将函数定义为宏的结果无法解释

Question

#include <stdio.h>

#define square(x) x*x

int main() {
    for (int i = 0, j = 0; i < 10; ++i, ++j) {
        int a = square(i + j);
        printf("square(%d + %d) = %d\n", i, j, a);
    }
    return 0;

}

在竞猜网站上观看了类似的代码后，我编写了上面的代码，上面的代码的输出确实很奇怪。如果您能解释发生了什么，将不胜感激。
在GCC编译器上，输出如下所示

square(0 + 0) = 0
square(1 + 1) = 3
square(2 + 2) = 8
square(3 + 3) = 15
square(4 + 4) = 24
square(5 + 5) = 35
square(6 + 6) = 48
square(7 + 7) = 63
square(8 + 8) = 80
square(9 + 9) = 99