hello, we obtain $\frac{1}{4} \left(2 n-2^t-2^t t\right) \left(2 n+2^t-2^t t\right)$ with $t=2014$ Sonnhard.

it's easy to see that the expression it's equal to (n- 2^2013 * 2015)(n- 2^2013 * 2013), if n E [2^2013 * 2013, 2^2013 * 2015], the expression it's negative or 0, so it's not prime, ‎if n- 2^2013 * 2015=1 => n- 2^2013 * 2013= 1+2^2014 but 3 divides it, the other cases do the factor negative so not prime and when n- 2^2013 * 2013=-1 => the expression is 1+2^2014 too, and the other cases do the expression negative so not prime, we're done.

Simply use square identities: (a-b )^2 = a^2 -2ab + b^2 & (x^2 - y^2) = (x-y)(x+y). Now the result is a product of two numbers and hence it is not a prime number.

Another solution: Suppose that $n^{2}-2^{2014}\cdot 2014n+4^{2013}\cdot (2014^{2}-1)=p$ for some prime $p$ and some positive integer $n$. then we have that $\sqrt{4^{2014}\cdot 2014^{2}-4^{2014}\cdot (2014^{2}-1)+4p}=2\cdot\sqrt{4^{2013}+p}=2k$ (substract $p$ and then apply Bhaskara), for some $k\in \mathbb{N}$. It follows that $4^{2013}+p=k^{2} \Rightarrow p=(k-2^{2013})(k+2^{2013})$. Then the only value for $k$ is $2^{2013}+1$, but then $p=2^{2014}+1$, which we can check it's a multiple of $5$.

anthony_ecu wrote: it's easy to see that the expression it's equal to (n- 2^2013 * 2015)(n- 2^2013 * 2013), if n E [2^2013 * 2013, 2^2013 * 2015], the expression it's negative or 0, so it's not prime, ‎if n- 2^2013 * 2015=1 => n- 2^2013 * 2013= 1+2^2014 but 3 divides it, the other cases do the factor negative so not prime and when n- 2^2013 * 2013=-1 => the expression is 1+2^2014 too, and the other cases do the expression negative so not prime, we're done. latex please

Let $P$ be a polynomial with $P(x)\in\mathbb{Z}[x]$ and $P(n)= n^2-2^{2014}\times 2014n+4^{2013}(2014^2-1)$. We see that this polynomial has degree 2 and can also be factorized. So, let's see: $P(x)= (x-r_1)(x-r_2)$, where $r_1$ and $r_2$ are the roots ($r_1>r_2$) of the polynomial. Let's find out these roots! $x^2-2^{2014}\cdot 2014x+4^{2013}(2014^2-1)=0$ $\implies x^2-2^{2014}\cdot 2\cdot 1007x+(2^2)^{2013}[(2\cdot 1007)^2-1]=0$ $\implies x^2- 2^{2015}\cdot 1007\cdot x+ 2^{4028}\cdot 1007- 2^{4026}= 0$ $\implies \Delta= 2^{4030}\cdot 1007^2-4\cdot 2^{4028}\cdot 1007^2+4\cdot 2^{4026}$ $\implies \Delta= 2^{4030}\cdot 1007^2-2^{4030}\cdot 1007^2+ 2^{4028}$ $\implies \Delta= 2^{4028}$ $\therefore x= \frac{2^{2015}\cdot 1007\pm 2^{2014}}{2}$ $\implies x= 2^{2014}\cdot 1007\pm 2^{1013}$ So, $r_1= 2^{2014}+2^{2013}$ and $r_2=2^{2014}-2^{2013}$ With that, $P(n)= (n-2^{2014}-2^{2013})(n-2^{2014}+2^{2013})$ As $P(n)$ is our number, we want to prove that $P(n)$ is not a prime. For that, it's necessary that $|n-2^{2014}\pm 2^{2013}|>1$ Suppose FTSOC that one of the factors is 1 or -1, then $n= 2^{2014}\pm 2^{2013}+1$ or $n= 2^{2014}\pm 2^{2013}-1$ Suppose WLOG $n= 2^{2014}\pm 2^{2013}+1$ , so one of the factors is 1 and the other is $2^{2014}+1$ or $-2^{2014}+1$ (the other case is analog) But notice that $2\equiv (-1)$ (mod 3) $\implies 2^{2014}\equiv (-1)^{2014}\equiv 1$ (mod 3) $\implies 3\mid -2^{2014}+1$, so if $-2^{2014}+1$ is a prime, then we must have $3= -2^{2014}+1$, which is impossible! Also, $2^2\equiv -1$ (mod 5) $\implies (2^2)^{1007}\equiv (-1)^{1007}\equiv -1$ (mod 5) , so $5\mid 2^{2014}+1$, and if $2^{2014}+1$ is a prime, we must have $2^{2014}+1= 5$, which is impossible! So, we have a contradiction, and then $P(n)$ can't be a prime, QED.