Proof: For all integers $x$ there exists an integer $y$ so that $3$ divides $x+y$ and $3$ divides $x-y$ We start of a proof by finding out if it is true or false, I am having trouble with this and I think it is true, because the negation (there exists and integer $x$ for all integers $y$ so that $3$ does not divides $x+y$ and $3$ does not divides $x-y$) says the same thing? So I tried to prove it for it being true.
First thing I tried was to use the definition of divides:
$3(k) = x+y$ and $3(p)=x - y$ for integers $k$ and $p$, 
then I isolated for $x$ in one of the equations: 
$3(k) - y = x$ so $3(p) = 3(k) - y - y$, 
then this would become:
$3(p) = 3(k) - 2(y)$ and since $k$ and $y$ are integers $p$ is also an integer. 
Would this reasoning be correct?
 A: Your reasoning is not quite correct because $3p = 3k-2y$ does not guarantee $p$ to be an integer since $p = k + \frac{2}{3}y$.
In fact, this statement is not actually true. Take some integer $x$. It is congruent to some value, let's call it $a$ modulo $3$. Take any other integer $y$, which is congruent to some $b$ modulo $3$. By some case analysis, we can see that our result is only true when $x\equiv 0\mod 3$:
If $x\equiv 0 \mod 3$ then choose $y=0$ and we're done. If $x\equiv 1 \mod 3$ then we must have $y\equiv 2 \mod 3$ to get $x+y\equiv 0 \mod 3$, but then we see $x-y\equiv 2\mod 3$. Likewise if $x\equiv 2 \mod 3$, we need $y\equiv 1 \mod 3$ to get $x+y\equiv 0 \mod 3$ but then $x-y\equiv 1 \mod 3$.
A: Hint:
It's simpler with congruences:
the required properties are that 
$$x+y\equiv 0\mod 3\quad\text{ and }\quad x-y\equiv 0\mod 3$$
Adding these congruences, we obtain
$$(x+y)+(x-y)=2x\equiv 0\mod 3,$$
in other words $3$ divides $2x$. However $3$ is prime – what can you deduce from this relation?
A: Remember that $(a+b)\pmod n = [a\pmod n + b\pmod n] \pmod n$, for any $a$ and $b$ positive or negative.
So, if $x \pmod 3 = 0$, we are done, and $y=0$ for instance.
If $x \pmod 3 = 1$, we may set $y=2$ (or any other $y$ such that $y \pmod 3 = 2$) so that $(x+y) \pmod 3 = 3 \pmod 3 = 0$. But $(x-y) \pmod 3 = (1 + 1) \pmod 3  = 2 \neq 0$!
So the statment is not true and we cannot satisfy both $(x+y) \pmod 3 = 0$ and $(x-y) \pmod 3 = 0$.
Note: $-2 \pmod 3 = 1$, not $-2$!
