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Section 8.7 Powers

Suppose \(\sqrt{n}\in\QQ\text{.}\) Thus \(\sqrt{n} = \frac{a}{b}\text{,}\) and \(\frac{a^2}{b^2} = n\text{.}\) By Lemma 8.6.4, we may assume that \(a\) and \(b\) are coprime and positive. Now since \(n = \frac{n}{1}\text{,}\) the uniqueness of Lemma 8.6.4 implies that \(b^2=1\text{.}\) Since \(b\) is positive, it follows that \(b = 1\text{.}\)

Below, we say an integer \(a\) is an \(n\)th power if \(a=b^n\) for some integer \(b\text{.}\)

If \(a=1\text{,}\) then we have \(ab=b\) is an \(n\)th power, so the theorem is trivial in this case, and similarly if \(b=1\text{,}\) so we can assume \(a,b \geq 2\text{,}\) so we can apply the FTA.

Let \(a=p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}\) and \(b=q_{1}^{s_{1}}\cdots q_{\ell}^{s_{\ell}}\) be the prime factorizations of \(a\) and \(b\text{.}\) Since \(a\) and \(b\) are coprime, they share no common prime factors, so \(p_{i}\neq q_{j}\) for all \(j\text{.}\) By assumption \(ab=c^n\) for some \(c\text{.}\) Let \(w_{1}^{t_{1}}\cdots w_{j}^{t_{j}}\) be the prime facorization for \(c\text{,}\) then

\begin{equation*} ab = p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}q_{1}^{s_{1}}\cdots q_{\ell}^{s_{\ell}} =c^n=w_{1}^{nt_{1}}\cdots w_{j}^{nt_{j}}\text{.} \end{equation*}

By the FTA, these prime factorizations are equal, which means for each \(i\text{,}\) \(p_{i}=w_{j}\) for some \(j\text{,}\) and \(r_{i}=nt_{j}\text{.}\) In particular, \(n\) divides \(r_{i}\) for all \(i\text{,}\) and so \(a\) is an \(n\)th power. The same holds for \(b\text{.}\)

Let's do an example of another diophantine equation. This one is from Liebeck (but is not proven correctly there).

Find all integer solutions to \(4x^2=y^3+1\text{.}\)

Reductions: If \((x,y)\) is a solution, then \((-x,y)\) is also a solution, so we can assume \(x\geq 0\text{.}\) If \(x=0\text{,}\) then \(y^3=-1\text{,}\) which is only possible if \(y=-1\text{.}\) So now we can assume \(x>0\text{.}\) But then \(y^3+1=4x^2\geq 4\text{,}\) so we must have \(y>0\) as well.

We are now left to finding all integer solutions \(x,y>0\text{.}\) Observe that

\begin{equation*} y^3=4x^2-1=(2x-1)(2x+1)\text{.} \end{equation*}

By The Linear Combination Lemma, the gcd of \(2x-1\) and \(2x+1\) must divide \(2x+1-(2x-1)=2\text{,}\) so the gcd is either \(1\) or \(2\text{.}\) However, since both of these numbers are odd, the gcd must actually be \(1\text{.}\) Hence, \(2x\pm 1\) are coprime.

By Theorem 8.7.2, \(2x\pm 1\) are both cubes. Thus, there are \(m,n\in\ZZ\) so that \(m^3=2x+1\) and \(n^3=2x-1\text{.}\) Since \(x\) is a positive integer, so are \(m^3\) and \(n^3\text{,}\) and thus so are \(m\) and \(n\text{.}\) Then

\begin{equation*} 2=2x+1-(2x-1)=m^3-n^3=(m-n)(m^2+mn+n^2)\text{,} \end{equation*}

so \(m-n\) divides \(2\text{,}\) hence it is either \(1\) or \(2\) (since \(m>n\)). If it is \(1\text{,}\) then \(m=n+1\text{,}\) and so by the above equation

\begin{equation*} 2=m^3-n^3=(n+1)^3-n^3=3n^2+3n+1\geq 3\cdot 1^2+3\cdot 1 + 1 = 7 \end{equation*}

which is a contradiction. The case that \(m-n=1\) can be handled similarly. Thus, there are no integer solutions to \(4x^2=y^3+1\) when \(x>0\text{.}\) Thus, the only integer solution is \((x,y)=(0,-1)\text{.}\)