Find integer solutions to $7236x + 3283y = 201$

Question

Find integer solutions to $$7236x + 3283y = 201$$ and show no solutions exist if $x \ge 0$ and $y \ge 0$.

This boils to $108x + 49y = 3$ Taking mods, we get:

$10x \equiv 3 \pmod{49}$, which means $x \equiv 15 \pmod{49}$ so $x = 15 + 49k$. That means $y = 3 - 1620 - 5292k = -1617 - 5292k$.

Thus, $(x, y) = (15 + 49k, -1617 - 5292k)$.

Suppose $x \ge 0$ then, $k \ge 0$, which implies that $-5292k < 0$ thus $-1617 - 5292k < 0$. Suppose $y \ge 0$ then $-5292k \ge 1617 \implies k \le -1$ which implies $15 + 49k < 0$.

Complete?

Answer 1

By your calculation, there is a suitable $y$ if $x=15$. Substituting in the equation $108x+49y=3$ we get $y=-33$.

You showed that the general solution for $x$ is $x=15+49k$, where $k$ ranges over the integers. The appropriate $y$ for $x=15+49k$ is $y=-33-108k$.

As to solutions with $x$ and $y$ both $\ge 0$, looking at the original equation shows us there cannot be any. For $(0,0)$ does not work, and if one or both of $x$ or $y$ is $\ge 1$, then $7236x+3283y$ is much larger than $201$.