>>947

The x at (aa x bb) is equal to c-a^2, though that's kind of the same result you already have.

>>1259

PMA, you probably already noticed, but it looks like some of your calcs are hitting 32-bit integer overflow limits in the c=6107 example.

>>1283

So it looks like the pattern is driven by x^2+e equaling 144 as a fixed constant. So n(d-x) = na = 72, and thus n and a can take on all the various factor combinations of 72, which also gives d with the offset of 11 from a. And b=d+x+2n.

>>1299

In general e is always going to be bounded by 2d, because (d+1)^2 = d^2 + 2d + 1.

That is, if the remainder e gets up to 2d+1, you just tick over to records using the next larger d value.

>>1309

You may want to check your code for why you didn't find any more in the pattern.

The next record in the sequence containing 225 is (0, 128, 465, 240, 225, 961), and there should be an infinite number of them afterwards.

>>1342

PMA, you may have a bug in your sqrt() routine, since for n=2 it's saying sqrt(1)=0 and sqrt(4)=1.

>>1369

For c=ab where a and b are primes, there are only two differences of squares that work, one trivial (corresponding to c=1c), and the other true solution with c=a*b.

If c has additional factors (like the aa*bb cases), there will be additional solutions, though.

>>1371

Yep, exactly. This is because you can transform a difference of squares into a two-part factorization and vice-versa using y^2-z^2 = (y+z)(y-z).

NIST recently posted a bunch of submissions for potential quantum-resistant crypto algorithms:

http:/ /csrc.nist.gov/Projects/Post-Quantum-Cryptography/Round-1-Submissions