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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.

Community
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awllower
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    I wonder if this is true when $\alpha$, $\beta$, and $\gamma$ all have a pairwise GCD of 1. – Ryan Aug 29 '13 at 07:19
  • @Ryan See the example by Robert Israel: the $\alpha,\beta,\gamma$ there are pairwise coprime, but the assertion is still false. – awllower Aug 29 '13 at 08:12

2 Answers2

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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.

Xoff
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  • Thanks for your interesting counter-example! – awllower Aug 28 '13 at 15:31
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    A proof that may be simpler: $6x+15xy+10y=1 \iff (3x+2)(5y+2)=5 \implies 5y+2|5 \implies 2|5$.Contradiction. – chubakueno Aug 28 '13 at 22:20
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    @chubakueno I think your last step could be modified: $$5y+2\mid 5\implies 5y+2\mid (5y+2)-5y=2\implies y=0\implies 2\mid 5.$$ Of course the key is the factorisation. :b – awllower Aug 29 '13 at 08:08
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$\alpha x + \beta x y + \gamma y = z$ 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 $P = 13$).

Robert Israel
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  • Thanks for this simple answer: it told me that simple congruence-considerations could rule out the solvability of this equation. So my conjecture is plain wrong... Thanks again. :P – awllower Aug 28 '13 at 15:32
  • From this calculation we can even deduce that, if $\gcd(\alpha,\beta)=1,$ then a sufficient condition for the solvability is that $\alpha+\beta y\mid \alpha\gamma+\beta z.$ – awllower Aug 29 '13 at 14:03
  • An amusing but not surprising fact: Denote $\alpha\gamma+\beta z=n_1, \alpha+\beta y=n_2.$ Then $n_2\equiv \alpha\pmod \beta, n_1\equiv \alpha\gamma\pmod\beta.$ So $(n_2\mid n_1)\implies \exists k\in\mathbb Z, (\alpha+\beta y)(\gamma+\beta k)=(\alpha\gamma+\beta z)\implies \alpha k+\gamma y+\beta yk=z$. Namely, this condition is really equivelent with our original equation. – awllower Aug 29 '13 at 14:17