An integer $n$ is a $k$'th power if $ a^k = nb^k\,$ for some integers $a,b$

Question

I was reading the proof of this theorem and have a little trouble understanding one part of it:

Theorem: If $k > 2$ and $n$ are natural numbers, then $n^{\frac{1}{k}}$ is irrational unless $n$ is a perfect $k$th power.

Proof: Assume the contrary: $a^{k} = nb^{k}$, and some prime divisor of $n$ has an exponent that is not a multiple of $k$. Let $p$ be such a prime and note that the exponent of $p$ in $a^k$ is a multiple of $k$, but the exponent of $p$ in $b^{k}n$ is not a multiple of $k$. This violates the Fundamental Theorem of Arithmetic, so our assumption that $n$ is not a perfect kth power and $n^{\frac{1}{k}}$ is rational must be false.

The bold part is the part that is not very clear to me.

Thanks.

Answer 1

If all prime divisors of $n$ have an exponent wich is as multiple of $k$ we have that $n$ is a perfect $k$-power. We are assuming that the theorem is wrong, so that $n$ is not a $k$-power (i.e. there is a prime $p$ that have an exponent is not a multilple of $k$ ) but it is rational. The decomposition in prime numbers of $a^k$ and $nb^k$ is the same, so the total exponent of $p$ must be the same. In $a^k$ the exponent of $p$ is a multiple of $k$, instead in $nb^k$ it is the sum of the exponent of $p$ in $n$ (called $e_1$) and the exponent of $p$ in $b^k$ (called $e_2$ that can be also $0$). Clearly $e_1+e_2$ it is not divisible by $k$ because we are summing a multiple of $k$ and a number not divisible by $k$, so you have the contraddiction.

Answer 2

Suppose $\,n^{1/k}\,$ is rational, so taking a reduced fraction we get:

$$n^{1/k}=\sqrt[k] n=\frac{a}{b}\Longrightarrow b^kn=a^k$$

Let

$$\,n_1=\prod_{i=1}^rp_i^{\gamma_1}\;\;,\;a=\prod_{i=1}^sq_i^{\alpha_i}\;\;,\;b=\prod_{i=1}^tm_i^{\beta_i}$$

be the prime decompositions of the numbers $\,n,a,b\,$ ,so that

$$a^k=\prod_{i=1}^sq_i^{\alpha_ik}=b^kn=\prod_{i=1}^tm_i^{\beta_ik}\prod_{i=1}^rp_i^{\gamma_i}$$

Since every prime dividing the RHS also divides the LHS and the other way around, we get a straighforward contradiction...can you see why? Try to round and complete the argument now.

Answer 3

If I understand correctly, in order to contradict the theorem, you need to find a number n such that its kth root is non-integer rational number. If that is the case, it can always be written in the form given. Also, for RHS, if some prime divisor of n was present with powers in multiple of k, it would be subsumed by b, hence it is safe to say that there exists some prime divisor of n whose power in n is not multiple of k. The proof then goes on to say that it is not possible if we consider LHS.

Answer 4

Suppose that $a^k=nb^k$. Let $a=p_1^{r_1}p_2^{r_2}\dots p_m^{r_m}$, where $p_1,\dots,p_m$ are distinct primes. Then $a^k=p_1^{kr_1}p_2^{kr_2}\dots p_m^{kr_m}$. This of course means that $nb^k=p_1^{kr_1}p_2^{kr_2}\dots p_m^{kr_m}$ as well.

Consider any prime factor $q$ of $nb^k$. Let $q^t$ be the highest power of $q$ that divides $n$, and let $q^s$ be the highest power of $q$ that divides $b$. Then the highest power of $q$ that divides $nb^k$ is $q^{t+ks}$: $nb^k$ gets $t$ factors of $q$ from $n$ and $s$ factors of $q$ from each of the $k$ factors of $b$ in $b^k$. By the fundamental theorem of arithmetic the prime factorization of $a^k=nb^k$ is unique, so $q^{t+sk}$ must be one of the factors $p_i^{kr_i}$, meaning that $q=p_i$ and $t+sk=kr_i$. But then $t=kr_i-ks=k(r_i-s)$ is a multiple of $k$.

In other words, if $n=q_1^{t_1}q_2^{t_2}\dots q_\ell^{t_\ell}$ is the prime factorization of $n$, then each $q_i$ is one of the primes $p_1,\dots,p_m$ and, more important, each $t_i$ is a multiple of $k$. Say $t_i=\alpha_ik_i$ for $i=1,\dots,\ell$; then $$n=q_1^{\alpha_1 t_1}q_2^{\alpha_2t_2}\dots q_\ell^{\alpha_\ell t_\ell}=\left(q_1^{\alpha_1}q_2^{\alpha_2}\dots q_\ell^{\alpha_\ell}\right)^k\;,$$

and $n$ is therefore a perfect $k$-th power.

The argument that you quoted is phrased as a proof by contradiction instead of a direct proof showing that if $n^{1/k}$ is rational, then $n$ is a perfect $k$-th power, but the reasoning is essentially the same.

Answer 5

Let the power of the prime $\rm\,p\,$ in $\rm\:A,B,N,\:$ be $\rm\:a,b,n\:$ resp. Comparing powers of $\rm\,p\,$ yields

$$\begin{eqnarray}\rm A^k &\,=\,&\rm N\times B^k\\ \Rightarrow\ \ \rm a\,k &\,=\,&\rm n\, +\, b\,k\\ \Rightarrow\quad\rm n &\,=\,&\rm (a\!-\!b)\,k \end{eqnarray} $$

So $\rm k\,$ divides $\rm\,n,\,$ i.e. all primes $\rm\,p\,$ dividing $\rm N$ occur to powers a multiple of $\rm k,\,$ so $\rm N $ is a $\rm\,k$-th power

$$\rm\, N\, =\, P^{\,JK}\cdots\, Q^{\,L K} =\, (P^{\,J}\cdots Q^{\,L})^K$$

Remark $ $ The argument depends crucially on the Fundamental Theorem of Arithmetic, i.e. the existence and uniqueness of prime factorizations of naturals. To be completely rigorous you should explicitly mention where it was implictly invoked (many times) in the above sketched proof.