I was trying to solve some algorithms when I found out that one could leverage a simple stupid linear Diophantine equation, but I cannot figure out how solve it (I'm pretty sure I would have been able to do that back in high-school but I guess I am getting rusted).
Basically, I ended up on this equation: $ax+by=c$, where $a=3$, $b=5$ and the triplet $\lbrace a, b, c\rbrace \in ℕ$
Which is to say the least the most "basic" Diophantine equation.
Obviously, I am trying to find out the value of $x$ and $y$ according to $c$ that are .
The equation has a solution upon the condition that $c$ is a multiple of $gcd(a,b)=gcd(3,5)=1$
I noticed that there is an obvious solution here, which is the couple: $\lbrace x=2c, y=-c\rbrace$
Supposedly other solutions are of the form: $\lbrace x + kv, y − ku\rbrace$ where $k$ is an arbitrary integer such as: $u=\frac{a}{gcd(a,b)}=\frac{3}{1}=3$ and $v=\frac{b}{gcd(a,b)}=\frac{5}{1}=5$.
Therefore other solutions are more specifically $\lbrace 2c = x + 5k, -c = y − 3k\rbrace$
Equivalent to: $\lbrace x = 2c - 5k, y = -c + 3k\rbrace$
At that point I cannot manage to find the value of $k$ that would make $x$ and / or $y$ positive.
The replacement strategy seem to lead to nothing.
I am pretty sure it might be really easy but cannot get my head around it.
Any idea?
[Edit]
My point is how to find out $k$. If I'm just replacing $x$ and $y$ with their respective values found above in the former equation, I am going to end up with:
$3(2c+5k) + 5(-c+3k) = c$
$\Rightarrow 6c + 15k -5c + 15k = c$
$\Rightarrow c + 30k = c$
$\Rightarrow 30k = c - c = 0...$
I don't think $k$ is equal to $0$... so I might be wrong.
