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I am trying to find all $x,y \in \mathbb{Z}$ such that $32x + 5y = 1$.

Here's how I see the situation. Since $(32,5) = 1$, we know there exists infinitely many integer solutions. Since $x_0 = -2$ and $y_0 = 13$ are a particular solution, we have $x = 5t - 2$; $y = -32t + 13$ given that $x = \dfrac{b}{d}t + x_0$ and $y = \dfrac{-a}{d}k + y_0$ for any solvable linear diophantine equation of the form $ax + by = 1; (a,b) = 1$.

Therefore we have: $32(5t -2) + 5(-32t + 13) = 160t - 64 - 160t + 65 = 1$ so $\{(x,y) | x = 5t -2, y = -32t + 13, t \in \mathbb{Z}\}$ is a complete set of solutions.

Is this reasoning sound?

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    $\begingroup$ It's ok for me, but I prefer to write the set of solutions as $\{\,(5t-2,-32t+13) \mid t\in\mathbb{Z}\,\}$. $\endgroup$ – Éric Guirbal Oct 19 '15 at 1:46
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    $\begingroup$ Once you know that $x=5t-2$ and $y=-32t+13$ satisfy your equation, you are essentially done, which brings you to Eric's answer...More important, I think you know how to tackle these kind of equations (called Diophantic) so +1 $\endgroup$ – imranfat Oct 19 '15 at 2:10
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You really doubt yourself too much, this one is reasonable as well!

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