Sums of Square Reciprocals

There are several ways to write the number $\dfrac{1}{2}$ as a sum of square reciprocals using distinct integers.
For instance, the numbers $\{2,3,4,5,7,12,15,20,28,35\}$ can be used:
$$\begin{align}\dfrac{1}{2} &= \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dfrac{1}{5^2} +\\
&\quad \dfrac{1}{7^2} + \dfrac{1}{12^2} + \dfrac{1}{15^2} + \dfrac{1}{20^2} +\\
&\quad \dfrac{1}{28^2} + \dfrac{1}{35^2}\end{align}$$
In fact, only using integers between $2$ and $45$ inclusive, there are exactly three ways to do it, the remaining two being: $\{2,3,4,6,7,9,10,20,28,35,36,45\}$ and $\{2,3,4,6,7,9,12,15,28,30,35,36,45\}$.
How many ways are there to write $\dfrac{1}{2}$ as a sum of reciprocals of squares using distinct integers between $2$ and $80$ inclusive?

To find the number of ways to write $\dfrac{1}{2}$ as a sum of reciprocals of squares using distinct integers between $2$ and $80$ inclusive, we can use a combination of brute force and a recursive approach.

First, let’s define a function called `find_ways` that takes in a target fraction, lower and upper bounds for the integers, and a current sum of squares.

“`python
def find_ways(target, lower, upper, current_sum):
# Base case: if the target is reached, return 1
if target == 0:
return 1
# Base case: if the target goes below 0 or the current integer exceeds the upper bound, return 0
if target < 0 or lower > upper:
return 0

# Recursive case: try adding the next integer square to the sum
include_next = find_ways(target – 1 / (lower ** 2), lower + 1, upper, current_sum + [lower])

# Recursive case: try excluding the next integer square from the sum
exclude_next = find_ways(target, lower + 1, upper, current_sum)

# Return the total number of ways by summing the include and exclude cases
return include_next + exclude_next
“`

Using this function, we can call `find_ways(1/2, 2, 80, [])` to find the number of ways to write $\dfrac{1}{2}$ as a sum of reciprocals of squares using distinct integers between $2$ and $80$ inclusive.

“`python
num_ways = find_ways(1/2, 2, 80, [])
print(num_ways)
“`

The output will be the total number of ways.

(Note: This solution may require a significant amount of time and computational resources for larger inputs, as it uses a brute force approach. Implementations of dynamic programming or optimizations such as memoization could significantly improve the efficiency of this solution.)

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