Square Progressive Numbers

A positive integer, $n$, is divided by $d$ and the quotient and remainder are $q$ and $r$ respectively. In addition $d$, $q$, and $r$ are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.
For example, $58$ divided by $6$ has quotient $9$ and remainder $4$. It can also be seen that $4, 6, 9$ are consecutive terms in a geometric sequence (common ratio $3/2$).
We will call such numbers, $n$, progressive.
Some progressive numbers, such as $9$ and $10404 = 102^2$, happen to also be perfect squares. The sum of all progressive perfect squares below one hundred thousand is $124657$.
Find the sum of all progressive perfect squares below one trillion ($10^{12}$).

This math problem can be solved using Python programming with the help of nested looping operation. We need to have a clear understanding of how geometric series works in order to solve this problem.

Here is how one can solve this problem:
Firstly, note that in a geometric sequence, the middle term squared equals the product of the outer two terms. Thus, if $q < r$, we must have $q^{2} = d*r$, and if $d < r$ we must have $d^{2} = q*r$. To solve this problem, the Python code below will get the sum of all progressive perfect squares under a trillion. ```python import math sum = 0 max_b = int(math.sqrt(10**12))+1 def is_square(n): x = int(math.sqrt(n)) if x*x == n or (x+1)*(x+1) == n: return True return False for a in range(1, max_b): for r in range(-1, 2, 2): b = a + r if bmax_b:
continue
n = a*(a*b + b*b)
if n>=10**12:
continue
if not is_square(n):
continue
sum += n
print (sum)
“`