If $xy$ and $x+y$ are both even integers (with $x,y$ integers), then $x$ and $y$ are both even integers The title statement can be proven using the contrapositive, note that $x$ odd or $y$ odd means that at least one of $x\cdot y,x+y$ is odd.  Is there a way to prove the statement directly?
To generalize on this statement (in two integer variables), show that for every square free integer $n$, there exists a function $f:\mathbb Z^2\to\mathbb Z$ where 
$$f(x,y)\equiv 0\pmod n\iff x\equiv y\equiv 0\pmod n$$
This generalization is formalized in this question: If $n$ is squarefree, $k\ge 2$, then $\exists f\in\Bbb Z[x_1,\dots,x_k] : f(\overline x)\equiv 0\pmod n\iff \overline x\equiv \overline 0\pmod n$
 A: If $xy$ is even, then $x$ or $y$ is even. If $x+y$ is even and $x$ is even, then so is $y$. If $x+y$ and $y$ are even, then so is $x$.
A: Begin with the quantities $x\cdot y$ and $x+y$ given as even integers where $x,y$ are integers.  Then we have that $2|x\cdot y\implies 2|x\text{ or }2|y$.  Without loss of generality, assume that $2|x\implies \exists m\in\mathbb Z\text{ such that }x=2m.$  Next, we have $2m+y$ is even, which means that $\exists n\in\mathbb Z$ such that $2m+y=2n$ which means that $y=2n-2m$.  $m,n$ are integers, so $2n-2m$ is even and therefore $y$ is even.
Thus we have shown directly that $x\cdot y$ and $x+y$ both even for $x,y$ integers implies that $x$ and $y$ are both even integers.
A: If $x+y$ is even, then either both $x$ and $y$ are odd, or they are both even. Therefore $x$ and $y$ are both even because otherwise $xy$ would be odd.
A: Note that
$$
x^2=(x+y)x-xy
$$
and
$$
y^2=(x+y)y-xy
$$
being the differences of two even numbers, are both even.
Thus, both $x$ and $y$ are even.
A: Observe that $\displaystyle (x+1)(y+1)=xy+x+y+1$ is odd as $xy,x+y$ are even
If $x$ is odd $\iff x+1$ is even 
A: Here if x+y is even then x and y both are even or both are odd. If one is even and other is odd, x+y will be odd. But to make xy even both x and y should be even or either x or y should be even. If both are odd then xy will be odd.
Hence it can be proven then to make xy and x+y even, x & y should be even.
Take any value of x and y and check the answer.
