Only One Equation Given?

The question almost tricked me, because only natural number solutions are needed.

$x+y+xy=20\quad \Rightarrow (x+1)(y+1)-1=20\quad \Rightarrow (x+1)(y+1)=21$

For integer solutions of $x$ and $y$ , we have:

$\begin{array} {rrrr} x+1 & y+1 & x & y \\ -21 & -1 & -22 & -2 \\ -7 & -3 & -8 & -4 \\ -3 & -7 & -4 & -8 \\ -1 & -21 & -2 & -22 \\ 1 & 21 & 0 & 20 \\ 3 & 7 & 2 & 6 \\ 7 & 3 & 6 & 2 \\ 21 & 1 & 20 & 0 \end{array}$

There are only $\boxed{2}$ natural ordered pairs of $(x,y)$ solutions, which are $(2,6)$ and $(6,2)$ .

We can rewrite the equation as $(x + 1)(y + 1) = 21.$ Now as both $x$ and $y$ must be natural numbers we are looking to factorize $21$ as a product of two integers each greater than $1$ . There are only two ordered ways of doing this, namely $3*7$ and $7*3$ . Thus there are just $\boxed{2}$ solution pairs $(x,y)$ , namely $(2,6)$ and $(6,2)$ .

Depending on whether or not you include or exclude 0 as a natural number and depending on whether you count inverted solutions as unique, you could have either 2 or 4 solutions. (2,6) and its inverse of (6,2) as well as (0, 20) and its inverse of (20,0).

Here's my C code

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#include<stdio.h>

void main()
{
    int i,j;
    int count = 0;
    for(i=1;i<=19;i++)
    {
        for(j=1;j<=19;j++)
        {
            if((i + j + i*j)==20)
            {
                printf("%d %d\n",i,j);
                count++;
            }
        }
    }
    printf("Count is : %d\n",count);
}

I know that this isn't very formal, but I found a solution of (2,6) right away just by guessing.

After guessing, and by the commutative property of addition and the associative property of multiplication, you can reverse x and y for the other solution (6,2).

first- x=0 then equation gives 20 second y=0 then equation gives 20