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?
 A: 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.
A: Your proof seems to confuse composite numbers and reducible polynomials (and in fact, $n^4+3$ is irreducible, so there is something wrong with your computation -- why is the leading coefficient of the second factor $x^2?$). Also, for $n=2,$ the number $2^4+3 = 19$ is not composite.
A: Mr. Igor Rivin has point out the error in you proof. We may approach for it by Fermat's Little Theorem.
We have $n^4 \equiv 1$ (mod $5$). Which gives $5| n^4 -1$, i.e. $n^4 -1$ is composite. Take $a \equiv 1$ (mod $5$). You shall get countably infinite $a$ s.t. $n^4 + a$ is a composite number.
A: I am sorry I had to rethink the solution. 
By the Division algorithm we can prove that the square of any integer is of the form $3k$ or $3k +1$. An extension would give that $n^4 = 3k $ or $3k+1$ for some integer $k$.
If $n^4 = 3k$ then for any $a$ in $\{ 3p \ |$   $p$ is an integer     $\}$ the expression $n^4 +a$ is composite given that $p$ is non-zero.
Similarly  $n^4 = 3k + 1$ then any $a$ in $\{ 3p +2 \ |$   $p$ is an integer     $\}$ will render the expression $n^4 +a$ composite.
Both sets mentioned above have countably infinite many elements. 
A: See here: This is obvious that n^4+4k^4=(n^2+2kn+2k^2)(n^2−2kn+2k^2)=((n+k)^2+k^2)((n−k)^2+k^2)
 the above written expression is always composite for all n and k>1. 
As you can see the expression :
                                                               n^4 + 4k^4 can be factorised in this manner....so for any values of k>1 the number is composite.
So, the form of the number 'a' should be of form :   4k^4 
And easily it is observed that there are infinitely many number which is of form :  4(k^4).
