Is the assertion about the form $\alpha x+\beta xy+\gamma y$ true?

Question

In my answer, I was led to conjecture the following:

Statement:
If $\gcd(\alpha,\beta,\gamma)=1,$ then every integer can be written as $\alpha x+\beta xy+\gamma y$ for integer $x$ and $y$.

If the above is true, my original question would be solved, but the validity of the original question does not imply the truth of the above assertion.
And what I can think about now is that this is just Bézout's identity, in the case $\beta=0$.
Thanks for any help, and, please, localise each inappropriate point that takes place here.

Answer 1

$\alpha x + \beta x y + \gamma y = z$ means $(\alpha + \beta y) x = z - \gamma y$ so either $z - \gamma y = 0$ (in which case you could take $x=0$) or $\alpha + \beta y \ne 0$ and $z - \gamma y$ is divisible by $\alpha + \beta y$. But then $\beta (z - \gamma y) + \gamma (\alpha + \beta y) = \beta z + \alpha \gamma$ must also be divisible by $\alpha + \beta y$. Suppose e.g. $\beta z + \alpha \gamma$ is a prime $P$. Then $\alpha + \beta y$ can only be $\pm 1$ or $\pm P$.
Now $P \equiv \alpha \gamma \mod \beta$, while $\alpha + \beta y \equiv \alpha \mod \beta$. So if $\alpha$ is not congruent to $\pm 1$ or $\pm \alpha \gamma$ mod $\beta$, you're out of luck.

For example, take $\beta = 7$, $\gamma = 2$, $\alpha = 3$,$z=1$ (so $P = 13$).

Answer 2

This is false. $\gcd(6,15,10)=1$, but there is no integer solution to $$6x+15xy+10y=1 $$

You can verify this easily by finding condition for $6x+15xy+10y>0$ and finding the local minima in those areas.

Thanks to chubakueno :

$$6x+15xy+10y+4=5 $$

$$(3x+2)(5y+2)=5 $$

So $(5y+2)$ must divides $5$, this is not possible.