Question about a kind of generalized Fermat numbers












1












$begingroup$


The present question is directly inspired by this one.



Let $alpha$ be a unit in the ring of quadratic integers of a real quadratic field, or, in less sophisticated words:
$$alpha=frac{apmsqrt{a^2pm4}}{2}, $$ for some natural number $a$.



Let $baralpha$ be the conjugate of $alpha$, that is: $overline{frac{apmsqrt{a^2pm4}}{2}}=frac{ampsqrt{a^2pm4}}{2} $.



Now, for $n$ a natural number, let $F_n(alpha)$ be a sort of generalized Fermat number, such that
$$ F_n(alpha):= alpha^{2^n}+baralpha^{2^n}. $$



$F_n(alpha)$ is a natural number, since in the expansion of $F_n(alpha)$ all the square roots cancel each other.




Let $p$ be an odd prime divisor of $F_n(alpha)$. Is it true that $$ p^2 equiv 1 pmod {2^{n+1} } $$
Edit 21/12/18: More pecisely, is it true that
$$ pequiv left(frac{a^2pm 4}{p}right) pmod {2^{n+1} } $$
where $ left(frac{cdot}{p}right)$ is the Legendre symbol.




As an example, with $alpha = 2+sqrt5$, we have:



$$ F_4(alpha)= 10749957122=2cdot769cdot2207cdot3167 $$
and
begin{align*} 769 &equiv left(frac{5}{769}right) = +1pmod {2^5}\
2207 &equiv left(frac{5}{2207}right)=-1 pmod {2^5}\
3167 &equiv left(frac{5}{3167}right)=-1 pmod {2^5}
end{align*}



I have checked many other $alpha$ and $n$, but I could not find a counterexample to this.



Edit 21/12/18: a proof should be like in the answer to the above related question.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I see: when you choose $frac{a+ sqrt {a^2 + 4}}{2},$ as with $a = 4,$ the first value is $a$ and the second value is the unusual $a^2 + 2.$ After that, you return to the expected $f(k) = k^2 - 2.$ So your example sequence is $$ 4,18, 322, 103682, 1074995712,... $$
    $endgroup$
    – Will Jagy
    Dec 11 '18 at 0:04










  • $begingroup$
    But in your example with $alpha = 2+sqrt5$ you have $a=4$, for which neither $a^2+4$ nor $a^2-4$ is squarefree...
    $endgroup$
    – Servaes
    Dec 11 '18 at 0:09












  • $begingroup$
    Rene, see my answer at the question you link.
    $endgroup$
    – Will Jagy
    Dec 11 '18 at 2:28










  • $begingroup$
    @Servaes You are right, the squarefree condition is not necessary. I removed it.
    $endgroup$
    – René Gy
    Dec 13 '18 at 17:44
















1












$begingroup$


The present question is directly inspired by this one.



Let $alpha$ be a unit in the ring of quadratic integers of a real quadratic field, or, in less sophisticated words:
$$alpha=frac{apmsqrt{a^2pm4}}{2}, $$ for some natural number $a$.



Let $baralpha$ be the conjugate of $alpha$, that is: $overline{frac{apmsqrt{a^2pm4}}{2}}=frac{ampsqrt{a^2pm4}}{2} $.



Now, for $n$ a natural number, let $F_n(alpha)$ be a sort of generalized Fermat number, such that
$$ F_n(alpha):= alpha^{2^n}+baralpha^{2^n}. $$



$F_n(alpha)$ is a natural number, since in the expansion of $F_n(alpha)$ all the square roots cancel each other.




Let $p$ be an odd prime divisor of $F_n(alpha)$. Is it true that $$ p^2 equiv 1 pmod {2^{n+1} } $$
Edit 21/12/18: More pecisely, is it true that
$$ pequiv left(frac{a^2pm 4}{p}right) pmod {2^{n+1} } $$
where $ left(frac{cdot}{p}right)$ is the Legendre symbol.




As an example, with $alpha = 2+sqrt5$, we have:



$$ F_4(alpha)= 10749957122=2cdot769cdot2207cdot3167 $$
and
begin{align*} 769 &equiv left(frac{5}{769}right) = +1pmod {2^5}\
2207 &equiv left(frac{5}{2207}right)=-1 pmod {2^5}\
3167 &equiv left(frac{5}{3167}right)=-1 pmod {2^5}
end{align*}



I have checked many other $alpha$ and $n$, but I could not find a counterexample to this.



