Optimus Prime

Since $N \; = \; n^2 - 2^{2014} \times 2014n + 4^{2013}(2014^2-1) \; = \;\big(n - 2^{2013}\times2014\big)^2 - 4^{2013} \; = \; \big(n - 2^{2013}\times2013\big)\big(n - 2^{2013}\times2015\big)$ the only way that $N$ can be prime is if one of these factors is $\pm1$ , and hence $n$ is one of $2^{2013}\times2013 + 1 \hspace{1cm} 2^{2013}\times2013-1 \hspace{1cm} 2^{2013}\times2015+1 \hspace{1cm} 2^{2013}\times2015-1$ The corresponding value of $N$ in each of these cases is $1 - 2^{2014} \hspace{2cm} -1 - 2^{2014} \hspace{2cm} 2^{2014} + 1 \hspace{1cm} 2^{2014} - 1$ respectively. But $3 = 2^2 - 1$ divides $2^{2014}-1$ , while $5 = 2^2 + 1$ divides $2^{2014} + 1$ . Thus no integer value of $n$ exists that makes $N$ prime.