Find the smallest positive integer $x$ satisfying $gcd(x^n+a,(x+1)^n+a)>1$












2












$begingroup$


Given positive integers $n$ and $a$, I'd like to ask how to find the smallest positive integer $x$ satisfying $gcd(x^n+a,(x+1)^n+a)>1$?



I try using the extended Euclidean algorithm on the two polynomials to find the greatest common divisor $frac{p}{q}$. If $p$ is small, then we just need to check the solutions to $x^n+a equiv 0 (mod d)$ where $d$ divides $p$. Otherwise, it seems much harder.



I also notice that $gcd(n,phi(d))$ should be greater than $1$ in order to have a solution, which helps filter many unfavorable $d$.






polynomials greatest-common-divisor







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$endgroup$












  • $begingroup$
    Did you encounter an example that seems computationally difficult?
    $endgroup$
    – metamorphy
    Jan 2 at 18:48










  • $begingroup$
    For example, $n=19$ and $a=2$. Both $p$ and $q$ are extremely large and cannot be factored easily. I believe the case is quite common for large $n$.
    $endgroup$
    – Hang Wu
    Jan 3 at 4:13












  • $begingroup$
    The procedure I've suggested below gives the resultant as $$1103 cdot 87211 cdot 31308253657 cdot 818790224665362763936013994101920313,$$ and trying $p=1103, 87211, ldots$ we find $xbmod p=473, 836, ldots$ with $473$ the solution.
    $endgroup$
    – metamorphy
    Jan 3 at 7:31










  • $begingroup$
    Funny - for $n=19$ and $a=6$ the resultant appears to be prime, thus the smallest (!!!) solution is $$x=1578270389554680057141787800241971645032008710129107338825798$$
    $endgroup$
    – metamorphy
    Jan 3 at 7:48
















2












$begingroup$


Given positive integers $n$ and $a$, I'd like to ask how to find the smallest positive integer $x$ satisfying $gcd(x^n+a,(x+1)^n+a)>1$?



I try using the extended Euclidean algorithm on the two polynomials to find the greatest common divisor $frac{p}{q}$. If $p$ is small, then we just need to check the solutions to $x^n+a equiv 0 (mod d)$ where $d$ divides $p$. Otherwise, it seems much harder.



I also notice that $gcd(n,phi(d))$ should be greater than $1$ in order to have a solution, which helps filter many unfavorable $d$.






polynomials greatest-common-divisor







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you encounter an example that seems computationally difficult?
    $endgroup$
    – metamorphy
    Jan 2 at 18:48










  • $begingroup$
    For example, $n=19$ and $a=2$. Both $p$ and $q$ are extremely large and cannot be factored easily. I believe the case is quite common for large $n$.
    $endgroup$
    – Hang Wu
    Jan 3 at 4:13












  • $begingroup$
    The procedure I've suggested below gives the resultant as $$1103 cdot 87211 cdot 31308253657 cdot 818790224665362763936013994101920313,$$ and trying $p=1103, 87211, ldots$ we find $xbmod p=473, 836, ldots$ with $473$ the solution.
    $endgroup$
    – metamorphy
    Jan 3 at 7:31










  • $begingroup$
    Funny - for $n=19$ and $a=6$ the resultant appears to be prime, thus the smallest (!!!) solution is $$x=1578270389554680057141787800241971645032008710129107338825798$$
    $endgroup$
    – metamorphy
    Jan 3 at 7:48














2












2








2


2



$begingroup$


Given positive integers $n$ and $a$, I'd like to ask how to find the smallest positive integer $x$ satisfying $gcd(x^n+a,(x+1)^n+a)>1$?



I try using the extended Euclidean algorithm on the two polynomials to find the greatest common divisor $frac{p}{q}$. If $p$ is small, then we just need to check the solutions to $x^n+a equiv 0 (mod d)$ where $d$ divides $p$. Otherwise, it seems much harder.



I also notice that $gcd(n,phi(d))$ should be greater than $1$ in order to have a solution, which helps filter many unfavorable $d$.






polynomials greatest-common-divisor







share|cite|improve this question











$endgroup$




Given positive integers $n$ and $a$, I'd like to ask how to find the smallest positive integer $x$ satisfying $gcd(x^n+a,(x+1)^n+a)>1$?



I try using the extended Euclidean algorithm on the two polynomials to find the greatest common divisor $frac{p}{q}$. If $p$ is small, then we just need to check the solutions to $x^n+a equiv 0 (mod d)$ where $d$ divides $p$. Otherwise, it seems much harder.



