4 ms·

You forgot to sum over N

• jordigh 9y ago
  Yeah, okay, I messed up. Let's argue in a different way: there are approximately sqrt(N) squares between 1 and N, so let's call sqrt(N)/N = 1/sqrt(N) the "probability" that N itself is square. That also goes to zero as N goes to infinity. If we pick higher powers of N, it goes to zero much quicker.
  • nshepperd 9y ago
    That's not what ykler meant. 1/sqrt(N) is the "probability" that N itself is square. We don't care about that, we want the probability that there is any solution to x^2 = N, which is sum{N=0..∞} (1/sqrt(N)). That does not go to zero. ETA: Well, actually the above is the expected number of solutions, so naturally it diverges because there are (infinite) solutions. A more proper way would be to calculate the probability that there are no solutions, which indeed goes to zero. But the probability of there being a solution, and the expected number of solutions, are the same if it is << 1.
    • deleted 9y ago
      [deleted]