Edit 21/12/18: a proof should be like in the answer to the above related question.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I see: when you choose $frac{a+ sqrt {a^2 + 4}}{2},$ as with $a = 4,$ the first value is $a$ and the second value is the unusual $a^2 + 2.$ After that, you return to the expected $f(k) = k^2 - 2.$ So your example sequence is $$ 4,18, 322, 103682, 1074995712,... $$
    $endgroup$
    – Will Jagy
    Dec 11 '18 at 0:04










  • $begingroup$
    But in your example with $alpha = 2+sqrt5$ you have $a=4$, for which neither $a^2+4$ nor $a^2-4$ is squarefree...
    $endgroup$
    – Servaes
    Dec 11 '18 at 0:09












  • $begingroup$
    Rene, see my answer at the question you link.
    $endgroup$
    – Will Jagy
    Dec 11 '18 at 2:28










  • $begingroup$
    @Servaes You are right, the squarefree condition is not necessary. I removed it.
    $endgroup$
    – René Gy
    Dec 13 '18 at 17:44














1












1








1


0



$begingroup$


The present question is directly inspired by this one.



Let $alpha$ be a unit in the ring of quadratic integers of a real quadratic field, or, in less sophisticated words:
$$alpha=frac{apmsqrt{a^2pm4}}{2}, $$ for some natural number $a$.



Let $baralpha$ be the conjugate of $alpha$, that is: $overline{frac{apmsqrt{a^2pm4}}{2}}=frac{ampsqrt{a^2pm4}}{2} $.



Now, for $n$ a natural number, let $F_n(alpha)$ be a sort of generalized Fermat number, such that
$$ F_n(alpha):= alpha^{2^n}+baralpha^{2^n}. $$



$F_n(alpha)$ is a natural number, since in the expansion of $F_n(alpha)$ all the square roots cancel each other.




Let $p$ be an odd prime divisor of $F_n(alpha)$. Is it true that $$ p^2 equiv 1 pmod {2^{n+1} } $$
Edit 21/12/18: More pecisely, is it true that
$$ pequiv left(frac{a^2pm 4}{p}right) pmod {2^{n+1} } $$
where $ left(frac{cdot}{p}right)$ is the Legendre symbol.




As an example, with $alpha = 2+sqrt5$, we have:



$$ F_4(alpha)= 10749957122=2cdot769cdot2207cdot3167 $$
and
begin{align*} 769 &equiv left(frac{5}{769}right) = +1pmod {2^5}\
2207 &equiv left(frac{5}{2207}right)=-1 pmod {2^5}\
3167 &equiv left(frac{5}{3167}right)=-1 pmod {2^5}
end{align*}



I have checked many other $alpha$ and $n$, but I could not find a counterexample to this.



Edit 21/12/18: a proof should be like in the answer to the above related question.










share|cite|improve this question











$endgroup$




The present question is directly inspired by this one.



Let $alpha$ be a unit in the ring of quadratic integers of a real quadratic field, or, in less sophisticated words:
$$alpha=frac{apmsqrt{a^2pm4}}{2}, $$ for some natural number $a$.



Let $baralpha$ be the conjugate of $alpha$, that is: $overline{frac{apmsqrt{a^2pm4}}{2}}=frac{ampsqrt{a^2pm4}}{2} $.



Now, for $n$ a natural number, let $F_n(alpha)$ be a sort of generalized Fermat number, such that
$$ F_n(alpha):= alpha^{2^n}+baralpha^{2^n}. $$



$F_n(alpha)$ is a natural number, since in the expansion of $F_n(alpha)$ all the square roots cancel each other.




Let $p$ be an odd prime divisor of $F_n(alpha)$. Is it true that $$ p^2 equiv 1 pmod {2^{n+1} } $$
Edit 21/12/18: More pecisely, is it true that
$$ pequiv left(frac{a^2pm 4}{p}right) pmod {2^{n+1} } $$
where $ left(frac{cdot}{p}right)$ is the Legendre symbol.




