Why does this algorithm to calculate number of pronic numbers in an interval work?

Question

I am given $A$ and $B$, where $A$ is less than or equal to $B$, and they are in the range of $[1, 100000000]$. I want to calculate the number of pronic numbers in that interval $[A, B]$. These numbers are numbers $n$ such that $k \times (k + 1) = n$ for some $k$. The following algorithm works, but I am having a lot of trouble understanding the intuition behind it. Can one explain to me why it works?

def pronic(num) :
 
    # Check upto sqrt N
    N = int(num ** (1/2));
 
    # If product of consecutive
    # numbers are less than equal to num
    if (N * (N + 1) <= num) :
        return N;
 
    # Return N - 1
    return N - 1;
 
# Function to count pronic
# numbers in the range [A, B]
def countPronic(A, B) :
     
    # Subtract the count of pronic numbers
    # which are <= (A - 1) from the count
    # f pronic numbers which are <= B
    return pronic(B) - pronic(A - 1);

Answer 1

First of all, denoting the number of pronic integers in $1,\ldots,m$ by $N_m$, the number of pronic integers in $A,\ldots,B$ is $N_B - N_{A-1}$: this counts the number of pronic integers in $1,\ldots,A-1,A,\ldots,B$, minus their number in $1,\ldots,A-1$.

Next, a formula for $N_m$. It is the unique integer $x$ such satisfying $x(x+1) \leq m < (x+1)(x+2)$: the pronic integers in $1,\ldots,m$ are $1\cdot(1+1),2\cdot(2+1),\ldots,x\cdot(x+1)$. Roughly speaking, $x \approx \sqrt{m}$, and this suggests checking whether $\lfloor \sqrt{m} \rfloor$ might work. The main observations are $$ m = \sqrt{m}^2 < (\lfloor \sqrt{m} \rfloor + 1)^2 < (\lfloor \sqrt{m} \rfloor + 1)(\lfloor \sqrt{m} \rfloor + 2) $$ and $$ m = \sqrt{m}^2 \geq \lfloor \sqrt{m} \rfloor^2 > \lfloor \sqrt{m} - 1 \rfloor \lfloor \sqrt{m} \rfloor. $$ If $\lfloor \sqrt{m} \rfloor (\lfloor \sqrt{m} \rfloor + 1) \leq m$ then $x = \lfloor \sqrt{m} \rfloor$ works, and otherwise $x = \lfloor \sqrt{m} \rfloor - 1$ does.