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In group theory the number of Sylow $p$-subgroups of a finite group $G$, is of the form $kp+1$.

So it is interesting to discuss about the divisors of this form. As I checked it seems that for an odd prime $p$, there is not any divisor $a$ of $p^4+1$, where $1<a<p^4+1$ and $a=kp+1$, for some $k>0$. Could you help me about this question? If it is true how we can prove it?

Also when I checked the same fact for $p^8+1$, we get many counterexamples. What is the difference between $4$ and $8$?

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  • $\begingroup$ You checked for $p^4+1$ until which bound? $\endgroup$ – YCor Jun 7 '15 at 10:10
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    $\begingroup$ for primes $p$ less than 20000 by GAP $\endgroup$ – BHZ Jun 7 '15 at 10:15
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    $\begingroup$ Now YCor has settled this, of course. Note, however, that we have the easy factorization $p^{4}+1 = (p^{2} - \sqrt{2}p +1)(p^{2}+\sqrt{2}p +1)$ in the Euclidean ring $\mathbb{Z}[\sqrt{2}]$ (both factors clearly congruent to $1$ (mod $p$), though $p$ need not be prime in $\mathbb{Z}[\sqrt{2}]$). This is not especially relevant to counting Sylow $p$-subgroups in groups of course! $\endgroup$ – Geoff Robinson Jun 8 '15 at 8:25
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    $\begingroup$ Yes, groups of order $p^{i}(p^{4}+1)$ are interesting by virtue of the answer to the question. $\endgroup$ – Geoff Robinson Jun 8 '15 at 10:38
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    $\begingroup$ Indeed, it implies that a group $G$ of order $p^i(p^4+1)$ has characteristic subgroups $N\subset M$ such that $G/M$ and $N$ are $p$-groups, and $M/N$ has order $p^4+1$. $\endgroup$ – YCor Jun 12 '15 at 16:23
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There's none indeed.

Lemma: if $1<m<n$ are coprime integers then $mn+1$ does not divide $n^4+1$.

First observe that for any $m,n$, of $mn+1$ divides $n^4+1$, then it divides $n^4m^4+m^4=((nm)^4-1)+1+m^4$, and since $mn+1$ clearly divides $((nm)^4-1)$, we deduce that it also divides $1+m^4$.

To prove the lemma, assume the contrary. As we have just seen, $mn+1$ divides $m^4+1$ as well. Write $(m^4+1)/(mn+1)=k$, and $k=(\ell m+r)$ with $0\le r\le m-1$. Then $m^4+1=(\ell m+r)((mn+1)=mN'+r$, so $r=1$. So $(\ell m+1)$ divides $m^4+1$. Then $m$ and $\ell$ are coprime: indeed, we have $(m\ell +1)(mn+1)=m^4+1$, so $\ell(mn+1)-m^3=-n$. If a prime $p$ were dividing both $m$ and $\ell$ then it would also divide $n$, contradicting that $m$ and $n$ are coprime, so $m$ and $\ell$ are indeed coprime. We have $\ell<n$ because otherwise $$m^4+1=(mn+1)(m\ell+1)\ge (mn+1)^2> (m^2+1)^2>m^4+1.$$ So we found a new pair $(m,\ell)$ with $\max(m,\ell)<n$, with $m\ell+1$ dividing both $m^4+1$ and $\ell^4+1$. So, assuming that $n$ is minimal, we're done unless $\ell=1$. This happens if $m+1$ divides $m^4+1$, and since $m+1$ divides $m^4-1$ as well, if this occurs then $m+1=2$, contradicting $m>1$.

(Note: without the coprime assumption the conclusion fails, as $(m,n)=(m,m^3)$ for $m\ge 2$, e.g. $(m,n)=(2,8)$, satisfies $mn+1|n^4+1$.)

Proposition If $p$ is prime then $p^4+1$ has no divisor of the form $kp+1$ except $1$ and $p^4+1$.

Proof: write $p^4+1=(kp+1)(k'p+\ell)$ with $0\le\ell\le p-1$; then $\ell=1$. So exchanging $k$ and $k'$ if necessary we can suppose $k\le k'$. If by contradiction $k$ is divisible by $p$ then $k'\ge k\ge p$, so $p^4+1\ge (p^2+1)^2>p^4+1$, contradiction. So $k$ and $p$ are coprime, and the lemma yields a contradiction.

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  • $\begingroup$ The proof is very nice, thanks YCor. $\endgroup$ – BHZ Jun 7 '15 at 11:35
  • $\begingroup$ Where do you use primality of $p$? For $p$ composite $p=4096$ divisors are: 65537, 4294901761 $\endgroup$ – joro Jun 7 '15 at 11:42
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    $\begingroup$ when we get that any integer less than $p$ is coprime with respect to it (in the last line of the proof). $\endgroup$ – BHZ Jun 7 '15 at 11:46
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    $\begingroup$ A great example of Vieta jumping! en.wikipedia.org/wiki/Vieta_jumping [This technique has been burned into my brain since 1988, when I failed to solve an IMO problem requiring this trick.] $\endgroup$ – Terry Tao Jun 7 '15 at 15:23
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    $\begingroup$ Actually, now that I think about it, the problem is in fact closely related to IMO 1988 Q6, since if $kp+1$ divides $p^4+1$, then it also divides $k^2+p^2$. $\endgroup$ – Terry Tao Jun 7 '15 at 15:33
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Let $p$ be prime. Assume that $q \mid p^4+1$, where $q = ap+1$ for some $a \in \mathbb{N}$.

Then $(p^4+1)/q \equiv 1 \!\! \mod p$, hence $p^4+1 = (ap+1)(bp+1)$ for some $b \in \mathbb{N}$. Now we have $p^4 = abp^2+(a+b)p$, respectively, $p^3-abp-a-b = 0$. Therefore, effectively we are looking for solutions to the diophantine equation $x^3-xyz-y-z=0$ in positive $x$, $y$, $z$, and prime $x$.

When allowing negative (or zero) $x, y, z$, we would get a lot of solutions, and for positive $x, y, z \leq 1000$, we still find (up to switching of $y$ and $z$) $$ (x,y,z) \in \{(8,30,2),(27,240,3),(30,112,8),(112,418,30)\}. $$ I don't know whether there is a solution with prime $x$ as required -- however we have a diophantine equation of total degree $3$ in $3$ variables, and such are (at least potentially) hard.

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