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How to check if a given number is Fibonacci number?

import math 
 
def isPerfectSquare(x):
    s = int(math.sqrt(x))
    return s*s == x
 
 
def isFibonacci(n):
 
    # n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or both
    # is a perfect square
    return isPerfectSquare(5*n*n + 4) or isPerfectSquare(5*n*n - 4)
 
 for i in range(1, 11):
    if (isFibonacci(i) == True):
        print(i, "is a Fibonacci Number")
    else:
        print(i, "is a not Fibonacci Number ")

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Check Armstrong Number

def power(x, y):
     
    if y == 0:
        return 1
    if y % 2 == 0:
        return power(x, y // 2) * power(x, y // 2)
         
    return x * power(x, y // 2) * power(x, y // 2)
 
def order(x):
 
    # Variable to store of the number
    n = 0
    while (x != 0):
        n = n + 1
        x = x // 10
         
    return n
 
def isArmstrong(x):
     
    n = order(x)
    temp = x
    sum1 = 0
     
    while (temp != 0):
        r = temp % 10
        sum1 = sum1 + power(r, n)
        temp = temp // 10
 
    # If condition satisfies
    return (sum1 == x)
 
# Driver code
x = 153
print(isArmstrong(x))
 
x = 1253
print(isArmstrong(x))

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Program to Find Area of a Circle

def findArea(r):
    PI = 3.142
    return PI * (r*r);
 
# Driver method
print("Area is %.6f" % findArea(5));

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Print All Prime Numbers in an Interval

def prime(x, y):
    prime_list = []
    for i in range(x, y):
        if i == 0 or i == 1:
            continue
        else:
            for j in range(2, int(i/2)+1):
                if i % j == 0:
                    break
            else:
                prime_list.append(i)
    return prime_list
 starting_range = 2
ending_range = 7
lst = prime(starting_range, ending_range)
if len(lst) == 0:
    print("There are no prime numbers in this range")
else:
    print("The prime numbers in this range are: ", lst)

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Cube Sum of First n Natural Numbers

n=5
s=0
for i in range(1,n+1):
    # adding cube sum using pow() function
    s=s+pow(i,3)
print(s)

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