Diophantine Equation. Prove that it only have one set of equation

Question

I have this equation $25x - 35y = 40$. I have found a set of solution for it. How do I prove that it only has one set of solution? Thank you so much.

Answer 1

It's $$5x-7y=8$$ or $$5(x-y)-2y=8.$$ Let $x-y=t$.

Thus, $$5t-2y=8$$ or $$t+2(2t-y)=8.$$ Let $2t-y=-m$, where $m\in\mathbb Z$.

Thus, $$t=2m+8,$$ which gives $$y=5m+16$$ and $$x=7m+24$$ or $$y=5(m+3)+1$$ and $$x=7(m+3)+3$$ and we got the answer: $$\{(7m+3,5m+1)|m\in\mathbb Z\}$$

Answer 2

Note that the given diophantine equation is equivalent to $5x - 7y = 8$. One "easy" solution is $(3,1)$. Let $(x,y)$ be another solution, then $$5(x-3)=7(y-1)$$ and since $\gcd(5,7)=1$, it follows that $y-1=5k$ and $x-3=7k$ for some $k\in\mathbb{Z}$. Moreover, it is straightforward to verify that $(7k+3,5k+1)$ with $k\in\mathbb{Z}$ is a solution of $5x - 7y = 8$.

Therefore, the whole set of solutions is $\{(7k+3,5k+1):k\in\mathbb{Z}\}.$

Answer 3

Solutions for linear diophantine eq such as ax + by = c
exist only if gcd (a,b) divides c. (all letters stand for integers)

Then the solutions are given by

   x= r + (b/gcd(a,b))t       and  y= s -  (a/gcd(a,b))t

where r and s are any specific (integer) solution , in this case r=3, s=1, and t is any integer.