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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.

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This type of equation is called Diophantine ? I didn't knew that. Is there a general method to solve this type of equation ? –  Pyrofoux Aug 14 '14 at 15:20
Disclaimer: I am by no means an expert on the subject, I just tried to point you in the right direction. But from the Wikipedia article: "In 1900, in recognition of their depth, David Hilbert proposed the solvability of all Diophantine problems as the tenth of his celebrated problems. In 1970, a novel result in mathematical logic known as Matiyasevich's theorem settled the problem negatively: in general Diophantine problems are unsolvable." –  MGA Aug 14 '14 at 15:25
There's a general method of solving Diophantine equations of the form $ax+by=c$. Here's a great M.SE post about it. –  user26486 Aug 14 '14 at 16:20

3 Answers 3

up vote 10 down vote accepted


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 : $ax+by=c$

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Just a note: In some parts of the world, $0\in \mathbb{N}$ is the dominant convention, so it might be good to add a remark concerning $(0,9)$ and $(9,0)$. –  Daniel Fischer Aug 14 '14 at 15:31
Indeed, where I live $0 \in N$. –  Pyrofoux Aug 14 '14 at 16:29

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.

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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:

$$a+b=9, 0<a<9,0<b<9$$

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