Proof Verification: there are infinite values of $a$ such that $n^4 + a$ is composite for all $n \in Z$ Proof Verification: there are infinite values of $a$ such that $n^4 + a$ is composite for all $n \in Z$
Let $n^4 + a = (n^2+xn+p)(x^2+yn+q)$. Knowing that the cubic and quadratic coefficients must be zero, we can show that $x = -y, p+q=x^2,$ and $pq = a$. 
One immediate solution is $p = 1, q = 3, (x, y) = \pm2,$ for which $a = 3.$
Assume that there exists a finite, but more than zero, number of $a's$ generated in this manner, and the highest $a$ is generated from values $p, q, x,$ and $y$.
Then $x = -y, p+q=x^2,$ and $pq = a_{highest}$
Let $p_1 = q_1 = 2x^2$. Then $p_1 + q_1 = (2x)^2$. Let $x_1 = 2x$ and $y_1 = -2x$. 
Since $p_1 > p$ and $ q_1 > q$, $a_{new} = p_1q_1 > a_{highest}$, which contradicts our assumption. Thus there must be an infinite number of $a$ satisfying the conditions.
Is my proof correct?

For any integer $k>1$ we have 
$$n^4+4k^4=(n^2+2kn+2k^2)(n^2-2kn+2k^2)=((n+k)^2+k^2)((n-k)^2+k^2),$$
which is composite for all $n$, being the product of two integers greater than one.
