Find all of the possible solutions of $250x+111y=7$, where both $x$ and $y$ are integers.

Question

This is my steps

1 = 28-1x27

  = 28-1x(111-3x28)

  = 4x28-1x111

  = 4x(250-2x111)-1x111

  = 4x250-9x111

x7=> 7 = 28x250-63x111

Then I don’t know why the next step is this

$ = 250(28-111k) + 111(-63+250k) , \forall k\in \mathbb{Z}$.

Answer:

$x = 28-111k, y = -63+250k$.

Answer 1

You have shown that $(x,y)=(28,-63)$ is a solution to $$250x+111y=7.$$ If $(x,y)=(m,n)$ is another solution to this equation, then $$250(28-m)+111(-63-n)=0,$$ or equivalently $$250(28-m)=111(63+n).$$ By unique factorization of integers, and because $250$ and $111$ are coprime, it follows that $28-m$ is a multiple of $111$ and that $-63-n$ is a multiple $250$. So for some integer $k$ we have $$28-m=111k\qquad\text{ and }\qquad 63+n=250k.$$ Equivalently, we see that $$(m,n)=(28-111k,-63+250k),$$ for some integer $k$. It is then easy to verify that every integer $k$ yields a solution.

Answer 2

$$x=250^{-1} \cdot 7=28^{-1} \cdot 7=4^{-1} \cdot 1=112^{-1} \cdot 28=$$ $$=1^{-1} \cdot 28=1\cdot 28=28 \left(\mod {111} \right).$$

$$x=111n+28$$

$$250 \left( 111n+28 \right)+111y=7<=>...$$