Python Find next perfect square not working in python

[shivajikobardan]

def find_next_square(sq):
    # Return the next square if sq is a square, -1 otherwise
    sq2=(sq**1/2)
    xyz=isinstance(sq2, int)
    if (xyz==True):
        print("Is perfect square")
        nextsq=sq+1
        print("Next perfect square=",nextsq**2)
    else:
        print("Not perfect square")
        return -1

n=int(input("Enter an integer"))
find_next_square(n)

Output-:
Enter an integer25
Not perfect square

Expected output-:
Enter an integer25
Next perfect square=36

 

[shivajikobardan]

My logic is that xyz checks if $(sq)^0.5$ is integer or not. If it is integer we find next perfect square, else we return -1

 

[shivajikobardan]

I think I found the problem

sq=25
sq2=(sq**(1/2))
print(sq2)
xyz=isinstance(sq2, int)
print(xyz)

This generates output as 5.0 and false. I need a way to get integer as sq2.

 

[shivajikobardan]

I need a way to do this-: If I input 25, answer should be 5 as integer. If I input 24 answer should be 4.898989486 float. Is this possible to do in python?

 

[skifli]

I think this is what you want.

def find_next_square(sq):
    # Get the square root of sq.
    square_root = sq ** 0.5

    print(f"The square root of {sq} is {square_root}. The type of the variable square_root is {type(square_root)}")

n = int(input("Enter an number: "))
find_next_square(n)