Exercise java recursion

I don't understand how this exercise return the number's square. In particular I don't understand the sixth line in which there is return statement and after this "+2*x-1". What is the program behavior in that call?

public class Es {
    public static int RecCalc(int x) {
        if (x==0) {
            return 0;
        }else {
            return RecCalc(x - 1) + 2 * x - 1;
        }
    }

    public static void main(String[] args) {
        System.out.println(RecCalc(3));
    }
}

We can see how this works with a little algebra:

(x-1)² + 2x - 1
== x² - 2x + 1 + 2x - 1
== x²

If you are unfamiliar with the formula for (x + y)² then you can do (x-1)² by writing it as (x-1)*(x-1) and using the FOIL method or the distributive property. This is left as an exercise for the reader.

If you have 4 things, you can make a square with side 2:

xx
xx

If you want to make a square with side 3, you need 9 things: add 2 things on each of the side and bottom, plus 1 for the corner:

xx.
xx.
..+

Or, to put it another way, add 3 things on each of the side and bottom, take away 1 for the corner.

Generalizing, if you have a square of side length (n-1), to make a square of side length (n), you have to add on 2 lots of (n-1) things, plus one; or 2 lots of (n) things, take away one.

Hence:

    number of things in a square of side length n

=  (number of things in a square of side length (n-1))
   + 2 * (n-1) + 1

=  (number of things in a square of side length (n-1))
   + 2 * n - 1

Let's step through, one call at a time.

The first call to kick it all off is:

RecCalc(3);

In Java, the return statement will take everything up to the semi-colon.
So, return 3 + 2 will return 5 to the caller.

RecCalc(3) will result in calling:
RecCalc(2) + 2*3 -1;

RecCalc(2) will result in calling:
RecCalc(1) + 2*2 -1;

RecCalc(1) will result in calling:
RecCalc(0) + 2*1 - 1;

RecCalc(0) will return 0.

Now we can work our way back up the call stack.

RecCalc(0) == 0
RecCalc(1) == RecCalc(0) + 2*1 -1 == (0) + 2*1 -1 == 1
RecCalc(2) == RecCalc(1) + 2*2 -1 == (1) + 2*2 -1 == 4
RecCalc(3) == RecCalc(2) + 2*3 -1 == (4) + 2*3 -1 == 9

This doesn't explain the math, but explains the recursion.

Let's look at the math.
As explained by @CodeApprentice, x² = (x-1)² + 2x -1
The real trick to this whole recursive scheme is the (x-1)².

We know that for x = 4, we can use (x-1)² plus some other junk to get the answer.
But that's just the square of 3 plus some other junk!

Now, to get the square of 3, we know that 3² = (x-1)² plus junk.
But that's just the square of 2 plus some other junk!

And so, we work our way down until we get to a trivial answer, where we return 0. (In fact, you could also return 1 for x=1).

I hope that explains it!

Perhaps if you put in a print statement it will help.

public static void main(String[] args) {
	System.out.println(RecCalc(5));
}
	
public static int RecCalc(int x) {
	if (x == 0) {
		return 0;
	} else {
		int v = RecCalc(x - 1) + 2 * x - 1;
		System.out.println((x-1) + " " + (2*x) + " " + (-1));
		return v;
	}
}

Prints

0 2 -1
1 4 -1
2 6 -1
3 8 -1
4 10 -1
25

Notice that the value of the sum of the last two columns in each line is an odd number. And the sum of any number of consecutive odd numbers starting with 1 is a perfect square. So essentially, this method just sums up the first x odd numbers.