Square Addible

Let't's first find what numbers are square addible by $7$ , then move on to $11$ , and finally move on to $13$ .

We have that $7\mid n+2014x^2$ This can also be represented as $n+2014x^2\equiv n+5x^2\equiv 0\pmod{7}$

The quadratic residues of $7$ , after checking squares, are $x^2\equiv 0,1,2,4$ . Thus, $n+5x^2\equiv n+5(0,1,2,4)\equiv n+(0,5,3,6)\equiv 0\pmod{7}$

Thus, $n\equiv 0,2,4,1\pmod{7}$

Now let's consider mod $11$ . We have that $n+2014x^2\equiv n+x^2\equiv 0\pmod{11}$

The quadratic residues mod $11$ are $x^2\equiv 0,1,3,4,5,9\pmod{11}$ . Thus, $n+x^2\equiv n+(0,1,3,4,5,9)\equiv 0\pmod{11}$

Therefore $n\equiv 0,10,8,7,6,2\pmod{11}$

Finally, we consider mod $13$ . We have $n+2014x^2\equiv n+12x^2\not\equiv 0\pmod{13}$

The quadratic residues mod $13$ are $x^2\equiv 0,1,3,4,9,10,12\pmod{13}$ . Thus, $n+12x^2\equiv n+12(0,1,3,4,9,10,12)\equiv n+(0,12,10,9,4,3,1)\not\equiv 0\pmod{13}$

Thus, $n\not\equiv 0,1,3,4,9,10,12 \pmod{13}$ so $n\equiv 2,5,6,7,8,11\pmod {13}$

Thus, we have the following set of modular equations: $\left\{\begin{array}{l}n\equiv 0,2,4,1\pmod{7}\\ n\equiv 0,10,8,7,6,2\pmod{11}\\ n\equiv 2,5,6,7,8,11\pmod {13}\end{array}\right.$

By the Chinese Remainder Theorem, there exists exactly one solution $\text{mod }7\cdot 11\cdot 13=1001$ for each possible choice of $\text{mod }7$ , $\text{mod }11$ , and $\text{mod }13$ . Thus, there are $4\times 6\times 6=144$ solutions $\text{mod }1001$ .

Therefore, there are $2\times 144=288$ solutions between $1$ and $2002$ . However, we also have to check $2003\to 2014$ to see if there are any more solutions. Taking mod $1001$ , we check the numbers $1\to 12$ .

We find that $2, 7, 8, 11$ are the numbers that satisfy our system of modular equations. Thus, there are $288+4=\boxed{292}$ solutions total.