ABC conjecture

2024-07-2813:33 Status:Complete

Suppose that $a+b=c$ where $a,b$ and $c$ are whole numbers and do not share any factors. In the example: \begin{gather}1024+81=1105\\ 2^{10} + 3^4 =5 \times 13 \times 17\end{gather} The general tendency is that is if $a$ and $b$ are primes to high powers, $c$ will generally be composed of primes. However, there are instances where this is not the case: \begin{gather}3 + 125=128 \\ 3+5^3 = 2^7\end{gather}

If you multiply only the factors of $a$ by $b$ by $c$ (e.g. $2\times 3\times 5\times 13\times 17=13260>1150$) it is the radical of $abc$ and it is usually greater than $c$. However, in cases like: $rad(3,125,128)=3\times 5\times 2=30<128$.

Thus, the ABC conjecture states that the $rad(abc{)}^k>c$. However, there are exceptions: When $k$ = 1, there are infinitely many exceptions. When $k>1$ then there are finite exceptions.

If this is proven it will prove several other theorems, (including Fermat’s final theorem, however this was solved before the ABC conjecture).

Source(s)

https://www.youtube.com/watch?v=RkBl7WKzzRw&pp=ygUPYWJjIGNvbmplY3R1cmUg