As an example, with $alpha = 2+sqrt5$, we have:



$$ F_4(alpha)= 10749957122=2cdot769cdot2207cdot3167 $$
and
begin{align*} 769 &equiv left(frac{5}{769}right) = +1pmod {2^5}\
2207 &equiv left(frac{5}{2207}right)=-1 pmod {2^5}\
3167 &equiv left(frac{5}{3167}right)=-1 pmod {2^5}
end{align*}



I have checked many other $alpha$ and $n$, but I could not find a counterexample to this.



Edit 21/12/18: a proof should be like in the answer to the above related question.







number-theory elementary-number-theory modular-arithmetic quadratic-residues fermat-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 21:01







René Gy

















asked Dec 10 '18 at 23:10









René GyRené Gy

1,143613




1,143613












  • $begingroup$
    I see: when you choose $frac{a+ sqrt {a^2 + 4}}{2},$ as with $a = 4,$ the first value is $a$ and the second value is the unusual $a^2 + 2.$ After that, you return to the expected $f(k) = k^2 - 2.$ So your example sequence is $$ 4,18, 322, 103682, 1074995712,... $$
    $endgroup$
    – Will Jagy
    Dec 11 '18 at 0:04










  • $begingroup$
    But in your example with $alpha = 2+sqrt5$ you have $a=4$, for which neither $a^2+4$ nor $a^2-4$ is squarefree...
    $endgroup$
    – Servaes
    Dec 11 '18 at 0:09












  • $begingroup$
    Rene, see my answer at the question you link.
    $endgroup$
    – Will Jagy
    Dec 11 '18 at 2:28










  • $begingroup$
    @Servaes You are right, the squarefree condition is not necessary. I removed it.
    $endgroup$
    – René Gy
    Dec 13 '18 at 17:44


















  • $begingroup$
    I see: when you choose $frac{a+ sqrt {a^2 + 4}}{2},$ as with $a = 4,$ the first value is $a$ and the second value is the unusual $a^2 + 2.$ After that, you return to the expected $f(k) = k^2 - 2.$ So your example sequence is $$ 4,18, 322, 103682, 1074995712,... $$
    $endgroup$
    – Will Jagy
    Dec 11 '18 at 0:04










  • $begingroup$
    But in your example with $alpha = 2+sqrt5$ you have $a=4$, for which neither $a^2+4$ nor $a^2-4$ is squarefree...
    $endgroup$
    – Servaes
    Dec 11 '18 at 0:09












  • $begingroup$
    Rene, see my answer at the question you link.
    $endgroup$
    – Will Jagy
    Dec 11 '18 at 2:28










  • $begingroup$
    @Servaes You are right, the squarefree condition is not necessary. I removed it.
    $endgroup$
    – René Gy
    Dec 13 '18 at 17:44
















$begingroup$
I see: when you choose $frac{a+ sqrt {a^2 + 4}}{2},$ as with $a = 4,$ the first value is $a$ and the second value is the unusual $a^2 + 2.$ After that, you return to the expected $f(k) = k^2 - 2.$ So your example sequence is $$ 4,18, 322, 103682, 1074995712,... $$
$endgroup$
– Will Jagy
Dec 11 '18 at 0:04




$begingroup$
I see: when you choose $frac{a+ sqrt {a^2 + 4}}{2},$ as with $a = 4,$ the first value is $a$ and the second value is the unusual $a^2 + 2.$ After that, you return to the expected $f(k) = k^2 - 2.$ So your example sequence is $$ 4,18, 322, 103682, 1074995712,... $$
$endgroup$
– Will Jagy
Dec 11 '18 at 0:04












$begingroup$
But in your example with $alpha = 2+sqrt5$ you have $a=4$, for which neither $a^2+4$ nor $a^2-4$ is squarefree...
$endgroup$
– Servaes
Dec 11 '18 at 0:09






$begingroup$
But in your example with $alpha = 2+sqrt5$ you have $a=4$, for which neither $a^2+4$ nor $a^2-4$ is squarefree...
$endgroup$
– Servaes
Dec 11 '18 at 0:09














$begingroup$
Rene, see my answer at the question you link.
$endgroup$
– Will Jagy
Dec 11 '18 at 2:28




$begingroup$
Rene, see my answer at the question you link.
$endgroup$
– Will Jagy
Dec 11 '18 at 2:28












$begingroup$
@Servaes You are right, the squarefree condition is not necessary. I removed it.
$endgroup$
– René Gy
Dec 13 '18 at 17:44




$begingroup$
@Servaes You are right, the squarefree condition is not necessary. I removed it.
$endgroup$
– René Gy
Dec 13 '18 at 17:44










1 Answer
1






active

oldest

votes


















0












$begingroup$

The conjecture is true.



