Finding integer square root of an integer hobby code... I'm not sure if it's actually efficient

Question

I am very new to this and I am trying to write a program that determines whether an integer has an integer root (or not). I have combined many different ways to narrow down the search range to find the integer root including checking the ending and binary searches. Here's the code:

def endingTest (num):
  print ("Now checking the input for ending that won't have integer roots")
  print ()
  endCheck = str(num)
  lastDigit = endCheck[-1]
  noIntegerSquareEnding = ["2", "3", "7", "8"]
  for digit in noIntegerSquareEnding:
    if lastDigit in noIntegerSquareEnding:
      print ("The input ends in",lastDigit,"which can't have any integer roots")
      print ("Number ending check is done \n")
      print ("No integer root found.")
      return False
    else:
      print ("Input does not end in 2, 3, 7, 8")
      print ("Input ending check is done")
      print ()
      return True
      
def binarySearch (num):    
  print ("Now checking for the range where integer root could be found")
  digits = str(num)
  count = 0
  for digit in digits:
    count += 1
  ceiling = 10**((count/2))
  floor = 10**((count/2)-1)
  high = int(ceiling)
  low = int(floor)
  print ("starting search range is between", low,"and",high)
  while high-low>10:
    mid = (low+high)//2
    if mid**2 == num:
      print ("Integer root found: ", mid, "\n")
      print (steps)
      return [False, None, None]
    elif mid**2 > num:
      high = mid
    elif mid**2 < num:
      low = mid
  print ()
  print ("Binary search is done. Search range is narrowed down to between", low, "and", high)
  return [True, low, high]

def squareTest (num, low, high):
  print ("Begin squaring every number from", low, "to", high)
  print ()
  for i in range (low, high+1):
    if i**2 == num:
      print ("Integer root found: ", i)
      return
    
    elif i**2 > num and i < high:
      print ("The root",i,"already produces a square of",i**2, "which is higher than", num)
      print ("All values from", i+1, "to", high,"are definitely out of range and therefore skipped")
      print ()
      print ("No integer root found")
      return
    
    elif i**2 < num and i < high:
      pass
    
    else:
      print ("No integer root found")
      return
      
    

def userInput ():
  try:
    userNumber = int(input("Enter an integer to find its integer root: "))
  except:
    print("That is an invalid entry. Try again.")
    return None, True
  else:
    print ("-----------------------------")
    return userNumber, False

def findIntegerRoot(testVal):
  print ("Searching for integer root of", testVal)
  print ("-----------------------------")
  if testVal < 100:
    print ("Because the number is really small, we will try squaring numbers to find root.\n")
    print ("If", testVal, "has an integer square root, it would fall between 0 and half of", testVal, "plus 1\n")
    squareTest (testVal, 1, (testVal//2)+1)
    return
  if endingTest(testVal) == False:
    return
  if testVal <= 2500:
    print ("Because the number is relatively small, we believe it's more efficient to skip binary search step.\n")
    print ("If", testVal, "has an integer square root, it would fall between 0 and half of", testVal, "plus 1\n")
    squareTest (testVal, 1, (testVal//2)+1)
    return
  proceed = binarySearch(testVal)
  if proceed[0] == False:
    return
  high = proceed[2]
  low = proceed[1]
  squareTest (testVal, low, high)
  

def main():
  repeat = True
  while repeat == True:
    testVal, repeat = userInput()
  findIntegerRoot(testVal)
  print ("----------------------------")
  print ("End of integer root search. \n")
  

main()

Since I am very new to programming and I'm just trying this out as a hobby, I am not sure if this program is actually efficient at finding integer root. I would imagine "efficient" means it takes the least number of steps and least amount of resources to determine the outcome. Can I please ask for some feedback?

Answer 1

The great thing about Python is that it offers so many built-in functions in a very simple way:

from math import sqrt

def has_integer_root(num):
    root = int(sqrt(num))
    return root ** 2 == num

Choosing coding as a hobby was a great decision, you can find out more here.