The relation F on Z is defined by F = {(n,m)| (2n+3m) is divisible by 5}
I need help proving this relation is symmetric.
I know I should choose two members of Z and assume F(x,y).
and I know I need to prove F(y,x).
But I keep getting stuck on the algebra. I'm not sure how to manipulate the equation to get 2y + 3x = 5*(some int)

Choose x,y ∈ Z and assume F(x,y)
So 5|2x+3y
Which means 2x+3y = 5 * k for some integer K
-1(2x + 3y) = -1(5k) (mult both sides by -1)
-2x - 3y = -5k (distribute the negative)
-2x + 5x - 3y + 5y = -5k + 5x + 5y (add 5x and 5y to both sides)
2y + 3x = 5(-k + 5x + 5y)
Since -k, 5, x, y ∈ ℤ then -k + 5x + 5y ∈ ℤ (since integers are closed under addition)
Therefore 5|2y + 3x


One observation is that


Since $5$ divides the left side always, it follows that $5\mid 2x+3y$ iff $5\mid 2y+3x$.


You should assume $F(x,y)$ and prove $F(y,x)$, not $F(y,z)$

There are only five cases for $x \bmod 5$. For each one, you can compute $y \bmod 5$. Now note that you have $F(y,x)$ as well.


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