discrete mathematics and proofs Let $a$ and $b$ be in the universe of all integers, so that $2a + 3b$ is a multiple of $17$.
Prove that $17$ divides $9a + 5b$. 
In my textbook they do $17|(2a+3b) \implies 17|(-4)(2a+3b)$.
They do this with the theorem of $a|b \implies a|bx$.
However, I don't know how the book got $x=-4$.
What is the math behind this?
This is just a section of the steps that complete the proof.
Once I know how the book figured out $x$ was $-4$ then i will be happy.

The author of the book "cheated" here.
We know: if $17$ divides $2a+3b$, then $17$ divides $k(2a+3b)$ for any integer $k$.
The author, aiming to write an interesting problem, would have chosen $k$ so that $(2k,3k) \text{ mod } 17$ didn't look like an obvious multiple of $(2,3)$ modulo $17$.  So the author of the question picked $-4$, so $(2k,3k) \text{ mod } 17=(9,5)$.  The author knew how to prove the claim, since the author chose $-4$.
When I first saw the problem (without knowing $-4$ in advance), it didn't strike me as obvious to multiply by $-4$.  As an attempt at a proof, I probably would have tried multiplying by $\{1,2,\ldots,16\}$ on a computer to see if any of them worked.  It would have found that $13$ works (which is $-4 \pmod {17}$).  Then I would have presented the proof without explaining the process of discovery (i.e., without mentioning the $15$ failed attempts).
