Factorization of $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$

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Prove that for any prime $q > 3$, $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$ is never prime (or disprove by counterexample). If this is true, is there a possible trivial factorization for $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$?



The first few examples appear to be



$q=5$ and $(5^5-1)/4=11*71$, $q=7$ and $(7^7+1)/8=113*911$, $q=11$ and $(11^11+1)/12=23*89*199*58367$, $q=13$ and $(13^13-1)/(13-1)=53*264031*1803647$.



Let $p*p_2$ be the non-trivial factorization of $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$. Then, for all odd primes $q$,



$q=3$, $p=1$, $p_2=7$



$q=5$, $p=11$, $p_2=71$



$q=7$, $p=113$, $p_2=911$



$q=11$, $p=58367$, $p_2=407353$



$q=13$, $p=1803647$, $p_2=13993643$



$p$ and $p_2$ seem to be of relative same size, and with $p_2$ being approximately $7.76p$.



Can anyone find the interesting pattern here?




prime-numbers factoring prime-factorization




share|cite|improve this question















  • 1




    With factordb and PARI/GP, I found no counterexample. But examples with large smallest prime factors exist, so there seems to be no reason to prevent the existence of such primes.
    – Peter
    Jul 16 at 13:41






  • 1




    For a counterexample, $q$ must be greater than $10 000$. Maybe, aurifeuillan factors could solve the problem, but unfortunately, I have no idea how. I came across the factors of $$frac521^521-1520$$ (a composite $706$ and a composite $707$ digit factor , see here : factordb.com/index.php?query=%28521%5E521-1%29%2F520) , so the question might be completely solveable. I asked a question, how the "monster factors" were found, but noone gave an enlightning answer yet.
    – Peter
    Jul 20 at 11:09







  • 1




    @Peter I think you solved the problem. Reading the wikipedia page about aurifeuillian factors (I had no idea about these types of factors), it states that the $n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=1pmod 4$ and is square-free, while the $2n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=2,3pmod 4$. For prime $n=1pmod 4$ and $x=1$, this amounts to $n^n-1$ having an aurifeuillan factorization, while for prime $n=3pmod 4$ and $x=1$ amounts to $n^n+1$ having an aurifeuillan factorization.
    – J. Linne
    Jul 21 at 7:23







  • 1




    The next interesting thing is, given this information, is there a way to find the two composite factors of $(n^n+-1)/(n+-1)$ (given that $(n^n+-1)/(n+-1)$ has two special factors)? Based on what I already know, I don't know of an efficient way of finding the two composite factors of $(n^n+-1)/(n+-1)$ for prime $n$ (the form in the title is omitted and shortened).
    – J. Linne
    Jul 21 at 7:28











  • Very good, this answer also this question : math.stackexchange.com/questions/2854280/…
    – Peter
    Jul 21 at 16:34














up vote
0
down vote

favorite












Prove that for any prime $q > 3$, $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$ is never prime (or disprove by counterexample). If this is true, is there a possible trivial factorization for $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$?



The first few examples appear to be



$q=5$ and $(5^5-1)/4=11*71$, $q=7$ and $(7^7+1)/8=113*911$, $q=11$ and $(11^11+1)/12=23*89*199*58367$, $q=13$ and $(13^13-1)/(13-1)=53*264031*1803647$.



Let $p*p_2$ be the non-trivial factorization of $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$. Then, for all odd primes $q$,



$q=3$, $p=1$, $p_2=7$



$q=5$, $p=11$, $p_2=71$



$q=7$, $p=113$, $p_2=911$



$q=11$, $p=58367$, $p_2=407353$



$q=13$, $p=1803647$, $p_2=13993643$



$p$ and $p_2$ seem to be of relative same size, and with $p_2$ being approximately $7.76p$.



Can anyone find the interesting pattern here?




prime-numbers factoring prime-factorization




share|cite|improve this question















  • 1




    With factordb and PARI/GP, I found no counterexample. But examples with large smallest prime factors exist, so there seems to be no reason to prevent the existence of such primes.
    – Peter
    Jul 16 at 13:41






  • 1




    For a counterexample, $q$ must be greater than $10 000$. Maybe, aurifeuillan factors could solve the problem, but unfortunately, I have no idea how. I came across the factors of $$frac521^521-1520$$ (a composite $706$ and a composite $707$ digit factor , see here : factordb.com/index.php?query=%28521%5E521-1%29%2F520) , so the question might be completely solveable. I asked a question, how the "monster factors" were found, but noone gave an enlightning answer yet.
    – Peter
    Jul 20 at 11:09







  • 1




    @Peter I think you solved the problem. Reading the wikipedia page about aurifeuillian factors (I had no idea about these types of factors), it states that the $n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=1pmod 4$ and is square-free, while the $2n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=2,3pmod 4$. For prime $n=1pmod 4$ and $x=1$, this amounts to $n^n-1$ having an aurifeuillan factorization, while for prime $n=3pmod 4$ and $x=1$ amounts to $n^n+1$ having an aurifeuillan factorization.
    – J. Linne
    Jul 21 at 7:23