I also notice that $gcd(n,phi(d))$ should be greater than $1$ in order to have a solution, which helps filter many unfavorable $d$.






polynomials greatest-common-divisor




polynomials greatest-common-divisor






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 12:46







Hang Wu

















asked Jan 2 at 10:36









Hang WuHang Wu

478311




478311












  • $begingroup$
    Did you encounter an example that seems computationally difficult?
    $endgroup$
    – metamorphy
    Jan 2 at 18:48










  • $begingroup$
    For example, $n=19$ and $a=2$. Both $p$ and $q$ are extremely large and cannot be factored easily. I believe the case is quite common for large $n$.
    $endgroup$
    – Hang Wu
    Jan 3 at 4:13












  • $begingroup$
    The procedure I've suggested below gives the resultant as $$1103 cdot 87211 cdot 31308253657 cdot 818790224665362763936013994101920313,$$ and trying $p=1103, 87211, ldots$ we find $xbmod p=473, 836, ldots$ with $473$ the solution.
    $endgroup$
    – metamorphy
    Jan 3 at 7:31










  • $begingroup$
    Funny - for $n=19$ and $a=6$ the resultant appears to be prime, thus the smallest (!!!) solution is $$x=1578270389554680057141787800241971645032008710129107338825798$$
    $endgroup$
    – metamorphy
    Jan 3 at 7:48


















  • $begingroup$
    Did you encounter an example that seems computationally difficult?
    $endgroup$
    – metamorphy
    Jan 2 at 18:48










  • $begingroup$
    For example, $n=19$ and $a=2$. Both $p$ and $q$ are extremely large and cannot be factored easily. I believe the case is quite common for large $n$.
    $endgroup$
    – Hang Wu
    Jan 3 at 4:13












  • $begingroup$
    The procedure I've suggested below gives the resultant as $$1103 cdot 87211 cdot 31308253657 cdot 818790224665362763936013994101920313,$$ and trying $p=1103, 87211, ldots$ we find $xbmod p=473, 836, ldots$ with $473$ the solution.
    $endgroup$
    – metamorphy
    Jan 3 at 7:31










  • $begingroup$
    Funny - for $n=19$ and $a=6$ the resultant appears to be prime, thus the smallest (!!!) solution is $$x=1578270389554680057141787800241971645032008710129107338825798$$
    $endgroup$
    – metamorphy
    Jan 3 at 7:48
















$begingroup$
Did you encounter an example that seems computationally difficult?
$endgroup$
– metamorphy
Jan 2 at 18:48




$begingroup$
Did you encounter an example that seems computationally difficult?
$endgroup$
– metamorphy
Jan 2 at 18:48












$begingroup$
For example, $n=19$ and $a=2$. Both $p$ and $q$ are extremely large and cannot be factored easily. I believe the case is quite common for large $n$.
$endgroup$
– Hang Wu
Jan 3 at 4:13






$begingroup$
For example, $n=19$ and $a=2$. Both $p$ and $q$ are extremely large and cannot be factored easily. I believe the case is quite common for large $n$.
$endgroup$
– Hang Wu
Jan 3 at 4:13














$begingroup$
The procedure I've suggested below gives the resultant as $$1103 cdot 87211 cdot 31308253657 cdot 818790224665362763936013994101920313,$$ and trying $p=1103, 87211, ldots$ we find $xbmod p=473, 836, ldots$ with $473$ the solution.
$endgroup$
– metamorphy
Jan 3 at 7:31




$begingroup$
The procedure I've suggested below gives the resultant as $$1103 cdot 87211 cdot 31308253657 cdot 818790224665362763936013994101920313,$$ and trying $p=1103, 87211, ldots$ we find $xbmod p=473, 836, ldots$ with $473$ the solution.
$endgroup$
– metamorphy
Jan 3 at 7:31












$begingroup$
Funny - for $n=19$ and $a=6$ the resultant appears to be prime, thus the smallest (!!!) solution is $$x=1578270389554680057141787800241971645032008710129107338825798$$
$endgroup$
– metamorphy
Jan 3 at 7:48




$begingroup$
Funny - for $n=19$ and $a=6$ the resultant appears to be prime, thus the smallest (!!!) solution is $$x=1578270389554680057141787800241971645032008710129107338825798$$
$endgroup$
– metamorphy
Jan 3 at 7:48










1 Answer
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$begingroup$

I don't quite understand what does $gcd(y^n+a,(y+1)^n+a)$ give you if it is computed over $mathbb{Z}[y]$ (if it is roughly what you do; otherwise, probably, we're talking about the same thing).



