Question

wap accept an integer and test wheater it is prefect square or not​

Answer 1

Answer:

import math

a = int(input("Enter an integer: "))

if(math.sqrt(a) % 1 != 0):

   print(a, "is not a perfect square")

else:

   print(a, "is a perfect square.")

Answer 2

Question:-

Write a program to accept an integer and test whether it is a perfect square or not.

Code:-

Here is your code, written in Java.

import java.util.*;

class Perfect_Square

{

public static void main(String args[])

{

Scanner sc = new Scanner(System.in);

System.out.print("Enter an integer.");

int n=sc.nextInt();

double s=Math.sqrt(n);

if(s==(int)s)

System.out.println("Number is a perfect square. ");

else

System.out.println("Number is not a perfect square. ");

}

}

How it works?

Square root of a perfect square integer is always an integer. Consider a number say 49

n=49

s=7.0

(int)s=7

So, 7.0==7 which is true.

So, it is a perfect square.

Again, consider a number, say 3.

n=3

s=1.732...(square root of 3)

(int)s=1

But, 1.732==1 is false.

So, 3 is not a perfect square.

In this way, the program works.