1. Divisibility 13

Show that:

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

2. Originally Posted by Sea
Show that:

$\displaystyle k\in\mathbf{Z^{+}}$ and $\displaystyle a|b \Rightarrow a^{k}|b^{k}$
a|b means that there exists an integer n such that $\displaystyle b=an$.
Hence $\displaystyle b^k=a^kn^k$

Therefore...

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

4. Originally Posted by JaneBennet
Are you trying to prove the converse, namely $\displaystyle a^k\mid b^k\ \Rightarrow\ a\mid b\,?$

I don't know......

But good question......

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

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

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

5. I would suggest maybe using prime factorization.

6. I would suggest maybe using prime factorization.

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

I think;

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

and

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

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

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

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

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

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

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

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

I think...?

7. 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.