  • 1




    The next interesting thing is, given this information, is there a way to find the two composite factors of $(n^n+-1)/(n+-1)$ (given that $(n^n+-1)/(n+-1)$ has two special factors)? Based on what I already know, I don't know of an efficient way of finding the two composite factors of $(n^n+-1)/(n+-1)$ for prime $n$ (the form in the title is omitted and shortened).
    – J. Linne
    Jul 21 at 7:28











  • Very good, this answer also this question : math.stackexchange.com/questions/2854280/…
    – Peter
    Jul 21 at 16:34












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Prove that for any prime $q > 3$, $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$ is never prime (or disprove by counterexample). If this is true, is there a possible trivial factorization for $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$?



The first few examples appear to be



$q=5$ and $(5^5-1)/4=11*71$, $q=7$ and $(7^7+1)/8=113*911$, $q=11$ and $(11^11+1)/12=23*89*199*58367$, $q=13$ and $(13^13-1)/(13-1)=53*264031*1803647$.



Let $p*p_2$ be the non-trivial factorization of $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$. Then, for all odd primes $q$,



$q=3$, $p=1$, $p_2=7$



$q=5$, $p=11$, $p_2=71$



$q=7$, $p=113$, $p_2=911$



$q=11$, $p=58367$, $p_2=407353$



$q=13$, $p=1803647$, $p_2=13993643$



$p$ and $p_2$ seem to be of relative same size, and with $p_2$ being approximately $7.76p$.



Can anyone find the interesting pattern here?




prime-numbers factoring prime-factorization




share|cite|improve this question











Prove that for any prime $q > 3$, $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$ is never prime (or disprove by counterexample). If this is true, is there a possible trivial factorization for $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$?



The first few examples appear to be



$q=5$ and $(5^5-1)/4=11*71$, $q=7$ and $(7^7+1)/8=113*911$, $q=11$ and $(11^11+1)/12=23*89*199*58367$, $q=13$ and $(13^13-1)/(13-1)=53*264031*1803647$.



Let $p*p_2$ be the non-trivial factorization of $(q^q - (-1)^(q-1)/2)/(q-(-1)^(q-1)/2)$. Then, for all odd primes $q$,



$q=3$, $p=1$, $p_2=7$



$q=5$, $p=11$, $p_2=71$



$q=7$, $p=113$, $p_2=911$



$q=11$, $p=58367$, $p_2=407353$



$q=13$, $p=1803647$, $p_2=13993643$



$p$ and $p_2$ seem to be of relative same size, and with $p_2$ being approximately $7.76p$.



Can anyone find the interesting pattern here?





prime-numbers factoring prime-factorization





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 16 at 1:31









J. Linne

742315




742315







  • 1




    With factordb and PARI/GP, I found no counterexample. But examples with large smallest prime factors exist, so there seems to be no reason to prevent the existence of such primes.
    – Peter
    Jul 16 at 13:41






  • 1




    For a counterexample, $q$ must be greater than $10 000$. Maybe, aurifeuillan factors could solve the problem, but unfortunately, I have no idea how. I came across the factors of $$frac521^521-1520$$ (a composite $706$ and a composite $707$ digit factor , see here : factordb.com/index.php?query=%28521%5E521-1%29%2F520) , so the question might be completely solveable. I asked a question, how the "monster factors" were found, but noone gave an enlightning answer yet.
    – Peter
    Jul 20 at 11:09







  • 1




    @Peter I think you solved the problem. Reading the wikipedia page about aurifeuillian factors (I had no idea about these types of factors), it states that the $n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=1pmod 4$ and is square-free, while the $2n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=2,3pmod 4$. For prime $n=1pmod 4$ and $x=1$, this amounts to $n^n-1$ having an aurifeuillan factorization, while for prime $n=3pmod 4$ and $x=1$ amounts to $n^n+1$ having an aurifeuillan factorization.
    – J. Linne
    Jul 21 at 7:23







  • 1




    The next interesting thing is, given this information, is there a way to find the two composite factors of $(n^n+-1)/(n+-1)$ (given that $(n^n+-1)/(n+-1)$ has two special factors)? Based on what I already know, I don't know of an efficient way of finding the two composite factors of $(n^n+-1)/(n+-1)$ for prime $n$ (the form in the title is omitted and shortened).
    – J. Linne
    Jul 21 at 7:28











  • Very good, this answer also this question : math.stackexchange.com/questions/2854280/…
    – Peter
    Jul 21 at 16:34












  • 1




    With factordb and PARI/GP, I found no counterexample. But examples with large smallest prime factors exist, so there seems to be no reason to prevent the existence of such primes.
    – Peter
    Jul 16 at 13:41