On the other hand, if $x$ is the solution, $d=gcd(x^n+a,(x+1)^n+a)$, and $p$ is a prime divisor of $d$, then $x$ is a common root over $mathbb{Z}/pmathbb{Z}$ of these two polynomials, and this implies that $p$ divides their resultant, which (up to sign, depending on the definition taken) is equal to
$$det{r_{i,j} : 1leqslant i,jleqslant n},qquad r_{i,j}=begin{cases}-abinom{n}{j-i},& i < j\ hfillbinom{n}{i-j},& igeqslant jend{cases}$$
(resembling the "binomial circulant" a.k.a. Wendt's determinant).



Moreover, let $n=p^k m$ with $pnmid m$. Then, over $mathbb{Z}/pmathbb{Z}$, we have
$$gcd(y^n+a,(y+1)^n+a)=g^{p^k}(y),quad g(y)=gcd(y^m+a,(y+1)^m+a).$$
As we must clearly have $pnmid a$, $x$ is a root of $g(y)$, and in fact a simple one (because $y^m+a$ has only simple roots).



This suggests the following idea. For each $p$ found, we solve $g(x)=0$ in $mathbb{Z}/pmathbb{Z}$ (I can only suggest general root-finding methods here, such as random splitting), and apply Hensel's lifting to get solutions in $mathbb{Z}/p^kmathbb{Z}$ (if needed). If $x$ exists at all, then its value modulo $p^k$ must stabilize eventually (for large enough $k$).



That's it. Definitely there are gaps to fill in - I will edit this answer along any progress.






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    $begingroup$

    I don't quite understand what does $gcd(y^n+a,(y+1)^n+a)$ give you if it is computed over $mathbb{Z}[y]$ (if it is roughly what you do; otherwise, probably, we're talking about the same thing).



    On the other hand, if $x$ is the solution, $d=gcd(x^n+a,(x+1)^n+a)$, and $p$ is a prime divisor of $d$, then $x$ is a common root over $mathbb{Z}/pmathbb{Z}$ of these two polynomials, and this implies that $p$ divides their resultant, which (up to sign, depending on the definition taken) is equal to
    $$det{r_{i,j} : 1leqslant i,jleqslant n},qquad r_{i,j}=begin{cases}-abinom{n}{j-i},& i < j\ hfillbinom{n}{i-j},& igeqslant jend{cases}$$
    (resembling the "binomial circulant" a.k.a. Wendt's determinant).



    Moreover, let $n=p^k m$ with $pnmid m$. Then, over $mathbb{Z}/pmathbb{Z}$, we have
    $$gcd(y^n+a,(y+1)^n+a)=g^{p^k}(y),quad g(y)=gcd(y^m+a,(y+1)^m+a).$$
    As we must clearly have $pnmid a$, $x$ is a root of $g(y)$, and in fact a simple one (because $y^m+a$ has only simple roots).



    This suggests the following idea. For each $p$ found, we solve $g(x)=0$ in $mathbb{Z}/pmathbb{Z}$ (I can only suggest general root-finding methods here, such as random splitting), and apply Hensel's lifting to get solutions in $mathbb{Z}/p^kmathbb{Z}$ (if needed). If $x$ exists at all, then its value modulo $p^k$ must stabilize eventually (for large enough $k$).



    That's it. Definitely there are gaps to fill in - I will edit this answer along any progress.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      I don't quite understand what does $gcd(y^n+a,(y+1)^n+a)$ give you if it is computed over $mathbb{Z}[y]$ (if it is roughly what you do; otherwise, probably, we're talking about the same thing).



      On the other hand, if $x$ is the solution, $d=gcd(x^n+a,(x+1)^n+a)$, and $p$ is a prime divisor of $d$, then $x$ is a common root over $mathbb{Z}/pmathbb{Z}$ of these two polynomials, and this implies that $p$ divides their resultant, which (up to sign, depending on the definition taken) is equal to
      $$det{r_{i,j} : 1leqslant i,jleqslant n},qquad r_{i,j}=begin{cases}-abinom{n}{j-i},& i < j\ hfillbinom{n}{i-j},& igeqslant jend{cases}$$
      (resembling the "binomial circulant" a.k.a. Wendt's determinant).



