## BdMO National 2012: Higher Secondary, Secondary 06

Moon
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### BdMO National 2012: Higher Secondary, Secondary 06

Problem:
Show that for any prime $p$, there are either infinitely many or no positive integer $a$, so that $6p$ divides $a^p + 1$. Find all those primes for which there exists no solution.
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photon
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### Re: BdMO National 2012: Higher Secondary, Secondary 06

6 divides $a^p+1$,so, a is odd,$a^p$ is in form 3n-1.so p cannot be 2.
so, a is in form $12n-1$ and $12n+5$ form.(respectively in form $4k-1,4k+1$)
(a,p)=1 so,
p divides $a^p-a$
p divides $a^p+1$ p divides $a+1$
if a is in form $12n-1$,$a+1=12n$,then p divides 12n p divides n
if a is in form $12n+5$,$a+1=6(2n+1)$ then p divides 2n+1
set of a={12n:p divides n} U {12n+5:p divides 2n+1}
except 2 any prime p it happens.
is it ok?(sorry for not using latex properly)
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FahimFerdous
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### Re: BdMO National 2012: Higher Secondary, Secondary 06

You're right.

I just took a=6kp-1. It kills the problem.

nafistiham
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### Re: BdMO National 2012: Higher Secondary, Secondary 06

when $p$ is not $2$, $p$ is odd.so,
$a+1|a^p+1$
so, $a$ has infinitely many solutions
when $p=2$, $a \neq 2k$ as $6p=12$ is even
if, $a=2k+1$
suppose,
$12|a^2+1$
$\rightarrow 12| 4k^2 + 4k + 1 + 1$
$\rightarrow 6 | 2(k^2+k)+1$

contradiction $\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
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abdullah al munim
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### Re: BdMO National 2012: Higher Secondary, Secondary 06

For p=2,there is no quadratic residue of a number modulo 12 which equals -1.for the odd primes, a=6p-1+6kp will satisfy the condition for every k.

nafistiham
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### Re: BdMO National 2012: Higher Secondary, Secondary 06

Masum vaia wrote: Right now, my number of posts is a palindromic cube of $3$ digits. Find the number of my posts.
I think this problem is even better. $4^3,100,10^3=1000$
trying $5,6,7,8,9$, the palinedrome is $343=7^3$
how can i do it without trial and error $\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.