How to random generate a number from 1-499 and have the number square rooted if there is no decimals output the number,square root, and description of it having no decimal places

Question

import math

import random

print(random.randint(1,499))

X = rand.randint(1,499)

math.sqrt(X)

print(X)

Answer 1

#########################################

# Here is the new one....

############################

import math

import random

x=random.randint(1,499)

print("The random number is " + str(x))

perfect_square_flag = False

done_flag = False

t=1

while (done_flag==False):

tSqr=t*t

if (tSqr==x):

perfect_square_flag=True

done_flag=True

break

if (tSqr<x):

t=t+1

else:

done_flag=True

if (perfect_square_flag):

print(str(x) + " is a perfect square with square root "+ str(int(math.sqrt(x))))

else:

print(str(x) + " is not a perfect square with square root " + str(round(math.sqrt(x),2)))

##################################################################

#------------------------- Here is the OLD one

############################################################

import math

import random

perfect_squares=[1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484]

x=random.randint(1,499)

perfect_square_flag=False

for n in perfect_squares:

if int(n)==int(x):

perfect_square_flag=True

break

if (perfect_square_flag):

print(str(x) + " is a perfect square with square root "+ str(int(math.sqrt(x))))

else:

print(str(x) + " is not a perfect square with square root " + str(round(math.sqrt(x),2)))