than a square by $2$



For example, take for a square $1,000,000$, from which, subtracted by two, remains $999,998$. I say that the given remainder cannot be divided by $11$ or by $83$, by $227$, and so on.



You can prove the same rule for odd squares and, if I wanted, I would give you the lovely and general rule; but I'm content with having only indicated it to you.




In other words, numbers of the form $x^2 -2$ are not divisible by primes of the form $a^2 + 2$, where $x$ and $a$ are integers.



Questions :



$1)$ What is the general rule Fermat is talking about?



$2)$ Are there any modern references to this problem?



$3)$ How would Fermat have proved it?






elementary-number-theory reference-request prime-numbers math-history







share|cite|improve this question











$endgroup$




In his letter to Frenicle, dated 18th October, 1640, Fermat states the following (Point 8, translated) :




If you subtract $2$ from a square, the remaining value cannot be divided by a prime which is greater than a square by $2$



For example, take for a square $1,000,000$, from which, subtracted by two, remains $999,998$. I say that the given remainder cannot be divided by $11$ or by $83$, by $227$, and so on.



You can prove the same rule for odd squares and, if I wanted, I would give you the lovely and general rule; but I'm content with having only indicated it to you.




In other words, numbers of the form $x^2 -2$ are not divisible by primes of the form $a^2 + 2$, where $x$ and $a$ are integers.



Questions :



$1)$ What is the general rule Fermat is talking about?



$2)$ Are there any modern references to this problem?



$3)$ How would Fermat have proved it?






elementary-number-theory reference-request prime-numbers math-history




elementary-number-theory reference-request prime-numbers math-history






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 13:19







Henry

















asked Dec 7 '18 at 4:07









HenryHenry

2,22211249




2,22211249








  • 1




    $begingroup$
    I believe he considered 3 cases : 1) $a^2+2$ is prime, 2) $a^2+2$ is odd, 3) $a^2+2$ is any integer (general case)
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 4:19








  • 1




    $begingroup$
    $x^2-2 = m*(a^2+2)$ has no solutions over positive integers, general case/rule
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 4:21












  • $begingroup$
    Well, $a^2+2$ primes are always $2$ or $4k+3$ primes (the converse is of course not true). Not sure how that helps though
    $endgroup$
    – YiFan
    Dec 7 '18 at 6:04






  • 1




    $begingroup$
    Could the general rule be that numbers of the form $x^2-n$ are not divisible by primes of the form $a^2+n$?
    $endgroup$
    – tyobrien
    Dec 7 '18 at 6:20














  • 1




    $begingroup$
    I believe he considered 3 cases : 1) $a^2+2$ is prime, 2) $a^2+2$ is odd, 3) $a^2+2$ is any integer (general case)
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 4:19








  • 1




    $begingroup$
    $x^2-2 = m*(a^2+2)$ has no solutions over positive integers, general case/rule
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 4:21












  • $begingroup$
    Well, $a^2+2$ primes are always $2$ or $4k+3$ primes (the converse is of course not true). Not sure how that helps though
    $endgroup$
    – YiFan
    Dec 7 '18 at 6:04






  • 1




    $begingroup$
    Could the general rule be that numbers of the form $x^2-n$ are not divisible by primes of the form $a^2+n$?
    $endgroup$
    – tyobrien
    Dec 7 '18 at 6:20








1




1




$begingroup$
I believe he considered 3 cases : 1) $a^2+2$ is prime, 2) $a^2+2$ is odd, 3) $a^2+2$ is any integer (general case)
$endgroup$
– rsadhvika
Dec 7 '18 at 4:19






$begingroup$
I believe he considered 3 cases : 1) $a^2+2$ is prime, 2) $a^2+2$ is odd, 3) $a^2+2$ is any integer (general case)
$endgroup$
– rsadhvika
Dec 7 '18 at 4:19






1




1




$begingroup$
$x^2-2 = m*(a^2+2)$ has no solutions over positive integers, general case/rule
$endgroup$
– rsadhvika
Dec 7 '18 at 4:21






$begingroup$
$x^2-2 = m*(a^2+2)$ has no solutions over positive integers, general case/rule
$endgroup$
– rsadhvika
Dec 7 '18 at 4:21














