Problem Set 3

[*Mahi*]

Now it's enough to show that $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor \ge p^a$

Actually, $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor = $ The highest power that divide $n!$. So if thought $P^a||m$ and $P^a||n!$ then you have proved that $a \geq P^a$. Actually it doesn't make any sense to me. May be i couldn't get your solution correctly or you just make a bug. Please explain your solution to us more briefly.

He said $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor \ge p^a$ or if $p^x||n!$ then $x \geq a$, hope it clear things up.

[Phlembac Adib Hasan]

As $2$ and $a$ are co-primes, $x|2$ or $x|a$

What if $x=2k$ where $(2,k)=1$ , $2|2$, $k|a$. Are you considering $x$ a prime?I didn't get your solution...

Thank you.I missed the case.Now the proof has been edited.

[Phlembac Adib Hasan]

No 8:

$2^{2a}\equiv1(modb)$
$2^{2b}\equiv1(modc)$
$2^{2c}\equiv1(moda)$

a,b,c wont be even . So that
If b \geq 3
$2^{(2a,\phi(b))} \equiv 2^2 mod \equiv 1(modb)$ (i)
else $b=1$ (ii)
let $a=2n+1$ then for (i)
$2^{2n+1}+1=2^a+1 \equiv 3 \equiv 0( mod b)$
or $b=3$

Thus we can say $c=1,3$,$a=1,3$ are possible values of $b,c$ too.

$ a,b,c$ be pairwisely co prime ,
So that all possible triple is $(1,1,1)$ and permutations of$(3,1,1)$

If $(a,b)=1$ then it's not always true $(2a,\phi (b) )=2$ because $\phi (b)$ may have such prime factors that $b$ doesn't have.

[sourav das]

Now it's enough to show that $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor \ge p^a$

Actually, $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor = $ The highest power that divide $n!$. So if thought $P^a||m$ and $P^a||n!$ then you have proved that $a \geq P^a$. Actually it doesn't make any sense to me. May be i couldn't get your solution correctly or you just make a bug. Please explain your solution to us more briefly.

He said $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor \ge p^a$ or if $p^x||n!$ then $x \geq a$, hope it clear things up.

I meant to actually he has to proof $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor=a \ge x$ where he proved that $a\geq P^x$

[Nadim Ul Abrar]

:'( .... ......

[Masum]

Still there are mistakes. You know that $b$ does not share any common factor with $a$ but $\varphi(b)$ may share, because there are some $p_i-1$'s in $\varphi(b)$ who may contain a prime factor of $a$.