2
$\begingroup$

Given natural numbers $m,n$ such that $m^2 + n^2 - m \equiv 0 \pmod{mn}$ and $p$ a prime dividing $m$, then I want to show that $p^2$ divides $m$.

I have tried multiple approaches: Euclidean division gives that $m = qp^2 + r$ for some $q, r \in \mathbb{N}$ where $0 \leq r < p^2$ and filling this in to show that $r = 0$. I also tried using that $p$ divides $m$ and $mn$ divides $m^2 + n^2 - m$ and again filling this in to find that $p^2$ has to divide $m$, but with no succes.

Any hints would be appreciated.

$\textbf{EDIT}$ Based on the given hint I have the following solution: Since $p$ divides $m$, we have that $p$ divides $mn$ and therefore $m^2 + n^2 - m$. Hence, there is some $k \in \mathbb{Z}$ such that $kp = m^2 + n^2 -m$ and therefore we have that $kp - m^2 + m = n^2$, so that $p$ divides $n^2$. Since $p$ is prime, it must divide $n$. Since $p$ divides both $m,n$ we have that $p^2$ divides $mn$ and therefore $m^2 + n^2 - m$. Also $p^2$ divides $m^2, n^2$ hence it must divide $m$.

$\endgroup$
  • $\begingroup$ Hint: $p$ divides $m^2+n^2-m$ (why?), so it also divides $n$. Thus $p^2$ divides $mn$... $\endgroup$ – Wojowu Feb 15 '17 at 14:15
  • $\begingroup$ @Wojowu Why does $p$ divide $n$? $\endgroup$ – tatan Feb 15 '17 at 14:20
  • $\begingroup$ @tatan Do you see why it divides $n^2$? $\endgroup$ – Wojowu Feb 15 '17 at 14:21
  • $\begingroup$ @Wojowu Got it...thanks!! $\endgroup$ – tatan Feb 15 '17 at 14:24
0
$\begingroup$

$p$ divides $m\implies p$ divides both $mn$ and $m^2-m=m(m-1)$.
$p$ divides $mn$ and $mn$ divides $m^2+n^2-m\implies p$ divides $m^2+n^2-m$.
$p$ divides both $m^2-m$ and $m^2+n^2-m\implies p$ divides their substraction $m^2+n^2-m-(m^2-m) = n^2$.
$p$ is prime and $p$ divides $n^2=n\cdot n\implies p$ divides $n$.


Conclusion:$p$ divides both $m^2$ and $n^2\implies p$ divides $m^2+n^2-(m^2+n^2-m) =m$.

$\endgroup$
  • $\begingroup$ wasn't it to show $p^2$ divides $m$? $\endgroup$ – miniparser Feb 15 '17 at 15:26
  • $\begingroup$ Yeah just noticed, gonna edit $\endgroup$ – Giulio Feb 15 '17 at 15:28
  • $\begingroup$ @miniparser This is usually expressed in more symmetric form about the prime divisors of values of quadratic forms, e.g. see my answer. $\endgroup$ – Bill Dubuque Feb 15 '17 at 15:39
  • $\begingroup$ @BillDubuque: i sort of understand what '$\equiv$' means. 'modular congruence'? it means that '$p$ divides $a$-$b$' for $a\equiv b$ (mod p) which also means $a$ mod $p = b$ mod $p$. i am a bit confused by cases of $n_1\equiv n_2\equiv n_3\equiv...\equiv n_k$ (mod p) though. is that the same as $n_1\equiv n_2$ (mod p) $\Longleftrightarrow n_1\equiv n_3$ (mod p) $\Longleftrightarrow...\Longleftrightarrow n_1\equiv n_k$ (mod p)? thanks. $\endgroup$ – miniparser Feb 15 '17 at 21:54
  • $\begingroup$ @miniparser $\ a\equiv b\equiv c\,$ means they all have the same remainder mod $p,\,$ or, equivalently, $p$ divides the difference of any two of them. So $\,m\equiv 0 \equiv a\,$ means $\,p\,$ divides both $m$ and $a\ \ $ $\endgroup$ – Bill Dubuque Feb 15 '17 at 22:11
1
$\begingroup$

By hypothesis $\,m = m^2+n^2-kmn\,$ for $\,\in\Bbb Z\,$ so it follows by this well known

Theorem $ $ If $\ a = m^2+n^2+kmn\,$ then prime $\,p\mid m,a\,\Rightarrow\, p^2\mid a$

Proof $\,\ {\rm mod}\ p\!:\,\ \color{#c00}m\equiv 0\equiv a\equiv \color{#c00}m^2+n^2+k\color{#c00}mn\equiv n^2\,$ so $\ p\mid n^2\Rightarrow\,p\mid n$

Therefore $\,p\mid m,n\,\Rightarrow\, p^2\mid m^2,n^2,mn\,\Rightarrow\, p^2\mid a$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.