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.
BdMO National 2012: Higher Secondary, Secondary 06
BdMO National 2012: Higher Secondary, Secondary 06
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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 3n1.so p cannot be 2.
so, a is in form $12n1$ and $12n+5$ form.(respectively in form $4k1,4k+1$)
(a,p)=1 so,
p divides $a^pa$
p divides $a^p+1$
p divides $a+1$
if a is in form $12n1$,$a+1=12n$,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)
so, a is in form $12n1$ and $12n+5$ form.(respectively in form $4k1,4k+1$)
(a,p)=1 so,
p divides $a^pa$
p divides $a^p+1$
p divides $a+1$
if a is in form $12n1$,$a+1=12n$,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=6kp1. It kills the problem.
I just took a=6kp1. It kills the problem.
Your hot head might dominate your good heart!
 nafistiham
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Re: BdMO National 2012: Higher Secondary, Secondary 06
when $p$ is not $2$, $p$ is odd.so,
\[a+1a^p+1\]
so, $a$ has infinitely many solutions
when $p=2$, $a \neq 2k$ as $6p=12$ is even
if, $a=2k+1$
suppose,
\[12a^2+1\]
\[\rightarrow 12 4k^2 + 4k + 1 + 1\]
\[\rightarrow 6  2(k^2+k)+1\]
contradiction
\[a+1a^p+1\]
so, $a$ has infinitely many solutions
when $p=2$, $a \neq 2k$ as $6p=12$ is even
if, $a=2k+1$
suppose,
\[12a^2+1\]
\[\rightarrow 12 4k^2 + 4k + 1 + 1\]
\[\rightarrow 6  2(k^2+k)+1\]
contradiction
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
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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=6p1+6kp will satisfy the condition for every k.
 nafistiham
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Re: BdMO National 2012: Higher Secondary, Secondary 06
$4^3,100,10^3=1000$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.
trying $5,6,7,8,9$, the palinedrome is $343=7^3$
how can i do it without trial and error
\[\sum_{k=0}^{n1}e^{\frac{2 \pi i k}{n}}=0\]
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