Specifically this is a general case of $alpha=frac 12 sqrt{a+2} pm frac 12 sqrt{a-2}$. We then have $alpha^2+alpha^{-2}=a$ and $alphaalpha'=1$.



From this we generate a series of $alpha^n+alpha^{-n}$, by the iteration of $A(n+1)=a.A(n)-A(n-1)$. When a<2 these represent the shortchords of polygone, the shortchord being the chord at the base of two edges.



We first see that A(-n)=A(n).



We can demonstrate that A(x+n)=A(n)A(x)-A(x-n). This corresponds to a polygon {p/n} can be inscribed in a polygon {p}.



Once this is done we consider the modular case, relative some prime $pmod{p}$



The trick here is to show that $A(p)=a pmod p$ which leads directly to A(p+1)=2 or A(p-1)=2.



This means that some series C(n), given as C(0)=0; C(1)=1; C(n+1)=a.C(n)-C(n-1), will represent a kind of repunit (eg 111111), and regular base theory applies. That is, if p | C(n), then p | C(mn). This is why the fibonacci series looks like a base.



We then note that if some p | C(n), but for no lesser values of n, then n | p-1 or n divides p+1, ie, $p mod n = pm 1$



The reason for going for fermat-style numbers is because $x^{n}-x^{-n}$ when n is a power of 2, has no other algebraic factors. But the condition is perfectly general and in keeping with ordinary base theory.



PS. I have done a general cunningham table for this sort of number as far as very large numbers, (14400, as far as 80-digit numbers).






share|cite|improve this answer









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    0












    $begingroup$

    The conjecture is true.



    Specifically this is a general case of $alpha=frac 12 sqrt{a+2} pm frac 12 sqrt{a-2}$. We then have $alpha^2+alpha^{-2}=a$ and $alphaalpha'=1$.



    From this we generate a series of $alpha^n+alpha^{-n}$, by the iteration of $A(n+1)=a.A(n)-A(n-1)$. When a<2 these represent the shortchords of polygone, the shortchord being the chord at the base of two edges.



    We first see that A(-n)=A(n).



    We can demonstrate that A(x+n)=A(n)A(x)-A(x-n). This corresponds to a polygon {p/n} can be inscribed in a polygon {p}.



    Once this is done we consider the modular case, relative some prime $pmod{p}$



    The trick here is to show that $A(p)=a pmod p$ which leads directly to A(p+1)=2 or A(p-1)=2.



    This means that some series C(n), given as C(0)=0; C(1)=1; C(n+1)=a.C(n)-C(n-1), will represent a kind of repunit (eg 111111), and regular base theory applies. That is, if p | C(n), then p | C(mn). This is why the fibonacci series looks like a base.



    We then note that if some p | C(n), but for no lesser values of n, then n | p-1 or n divides p+1, ie, $p mod n = pm 1$



    The reason for going for fermat-style numbers is because $x^{n}-x^{-n}$ when n is a power of 2, has no other algebraic factors. But the condition is perfectly general and in keeping with ordinary base theory.



    PS. I have done a general cunningham table for this sort of number as far as very large numbers, (14400, as far as 80-digit numbers).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The conjecture is true.



      Specifically this is a general case of $alpha=frac 12 sqrt{a+2} pm frac 12 sqrt{a-2}$. We then have $alpha^2+alpha^{-2}=a$ and $alphaalpha'=1$.



      From this we generate a series of $alpha^n+alpha^{-n}$, by the iteration of $A(n+1)=a.A(n)-A(n-1)$. When a<2 these represent the shortchords of polygone, the shortchord being the chord at the base of two edges.



      We first see that A(-n)=A(n).



      We can demonstrate that A(x+n)=A(n)A(x)-A(x-n). This corresponds to a polygon {p/n} can be inscribed in a polygon {p}.



