* We have that x / 2^shift < 2^54. Our initial approximation to sqrtFloor(x) will be
       * 2^(shift/2) * sqrtApproxWithDoubles(x / 2^shift).
       */
      sqrt0 = sqrtApproxWithDoubles(x.shiftRight(shift)).shiftLeft(shift >> 1);
    }
    BigInteger sqrt1 = sqrt0.add(x.divide(sqrt0)).shiftRight(1);
    if (sqrt0.equals(sqrt1)) {
      return sqrt0;
    }
    do {
      sqrt0 = sqrt1;