Let's consider all integers less than 50000 and identify how many are semiprimes. First we will eliminate all evens and all primes. That leaves 24330 composite candidates. Next we must eliminate composites with more than two factors. This leaves 11740 candidate semiprimes. Finally, we will eliminate the perfect squares (though technically semiprimes). This leaves a total of 11576.

Of these semiprimes, how many conform to the type that the next perfect square minus N equals any perfect square (x²)? That is:

     (floor(√(N))+1)² - N = x²

We know that for every semiprime there is a y such that (N+y)² - N = x² where √y is not greater than about N/6. But the problem is that, as N becomes bigger, the N/6 ceiling becomes a relatively much bigger number - a greater number of perfect squares - than the quantity of odd numbers less than the √N - and therefore a far less efficient way to find a factor than by trial division.*

However, we are considering here only the difference between N and the closest perfect square greater than N. If this difference is a perfect square, we know we can derive its factors instantly, using the procedure in the left column.

So it doesn't matter how large N is in this special case - it could be 200 digits - if it conforms to this simple formula, it can be factored instantly. There are 228 such semiprimes under 50,000, and they are shown below.

Not suprisingly, these semiprimes are the symmetrical kind - in fact, superficially, without knowing this trick, they would appear to be the 'hardest' kind for oversized Ns - where the factors are fairly close to the square root.