Elementary number theory exponent problem (Peter Hackman book)

Question

Let $a$,$b$ be positive integers satisfying $a^j$ = $b^k$ for positive and relatively prime integers j,k.

Show that a = $r^k$, b = $r^j$ for some positive integer r.

Can you help me to figure it out ?

Answer 1

Since $a^j=b^k$, the exponents in its prime factorization are all multiples of $k$. Suppose $p^i\|a$, then we have that $k|ij$. Since $k$ and $j$ are relatively prime, then $k|i$. This is true for each prime dividing $a$, so $a$ is a $k$th power. A symmetric argument shows that $b$ is a $j$th power.

(The notation $p^i\|a$ means that $p^i|a$ and $p^{i+1}\not| a$.)

Answer 2

Notation: for a prime $p$ and a positive integer $n$ let $m_p(n)$ be the greatest integer such that $p^{m_p(n)}\mid n$.

If $a=1$ then $b=1$ and the statement is obvious.

Otherwise, suppose that a prime $p$ divides $a$ and that $s=m_p(a)$.

Then $p^{js}$ divides $b^k$ and $js=m_p(b^k)$. Let $t=m_p(b)$. Then $js=kt$. Since $\gcd(j,k)=1$, $j\mid t$ and $k\mid s$. Say $t=d_1j$ and $s=d_2k$. We get

$$jd_2k=kd_1j$$

Therefore, $d_1=d_2$ (call it $d_p$) and $p^s=(p^{d_p})^k$, $p^t=(p^{d_p})^j$.

Now use this reasoning for every prime factor of $a$. You can define $d_p=m_p(a^j)/k$ for each prime factor of $a$ to get

$$a=\left(\prod_{p\mid a} p^{d_p}\right)^k$$

$$b=\left(\prod_{p\mid a} p^{d_p}\right)^j$$

Answer 3

I'm going to try to make use of this:

Since $j$ and $k$ are relatively prime, there are integers $u$ and $v$ such that $uj-vk = 1$.

After some experimentation, this looks promising:

If $a^j = b^k$, then $a^{uvj} = b^{uvk}$.

We will need the result that if a rational number raised to an integer power is an integer, then the rational number must be an integer.

$b^{uvk} =a^{uvj} = (a^v)^{uj} = (a^v)^{vk+1} = (a^v)(a^v)^{vk} $ so $a^v =\frac{b^{uvk}}{(a^v)^{vk}} =(\frac{b^{u}}{a^v})^{vk} $ or $a =(\frac{b^{u}}{a^v})^{k} $.

For $(\frac{b^{u}}{a^v})^{k}$ to be an integer, $\frac{b^{u}}{a^v}$ must be an integer, which proves the result for $a$.

Similarly $a^{uvj} =b^{uvk} =(b^u)^{vk} =(b^u)^{uj-1} $ so $b^ua^{uvj} =(b^u)^{uj} $ or $ba^{vj} =b^{uj} $ or $b =(\frac{b^u}{a^v})^j $.

Therefore $r = \frac{b^u}{a^v}$ works.