diff --git a/src/Python/Problems001-050/Problem044.py b/src/Python/Problems001-050/Problem044.py
index 5b2b1be..60a6689 100644
--- a/src/Python/Problems001-050/Problem044.py
+++ b/src/Python/Problems001-050/Problem044.py
@@ -1,4 +1,4 @@
-#!/usr/bin/env python3
+#!/usr/bin/env python
 """
 Created on 30 Aug 2021
 
@@ -11,38 +11,42 @@ https://projecteuler.net/problem=44
 
 from itertools import combinations
 from operator import add, sub
+
 from utils import timeit
 
+
 def pentagonal(n):
-    return int(n*(3*n-1)/2)
+    return int(n * (3 * n - 1) / 2)
+
 
 @timeit("Problem 44")
 def compute():
     """
-    Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2.
+    Pentagonal numbers are generated by the formula, Pn=n(3n-1)/2.
     The first ten pentagonal numbers are:
 
         1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
 
     It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their
-    difference, 70 − 22 = 48, is not pentagonal.
+    difference, 70 - 22 = 48, is not pentagonal.
 
     Find the pair of pentagonal numbers, Pj and Pk, for which their
-    sum and difference are pentagonal and D = |Pk − Pj| is minimised.
-    
+    sum and difference are pentagonal and D = |Pk - Pj| is minimised.
+
     What is the value of D?
     """
-    dif = 0
-    pentagonal_list = set(pentagonal(n) for n in range(1,2500))
+
+    ans = 0
+    pentagonal_list = set(pentagonal(n) for n in range(1, 2500))
     pairs = combinations(pentagonal_list, 2)
     for p in pairs:
         if add(*p) in pentagonal_list and abs(sub(*p)) in pentagonal_list:
-            dif = (abs(sub(*p)))
+            ans = abs(sub(*p))
             # the first one found would be the smallest
             break
-                
-    return dif
+
+    return ans
+
 
 if __name__ == "__main__":
-
-    print(f"Result for Problem 44: {compute()}")
\ No newline at end of file
+    print(f"Result for Problem 44 is {compute()}")