# Prove that for all whole numbers $x, y, z$, there exists $\big(\frac{d}{2b} + 1\big)^2$ solutions for $x + by + cz = d$.

Prove that: $$\forall \{x, y, z\} \subset \mathbb{W}, \ \exists \bigg(\frac{d}{2b} + 1\bigg)^2 \text{ solutions for } x + by + cz = \left\{d : \frac{d}{2b} \in \mathbb{Z}\right\}$$

$\mathbb{W}$ denotes the set of whole numbers, which is the set of numbers from $0, 1, 2, \ldots , \infty$. Simultaneously we could say $x, y, z$ are non-negative integers $($thus $\mathbb{W} = \mathbb{Z}^+)$.

I came across this in another post and it looked very interesting, but what was not mentioned was why it is true, hence my question. How can I prove this?

My attempt to find solutions for $b, c, d$ when $x, y, z$ have given values depends solely on the values of $x, y, z$, and I do not know any general formula for this problem. I believe there are questions similar to this, but not asking for this specific proof. Also, is this a Diophantine Equation by the way?

• The result as stated is not true, as it should be symmetric across $b$ and $c$, since the problem is symmetric across both $b$ and $c$. Also, what if $d/2b$ is not an integer? Then you are stating that there exists a non-integer amount of solutions which makes no sense. – Isaac Browne Oct 21 '17 at 23:31