  • 1




    For a counterexample, $q$ must be greater than $10 000$. Maybe, aurifeuillan factors could solve the problem, but unfortunately, I have no idea how. I came across the factors of $$frac521^521-1520$$ (a composite $706$ and a composite $707$ digit factor , see here : factordb.com/index.php?query=%28521%5E521-1%29%2F520) , so the question might be completely solveable. I asked a question, how the "monster factors" were found, but noone gave an enlightning answer yet.
    – Peter
    Jul 20 at 11:09







  • 1




    @Peter I think you solved the problem. Reading the wikipedia page about aurifeuillian factors (I had no idea about these types of factors), it states that the $n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=1pmod 4$ and is square-free, while the $2n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=2,3pmod 4$. For prime $n=1pmod 4$ and $x=1$, this amounts to $n^n-1$ having an aurifeuillan factorization, while for prime $n=3pmod 4$ and $x=1$ amounts to $n^n+1$ having an aurifeuillan factorization.
    – J. Linne
    Jul 21 at 7:23







  • 1




    The next interesting thing is, given this information, is there a way to find the two composite factors of $(n^n+-1)/(n+-1)$ (given that $(n^n+-1)/(n+-1)$ has two special factors)? Based on what I already know, I don't know of an efficient way of finding the two composite factors of $(n^n+-1)/(n+-1)$ for prime $n$ (the form in the title is omitted and shortened).
    – J. Linne
    Jul 21 at 7:28











  • Very good, this answer also this question : math.stackexchange.com/questions/2854280/…
    – Peter
    Jul 21 at 16:34







1




1




With factordb and PARI/GP, I found no counterexample. But examples with large smallest prime factors exist, so there seems to be no reason to prevent the existence of such primes.
– Peter
Jul 16 at 13:41




With factordb and PARI/GP, I found no counterexample. But examples with large smallest prime factors exist, so there seems to be no reason to prevent the existence of such primes.
– Peter
Jul 16 at 13:41




1




1




For a counterexample, $q$ must be greater than $10 000$. Maybe, aurifeuillan factors could solve the problem, but unfortunately, I have no idea how. I came across the factors of $$frac521^521-1520$$ (a composite $706$ and a composite $707$ digit factor , see here : factordb.com/index.php?query=%28521%5E521-1%29%2F520) , so the question might be completely solveable. I asked a question, how the "monster factors" were found, but noone gave an enlightning answer yet.
– Peter
Jul 20 at 11:09





For a counterexample, $q$ must be greater than $10 000$. Maybe, aurifeuillan factors could solve the problem, but unfortunately, I have no idea how. I came across the factors of $$frac521^521-1520$$ (a composite $706$ and a composite $707$ digit factor , see here : factordb.com/index.php?query=%28521%5E521-1%29%2F520) , so the question might be completely solveable. I asked a question, how the "monster factors" were found, but noone gave an enlightning answer yet.
– Peter
Jul 20 at 11:09





1




1




@Peter I think you solved the problem. Reading the wikipedia page about aurifeuillian factors (I had no idea about these types of factors), it states that the $n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=1pmod 4$ and is square-free, while the $2n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=2,3pmod 4$. For prime $n=1pmod 4$ and $x=1$, this amounts to $n^n-1$ having an aurifeuillan factorization, while for prime $n=3pmod 4$ and $x=1$ amounts to $n^n+1$ having an aurifeuillan factorization.
– J. Linne
Jul 21 at 7:23





@Peter I think you solved the problem. Reading the wikipedia page about aurifeuillian factors (I had no idea about these types of factors), it states that the $n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=1pmod 4$ and is square-free, while the $2n$-th cyclotomic polynomial evaluated at $n*x^2$ is reducible if and only if $n=2,3pmod 4$. For prime $n=1pmod 4$ and $x=1$, this amounts to $n^n-1$ having an aurifeuillan factorization, while for prime $n=3pmod 4$ and $x=1$ amounts to $n^n+1$ having an aurifeuillan factorization.
– J. Linne
Jul 21 at 7:23





1




1




The next interesting thing is, given this information, is there a way to find the two composite factors of $(n^n+-1)/(n+-1)$ (given that $(n^n+-1)/(n+-1)$ has two special factors)? Based on what I already know, I don't know of an efficient way of finding the two composite factors of $(n^n+-1)/(n+-1)$ for prime $n$ (the form in the title is omitted and shortened).
– J. Linne
Jul 21 at 7:28





The next interesting thing is, given this information, is there a way to find the two composite factors of $(n^n+-1)/(n+-1)$ (given that $(n^n+-1)/(n+-1)$ has two special factors)? Based on what I already know, I don't know of an efficient way of finding the two composite factors of $(n^n+-1)/(n+-1)$ for prime $n$ (the form in the title is omitted and shortened).
– J. Linne
Jul 21 at 7:28













Very good, this answer also this question : math.stackexchange.com/questions/2854280/…
– Peter
Jul 21 at 16:34




Very good, this answer also this question : math.stackexchange.com/questions/2854280/…
– Peter
Jul 21 at 16:34















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