How to find natural solutions of an equation? When I'm solving problems, I'm often confronted to solving equations, and when I'm solving equations, I'm often confronted to find the natural solutions of these equations.
My actual personal insolvable problem is : 
How to determine the natural solutions of an equation ?  For example, let's take this rather simple equation : 
$9 = a+b$
I want to solve it, knowing that $a$ and $b$ are natural numbers. How can I get all the possible answers ? Am I obliged to test every possible value for $a$ and $b$ ? I need Mathematics wisdom. 
 A: The equation $a+b=9$ represents a line $L$ in the Euclidean plane with horizontal coordinate $a$ and vertical coordinate $b$. Graph that line $L$. Then look for those integer lattice points that are contained in the line $L$ and in the interior of the first quadrant. 
A: $a+b=9$
Notice that $(a,b)=(1,8)$ is one particular solution to the equation, and the null solution is $(-t, t)$. All the solutions in integers can be given by: $\text{particular solution + null solution} = (1,8)+ (t,-t) = \color{red}{(1+t, 8-t)}$
Since you want solutions in naturals:
$1+t > 0$ and  $8-t > 0$
which is same as $\color{Red}{-1\lt t\lt 8}$
so these $t$ values give you all the solutions in naturals  : $\{(1+0,8-0), (1+1,8-1), \cdots  (1+7,8-7)\} \\\{(1,8), (2,7), \cdots  (8,1)\} $
That's it for the example problem. Next, see if you can extend this to solving $a+b= \text{any number}$, and finally try solving the linear diophantine equation : $ax+by=c$
A: The solutions $\in \mathbb{N}$ , so $a,b>0$
$$a+b=9 \Rightarrow b=9-a$$
$$b>0 \Rightarrow 9-a>0 \Rightarrow a<9$$
$$$$
$$a+b=9 \Rightarrow a=9-b$$
$$a>0 \Rightarrow 9-b>0 \Rightarrow b<9$$
So,these conditions must be satisfied:
$$a+b=9, 0<a<9,0<b<9$$