      Moreover, let $n=p^k m$ with $pnmid m$. Then, over $mathbb{Z}/pmathbb{Z}$, we have
      $$gcd(y^n+a,(y+1)^n+a)=g^{p^k}(y),quad g(y)=gcd(y^m+a,(y+1)^m+a).$$
      As we must clearly have $pnmid a$, $x$ is a root of $g(y)$, and in fact a simple one (because $y^m+a$ has only simple roots).



      This suggests the following idea. For each $p$ found, we solve $g(x)=0$ in $mathbb{Z}/pmathbb{Z}$ (I can only suggest general root-finding methods here, such as random splitting), and apply Hensel's lifting to get solutions in $mathbb{Z}/p^kmathbb{Z}$ (if needed). If $x$ exists at all, then its value modulo $p^k$ must stabilize eventually (for large enough $k$).



      That's it. Definitely there are gaps to fill in - I will edit this answer along any progress.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        I don't quite understand what does $gcd(y^n+a,(y+1)^n+a)$ give you if it is computed over $mathbb{Z}[y]$ (if it is roughly what you do; otherwise, probably, we're talking about the same thing).



        On the other hand, if $x$ is the solution, $d=gcd(x^n+a,(x+1)^n+a)$, and $p$ is a prime divisor of $d$, then $x$ is a common root over $mathbb{Z}/pmathbb{Z}$ of these two polynomials, and this implies that $p$ divides their resultant, which (up to sign, depending on the definition taken) is equal to
        $$det{r_{i,j} : 1leqslant i,jleqslant n},qquad r_{i,j}=begin{cases}-abinom{n}{j-i},& i < j\ hfillbinom{n}{i-j},& igeqslant jend{cases}$$
        (resembling the "binomial circulant" a.k.a. Wendt's determinant).



        Moreover, let $n=p^k m$ with $pnmid m$. Then, over $mathbb{Z}/pmathbb{Z}$, we have
        $$gcd(y^n+a,(y+1)^n+a)=g^{p^k}(y),quad g(y)=gcd(y^m+a,(y+1)^m+a).$$
        As we must clearly have $pnmid a$, $x$ is a root of $g(y)$, and in fact a simple one (because $y^m+a$ has only simple roots).



        This suggests the following idea. For each $p$ found, we solve $g(x)=0$ in $mathbb{Z}/pmathbb{Z}$ (I can only suggest general root-finding methods here, such as random splitting), and apply Hensel's lifting to get solutions in $mathbb{Z}/p^kmathbb{Z}$ (if needed). If $x$ exists at all, then its value modulo $p^k$ must stabilize eventually (for large enough $k$).



        That's it. Definitely there are gaps to fill in - I will edit this answer along any progress.






        share|cite|improve this answer











        $endgroup$



        I don't quite understand what does $gcd(y^n+a,(y+1)^n+a)$ give you if it is computed over $mathbb{Z}[y]$ (if it is roughly what you do; otherwise, probably, we're talking about the same thing).



        On the other hand, if $x$ is the solution, $d=gcd(x^n+a,(x+1)^n+a)$, and $p$ is a prime divisor of $d$, then $x$ is a common root over $mathbb{Z}/pmathbb{Z}$ of these two polynomials, and this implies that $p$ divides their resultant, which (up to sign, depending on the definition taken) is equal to
        $$det{r_{i,j} : 1leqslant i,jleqslant n},qquad r_{i,j}=begin{cases}-abinom{n}{j-i},& i < j\ hfillbinom{n}{i-j},& igeqslant jend{cases}$$
        (resembling the "binomial circulant" a.k.a. Wendt's determinant).



        Moreover, let $n=p^k m$ with $pnmid m$. Then, over $mathbb{Z}/pmathbb{Z}$, we have
        $$gcd(y^n+a,(y+1)^n+a)=g^{p^k}(y),quad g(y)=gcd(y^m+a,(y+1)^m+a).$$
        As we must clearly have $pnmid a$, $x$ is a root of $g(y)$, and in fact a simple one (because $y^m+a$ has only simple roots).



        This suggests the following idea. For each $p$ found, we solve $g(x)=0$ in $mathbb{Z}/pmathbb{Z}$ (I can only suggest general root-finding methods here, such as random splitting), and apply Hensel's lifting to get solutions in $mathbb{Z}/p^kmathbb{Z}$ (if needed). If $x$ exists at all, then its value modulo $p^k$ must stabilize eventually (for large enough $k$).



        That's it. Definitely there are gaps to fill in - I will edit this answer along any progress.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 12:25

























        answered Jan 2 at 16:38









        metamorphymetamorphy

        3,7021621




        3,7021621






























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