Divisibility 13

**Sea** · December 20th 2008, 09:39 AM

Show that:

$k\in\mathbf{Z^{+}}$ and $a|b \Rightarrow a^{k}|b^{k}$

**Moo** · December 20th 2008, 10:14 AM

Originally Posted by Sea

Show that:

$k\in\mathbf{Z^{+}}$ and $a|b \Rightarrow a^{k}|b^{k}$

a|b means that there exists an integer n such that $b=an$ .
Hence $b^k=a^kn^k$

Therefore...

**JaneBennet** · December 23rd 2008, 02:24 PM

Are you trying to prove the converse, namely $a^k\mid b^k\ \Rightarrow\ a\mid b\,?$

**Sea** · December 23rd 2008, 03:08 PM

Originally Posted by JaneBennet

Are you trying to prove the converse, namely $a^k\mid b^k\ \Rightarrow\ a\mid b\,?$

I don't know...

...

But good question...

...

$a^3|b^3 \Rightarrow a|b$ (Why?) ...write...

http://www.mathhelpforum.com/math-help/number-theory/65693-divisibility-14-a.html

I think...Maybe use binomial theorem...

**JaneBennet** · December 23rd 2008, 03:20 PM

I would suggest maybe using prime factorization.

**Sea** · December 23rd 2008, 05:23 PM

I would suggest maybe using prime factorization.

$ a^k|b^k (\Rightarrow):$

I think;

$(p_1 \times p_2 \times p_3 \times ....\times p_m,N)=1$

and

$(p_1 , p_2 , p_3 , ... , p_m)=1$

$a^k=(p_1^{y_1}\times p_2^{y_2}\times p_3^{y_3} \times .......p_m^{y_S})^k$

$=(p_1^{y_1})^k\times (p_2^{y_2})^k\times (p_3^{y_3})^k\times .......(p_m^{y_S})^k$

$ =(p_1^{ky_1}\times p_2^{ky_2}\times p_3^{ky_3} \times .......p_m^{ky_S}) $

$ b^k=(p_1^{z_1}\times p_2^{z_2}\times p_3^{z_3} \times .......p_m^{z_S})^k \times N$

$ =(p_1^{z_1})^k\times (p_2^{z_2})^k\times (p_3^{z_3})^k \times .......(p_m^{z_S})^k \times N $

$=p_1^{kz_1}\times p_2^{kz_2}\times p_3^{kz_3} \times .......p_m^{kz_S} \times N$

$a^k|b^k \Rightarrow p_i^{ky_i} | p_i^{kz_i}$ and $i=1,2,3,....,s$ $\Rightarrow ky_i \leq kz_i \Rightarrow y_i \leq z_i , k\in Z^+ \Rightarrow p_i^{y_i} | p_i^{z_i}$

I think...?

**InvisibleMan** · January 16th 2009, 03:53 AM

If your'e trying to prove that a^k|b^k => a|b then
a^k=nb^k
(a/b)^k=n
a/b=n^1/k
Now a|b iff n=m^k for some integer m, the problem is to show that it is so.

You can assume by contradiction that a!|b which means there exists p and r such that:
a=pb+r r<=b
Now write a^k=(pb+r)^k and see that it contradicts our assumption.

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