Python program to find if any number is a perfect square or not without using math module.

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Length of int(b**0.5) is always coming to be 1, this is the problem.

If the no. Will be perfect it's root length will be small and if it is not a perfect square then it will be in decimal and it's root length will be big.

What I have tried:

Python

Def square():
    a=int(input("Enter the no."))

    b=int(a**0.5)
    If Len(b)==1 or Len(b)==2:
        Print("the number is a perfect square")

    Elif Len(b)>3:
        Print("is not perfect square") 

#Its not working.

Posted 18-Aug-22 6:58am

Akshat Parmar

Updated 18-Aug-22 22:12pm

Patrice T

v3

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Patrice T 18-Aug-22 14:02pm

How "Its not working.", give details

Richard Deeming 19-Aug-22 3:47am

"if it is not a perfect square then it will be in decimal"

So why are you casting the square root to an int then?

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Richard MacCutchan · Answer 1 · 2022-08-18T22:12:00

Pretty much every line of code is incorrect. It looks as if you have been trying to write VB.NET statements with Python expressions. Python is a case sensitive language and all statements and builtin functions use lower case. You also do not need to check the length of b, but its value from the previous expression. You also do not need to check if it is greater than 3 (actually should be greater than 2) since if it is not 1 or 2 then it must be greater than 2. So go to The Python Tutorial — Python 3.10.6 documentation[^] and study the proper syntax of the language. Then think about what you need to do to correct your code.