Fermat Factorisation

Used when p and q are numericaly close

If $p$ and $q$ are numerically close, we can use Fermat Factorisation.

During Fermat Factorisation, we hope to find $a$ and $b$ such that

$$a^2 - b^2 = N$$

Because that then means we can factorise the left-hand expression into

$$(a+b)(a-b)=N$$

As thus we get the two factors of $N$ as $(a+b)$ and $(a-b)$.

The reason we use this when $p$ and $q$ are numerically close is because the closer they are to each other the closer they are to $\sqrt{N}$. If we say $a=\sqrt{N}$ rounded up to the nearest number, we can calculate $b^2 = a^2-N$ (as rearranged from before) until $b$ is a whole number, after which we've solved the equation.

An example of this attack can be found in this writeup, which may make it a bit clearer.