How can I convert this iterative statement into a recursive statement?

So I have this programming project in which I need to create a program that determines if a number is a perfect square, and if so, write it into a .txt document. This is very easy and effective to do with a for loop, however, the instructions for the assignment say that the program should accomplish this using recursion. This is the iterative statement I came up with:

double division;
		for (int i = 0; i < inputs.size(); i++) {
			division = (Math.sqrt(inputs.get(i)));
			if (division == (int)division) {
				pw.println(inputs.get(i));
			     }
			}

Where inputs is an ArrayList that is created by reading the inputs of the user.
This solves the problem, but like I said, it needs to be a recursive statement. I know that for recursion I need a base case that will eventually make the method stop calling itself, but I can't figure out what the base case would be. Also, I've seen several examples of converting from iteration to recursion, but all of these examples use a single int variable, and in my case I need to do it with an ArrayList.
Any help would be greatly appreciated

For recursive function, you can use bynary search algorithm:

 int checkPerfectSquare(long N,  
                              long start, 
                              long last) 
{ 
    // Find the mid value 
    // from start and last 
    long mid = (start + last) / 2; 
  
    if (start > last) 
    { 
        return -1; 
    } 
  
    // Check if we got the number which 
    // is square root of the perfect 
    // square number N 
    if (mid * mid == N) 
    { 
        return (int)mid; 
    } 
  
    // If the square(mid) is greater than N 
    // it means only lower values then mid 
    // will be possibly the square root of N 
    else if (mid * mid > N) 
    { 
        return checkPerfectSquare(N, start,  
                                  mid - 1); 
    } 
  
    // If the square(mid) is less than N 
    // it means only higher values then mid 
    // will be possibly the square root of N 
    else 
    { 
        return checkPerfectSquare(N, mid + 1,  
                                  last); 
    } 
} 

You could use the fact that a square number is the sum of the odd integers. E.g.

1+3 = 4 = 2^2

1+3+5 = 9 = 3^2

1+3+5+7 = 16 = 4^2, etc

public static void main(String[] args) {
    for (int i = 1; i < 1000; i++) {
      if (isSquare(i)) System.out.println(i);     
    }
  }
  public static boolean isSquare(int n) {
    if (n==0 || n==1) return true;
    return isSquare(n,1,1);
  }

  private static boolean isSquare(int n, int sum, int odd) {
    if (n==sum) return true;
    if (n < sum) return false;
    odd += 2;
    sum += odd;
    return isSquare(n, sum, odd);
  }

output:

1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
256
289
324
361
400
441
484
529
576
625
676
729
784
841
900
961

You could recursively check if the square of any smaller int is equal to your input.

public static boolean isSquare(int n) {
  if (n==0 || n==1) return true;
  return isSquare(n, 1);
}

private static boolean isSquare(int n, int i) {
  if (i*i == n) return true;
  if (i*i > n) return false;
  return isSquare(n,i+1);
}