If $x$ and $y$ are odd, prime integers, then $xy+x+y$ is not necessarily a prime number? Prove the theorems:

1) If $x$ and $y$ are odd, prime integers, then $xy+x+y$ is not necessarily a prime number.

2) If $x$ is a positive odd integer, then $x^4$ can be written in the form $8m+1$ where $m$ is an integer. Hint: $x$ can be written in the form $8i+j$ where $j$ is one of $1$, $3$, $5$ or $7$.

Thanks in advance!!
For (1), find a counterexample. You shouldn't have to look too long.

For (2), use the hint. Note in particular that $$x^4=(8i+j)^4=(8i)^4+4(8i)^3j+6(8i)^2j^2+4(8i)j^3+j^4.$$ You can factor an $8$ out of every term except the $j^4$ term, which lets you rewrite it in the form $$x^4=8(\text{a bunch of stuff})+j^4.$$ If you can show for each $j=1,3,5,7$ that there is some $n$ such that $j^4=8n+1$ (prove by cases), then it follows that $$x^4=8(\text{a bunch of stuff}+n)+1,$$ which has the right form.

A better approach for (2)--that completely avoids the need for a case-by-case proof--is to note that $x=2k+1$ for some integer $k$. Then $$x^4=(2k+1)^4=(2k)^4+4(2k)^3+6(2k)^2j+4(2k)+1.$$ There ends up being a factor of $8$ in each term but the last, so we can rewrite it as $$x^4=8(\text{different stuff})+1$$ right away.
For number 1, I see nothing saying $x$ and $y$ need to be distinct.  In this case, the number can be rewritten as $x(x+2)$.  Finding a counterexample is a little more complicated if $x$ and $y$ must be distinct.  Since $x$ and $y$ are odd, $x+1$ and $y+1$ are even, $(x+1)(y+1)$ is divisible by 4, and $xy+x+y=(x+1)(y+1)-1\equiv 3\pmod4$.  Checking composite numbers congruent to 3 mod 4, the first counterexample I can find is

$$55=5\times11=(3+1)(13+1)-1$$