As easy as ABC

I haven't done a blog post for a very longtime so I think I should return with something I think is special. I.e. THE ABC CONJECTURE!

The abc conjecture is a quite famous (and possibly solved but im not sure on the current status of that) problem in number theory. Before I explain the problem we need to learn some definitions.

Let $x$ be an integer. We denote the square free part of $x$ as $rad(x).$ i.e. if $x$ has prime factorisation, $x=p_{a_{1}}^{n_{1}}p_{a_{2}}^{n_{2}}...p_{a_{r}}^{n_{r}},$ then $rad(x)=p_{1}...p_{r}.$

Let $a,b,c\in\mathbb{N}$ such that the greatest common divisor of $a$ and $b$ is one and such that $a+b=c.$  We say that $(a,b,c)$ is an abc triple.

Now let's get to the conjecture.

Let $\epsilon>0.$ The abc conjecture states that there are only finitely many abc triples such that
\[rad(abc)^{1+\epsilon}<c. \]

So most of the time $c$ is less than the product of the dinstinct primes of $a$,$b$ and $c$.

This doesn't look too obvious or even apealing at first so why do we care about? Well it turns out that lots of interesting things follow very easily. Here is an example on Fermat's last theorem:

Let $n\geq 6$ and $x,y,z\in\mathbb{N}$ such that $x$ and $y$ are coprime and
\[x^n+y^n=z^n\].

Let $a=x^n,b=y^n$ and $c=z^n.$ Then

\[rad(abc)^2=rad(x^{n}y^{n}z^{n})^{2}\leq rad(x^{n})^{2}rad(y^{n})^{2}rad(z^{n})^{2} \]
\[= rad(x)^{2}rad(y)^{2}rad(z)^{2}\leq x^{2}y^{2}z^{2}<z^{6}\leq z^{n}=c. \]

And so if the abc conjecture holds there are only finitely many solutions to Fermat's last theorem for $n\geq 6.$ Yes FLT is already proved in full but this is a nice simple example of its use.

Now lets get to a result I discovered assuming the abc conjecture. I concerns arithmetic progressions and powerful numbers. An integer $x$ is powerful if it has the property that $p^{2}|x$ whenever $p|x$ from all primes $p$. Note that all powerful numbers can be expressed as the product of a square and a cube.

Let $d$ be an integer and let $(a_{n})_{n=1}^{\infty}$ be an arithmetic progression with common difference $d$ such that $a_{1}$ and $d$ are coprime. If $a_{k}, a_{k+1}$ and $a_{k+2}$ are all powerful then we say that they are a powerful triple. Suppose that the abc conjecture holds. Then $(a_{n})$ contains finitely many powerful triples.

PROOF: Let $rad(d)=N.$ We prove that there are only finitely many powerful triples $a_{k}, a_{k+1}$ and $a_{k+2}$  such that $a_{k}>N^{5}.$ Suppose that (a_{k},a_{k+1},a_{k+2}) is a powerful triple. Since they are powerful let
\[a_{k}=e^{2}x^{3}, a_{k+1}=f^{2}y^{3}, a_{k+2}=g^{2}z^{3}.\]
 Then let $a=d,$ $b=a_{k}a_{k+2}$ and $c=(a_{k+1}).$ We we wish to show that there exists an epislon such that

\[rad(abc)^{1+\epsilon}<c. \]

I claim that $\epsilon=\frac{1}{6}$ works. Since
\[rad(abc)^{\frac{7}{6}}=rad(da_{k}a_{k+1}^{2}a_{k+2})^{\frac{7}{6}}\leq N^{\frac{7}{6}}rad(e^{2}f^{2}g^{2}x^{3}y^{3}z^{3})^{\frac{7}{6}}=N^{\frac{7}{6}}rad(efgxyz)^{\frac{7}{6}}\]
\[\leq N^{\frac{7}{6}}(efgxyz)^{\frac{7}{6}}<N^{\frac{7}{6}}(a_{k}a_{k+1}a_{k+2})^{\frac{7}{6}}<N^{\frac{7}{6}}a_{k+1}^{\frac{7}{4}}<a_{k+1}^{2}=c .\]

Which concludes the proof.

As a side note I'm pretty sure there will be infinitely many pairs of powerful numbers. You have to solve a Diophantine equation that looks a bit like Pell's equation so there are bound to be infinitely many solutions(maybe).