$begingroup$
Well, $a^2+2$ primes are always $2$ or $4k+3$ primes (the converse is of course not true). Not sure how that helps though
$endgroup$
– YiFan
Dec 7 '18 at 6:04




$begingroup$
Well, $a^2+2$ primes are always $2$ or $4k+3$ primes (the converse is of course not true). Not sure how that helps though
$endgroup$
– YiFan
Dec 7 '18 at 6:04




1




1




$begingroup$
Could the general rule be that numbers of the form $x^2-n$ are not divisible by primes of the form $a^2+n$?
$endgroup$
– tyobrien
Dec 7 '18 at 6:20




$begingroup$
Could the general rule be that numbers of the form $x^2-n$ are not divisible by primes of the form $a^2+n$?
$endgroup$
– tyobrien
Dec 7 '18 at 6:20










2 Answers
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$begingroup$

Ad 3): If $p = a^2 + 2$ divides $x^2 - 2$, then $p$ divides $x^2 - 2 + p = x^2 + a^2$. But primes $p = 4n+3$ cannot divide sums of two squares without dividing the squares themselves. This observation is due to Weil.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Let $p=a^2+2$ be a prime. Suppose $a>0$ as for $a=0$, the statement clearly does not hold. Therefore, $pequiv 3 pmod 8$. Thus, the congruence $x^2equiv 2pmod p$ has no solutions$($$2$ is a quadratic residue modulo $p$ if and only if $pequiv ±1pmod 8$$)$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      2












      $begingroup$

      Ad 3): If $p = a^2 + 2$ divides $x^2 - 2$, then $p$ divides $x^2 - 2 + p = x^2 + a^2$. But primes $p = 4n+3$ cannot divide sums of two squares without dividing the squares themselves. This observation is due to Weil.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Ad 3): If $p = a^2 + 2$ divides $x^2 - 2$, then $p$ divides $x^2 - 2 + p = x^2 + a^2$. But primes $p = 4n+3$ cannot divide sums of two squares without dividing the squares themselves. This observation is due to Weil.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Ad 3): If $p = a^2 + 2$ divides $x^2 - 2$, then $p$ divides $x^2 - 2 + p = x^2 + a^2$. But primes $p = 4n+3$ cannot divide sums of two squares without dividing the squares themselves. This observation is due to Weil.






          share|cite|improve this answer











          $endgroup$



          Ad 3): If $p = a^2 + 2$ divides $x^2 - 2$, then $p$ divides $x^2 - 2 + p = x^2 + a^2$. But primes $p = 4n+3$ cannot divide sums of two squares without dividing the squares themselves. This observation is due to Weil.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 10:22









          Henry

          2,22211249




          2,22211249










          answered Dec 8 '18 at 15:19









          franz lemmermeyerfranz lemmermeyer

          7,07522047




          7,07522047























              1












              $begingroup$

              Let $p=a^2+2$ be a prime. Suppose $a>0$ as for $a=0$, the statement clearly does not hold. Therefore, $pequiv 3 pmod 8$. Thus, the congruence $x^2equiv 2pmod p$ has no solutions$($$2$ is a quadratic residue modulo $p$ if and only if $pequiv ±1pmod 8$$)$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let $p=a^2+2$ be a prime. Suppose $a>0$ as for $a=0$, the statement clearly does not hold. Therefore, $pequiv 3 pmod 8$. Thus, the congruence $x^2equiv 2pmod p$ has no solutions$($$2$ is a quadratic residue modulo $p$ if and only if $pequiv ±1pmod 8$$)$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $p=a^2+2$ be a prime. Suppose $a>0$ as for $a=0$, the statement clearly does not hold. Therefore, $pequiv 3 pmod 8$. Thus, the congruence $x^2equiv 2pmod p$ has no solutions$($$2$ is a quadratic residue modulo $p$ if and only if $pequiv ±1pmod 8$$)$.






                  share|cite|improve this answer









                  $endgroup$



                  Let $p=a^2+2$ be a prime. Suppose $a>0$ as for $a=0$, the statement clearly does not hold. Therefore, $pequiv 3 pmod 8$. Thus, the congruence $x^2equiv 2pmod p$ has no solutions$($$2$ is a quadratic residue modulo $p$ if and only if $pequiv ±1pmod 8$$)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 7 '18 at 7:13









                  Anubhab GhosalAnubhab Ghosal

                  1,01518




                  1,01518






























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