      Once this is done we consider the modular case, relative some prime $pmod{p}$



      The trick here is to show that $A(p)=a pmod p$ which leads directly to A(p+1)=2 or A(p-1)=2.



      This means that some series C(n), given as C(0)=0; C(1)=1; C(n+1)=a.C(n)-C(n-1), will represent a kind of repunit (eg 111111), and regular base theory applies. That is, if p | C(n), then p | C(mn). This is why the fibonacci series looks like a base.



      We then note that if some p | C(n), but for no lesser values of n, then n | p-1 or n divides p+1, ie, $p mod n = pm 1$



      The reason for going for fermat-style numbers is because $x^{n}-x^{-n}$ when n is a power of 2, has no other algebraic factors. But the condition is perfectly general and in keeping with ordinary base theory.



      PS. I have done a general cunningham table for this sort of number as far as very large numbers, (14400, as far as 80-digit numbers).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The conjecture is true.



        Specifically this is a general case of $alpha=frac 12 sqrt{a+2} pm frac 12 sqrt{a-2}$. We then have $alpha^2+alpha^{-2}=a$ and $alphaalpha'=1$.



        From this we generate a series of $alpha^n+alpha^{-n}$, by the iteration of $A(n+1)=a.A(n)-A(n-1)$. When a<2 these represent the shortchords of polygone, the shortchord being the chord at the base of two edges.



        We first see that A(-n)=A(n).



        We can demonstrate that A(x+n)=A(n)A(x)-A(x-n). This corresponds to a polygon {p/n} can be inscribed in a polygon {p}.



        Once this is done we consider the modular case, relative some prime $pmod{p}$



        The trick here is to show that $A(p)=a pmod p$ which leads directly to A(p+1)=2 or A(p-1)=2.



        This means that some series C(n), given as C(0)=0; C(1)=1; C(n+1)=a.C(n)-C(n-1), will represent a kind of repunit (eg 111111), and regular base theory applies. That is, if p | C(n), then p | C(mn). This is why the fibonacci series looks like a base.



        We then note that if some p | C(n), but for no lesser values of n, then n | p-1 or n divides p+1, ie, $p mod n = pm 1$



        The reason for going for fermat-style numbers is because $x^{n}-x^{-n}$ when n is a power of 2, has no other algebraic factors. But the condition is perfectly general and in keeping with ordinary base theory.



        PS. I have done a general cunningham table for this sort of number as far as very large numbers, (14400, as far as 80-digit numbers).






        share|cite|improve this answer









        $endgroup$



        The conjecture is true.



        Specifically this is a general case of $alpha=frac 12 sqrt{a+2} pm frac 12 sqrt{a-2}$. We then have $alpha^2+alpha^{-2}=a$ and $alphaalpha'=1$.



        From this we generate a series of $alpha^n+alpha^{-n}$, by the iteration of $A(n+1)=a.A(n)-A(n-1)$. When a<2 these represent the shortchords of polygone, the shortchord being the chord at the base of two edges.



        We first see that A(-n)=A(n).



        We can demonstrate that A(x+n)=A(n)A(x)-A(x-n). This corresponds to a polygon {p/n} can be inscribed in a polygon {p}.



        Once this is done we consider the modular case, relative some prime $pmod{p}$



        The trick here is to show that $A(p)=a pmod p$ which leads directly to A(p+1)=2 or A(p-1)=2.



        This means that some series C(n), given as C(0)=0; C(1)=1; C(n+1)=a.C(n)-C(n-1), will represent a kind of repunit (eg 111111), and regular base theory applies. That is, if p | C(n), then p | C(mn). This is why the fibonacci series looks like a base.



        We then note that if some p | C(n), but for no lesser values of n, then n | p-1 or n divides p+1, ie, $p mod n = pm 1$



        The reason for going for fermat-style numbers is because $x^{n}-x^{-n}$ when n is a power of 2, has no other algebraic factors. But the condition is perfectly general and in keeping with ordinary base theory.



        PS. I have done a general cunningham table for this sort of number as far as very large numbers, (14400, as far as 80-digit numbers).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 1:59









        wendy.kriegerwendy.krieger

        5,78711426




        5,78711426






























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