Given an integral number, determine if it's a square number

Question

I wrote this code below but it show error for no. 4 and 25
Error shown:-
(4 is a square number: False should equal True
25 is a square number: False should equal True )
Couldn't understand why?

code:-
def is_square(n):
i=1
if (n>0):
while (i<=n):
if (n%i==0):
if (n==(i**2)):
return True
else:
return False
else:
i=i+1

elif(n==0):
return True
else:
return False

Answer 1

I found your logic a bit confusing and did a try like that:

def is_sqlr(n):
if n == 0:
return False
for i in range(1,n + 1):
if i ** 2 == n:
return True
return False

print(is_sqlr(25))

Answer 2

while i**2<=n:
if n==i**2:
return True
i += 1 # i = i+1
return False

Answer 3

ive made it as identical to your code as possible

def is_square(n):
i=1
if (n>0):
while (i<=n):
if (n%i==0):
if (n==(i**2)):
return True
i=i+1
return False
elif(n==0):
return True
else:
return False

Answer 4

I didn't check the rest of the code, but what you did wrong here is that you started from i=1, literally any number%1=0, so whatever number you try will give False, change i=1 to i=2

Answer 5

As Aymane Boukrouh pointed out, your only error is in the second line i=1.
Replace it with i=2 and your code will work.

Another approach using a simple generator (TWOLINER)

https://code.sololearn.com/cFx04Symz9ly/?ref=app

Or using a list comprehension and a ternary conditional operator (ONELINER)

https://code.sololearn.com/cpb1l46tjwc2/?ref=app