jmc

For $1\le a\le 7,$ we give a value of $x$ for which $x^4+a^2$ is prime: $$\begin{array}{ccc}a& x& a^4+x^2\\ 1& 1& 2\\ 2& 1& 5\\ 3& 10& 10009\\ 4& 1& 17\\ 5& 2& 41\\ 6& 1& 37\\ 7& 20& 160049\end{array}$$For $a=8,$ $$\begin{array}{rl}{x}^4+a^2& =x^4+64\\ & =x^4+16x^2+64-16x^2\\ & =(x^2+8{)}^2-(4x{)}^2\\ & =(x^2+4x+8)(x^2-4x+8).\end{array}$$For any positive integer, both factors $x^2+4x+8$ and $x^2-4x+8$ are greater than 1, so $x^4+64$ is always composite. Thus, the smallest such $a$ is $\boxed{8}.$