3
$\begingroup$

Let $n$ be a positive integer with $k$ distinct prime divisors. Prove that there exists a positive integer $a$ with $1<a<\frac{n}k+1$ such that $n|a^2-a$

I can't solve it. Could someone help me?

$\endgroup$
  • $\begingroup$ Trivial observation: if $n$ is a prime power then we can take $a=n$. $\endgroup$ – Dietrich Burde Jan 5 '18 at 19:18
  • $\begingroup$ @DietrichBurde Now Chinese remainder. $\endgroup$ – orole Jan 5 '18 at 19:19
2
$\begingroup$

Let $n=\prod _{i=1}^kp_i^{c_i}$ with $c_i\ge 1$. Fix $j\in\{0,\ldots,k\}$. Then we can find $a_j\in\{1,\ldots,n\}$ with $a_j\equiv 1\pmod {p_i^{c_i}}$ for $i\le j$ and $a_j\equiv 0\pmod {p_j^{c_j}}$ for $i>j$.

If we order the $k+1$ distinct numbers $a_j$ in ascending order (apparently staring with $a_k=1$ and ending with $a_0=n$), there must be two consecutive terms $a_r,a_s$ in this sequence with $0<a_s-a_r\le\frac{n-1}{k}$. Let $\tilde a=a_s-a_r$.

First assume $s>r$. Then neither of $r,s$ is $=0$, hence $p_1^{c_1}\mid \tilde a$. In particular, $\tilde a>1$. We have $\tilde a\equiv 1\pmod {p_i^{c_i}}$ for $r<i\le s$ and $\tilde a\equiv 0\pmod {p_i^{c_i}}$ otherwise. We conclude that $\tilde a(\tilde a-1)\equiv 0\pmod n$. As $\frac{n-1}k<\frac nk+1$, we find a solution by letting $a=\tilde a$.

Next assume $s<r$. As above, we have $\tilde a\equiv 0\text{ or }{-1}\pmod {p_i^{c_i}}$ for all $i$, hence $\tilde a(\tilde a+1)\equiv 0\pmod n$. So if we let $a=\tilde a+1>1$, we have $a(a-1)\equiv 0\pmod n$. Also, $a\le \frac{n-1}k+1<\frac nk+1$, hence this $a$ has the desired properties.

$\endgroup$

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.