Find integer solution of a system of equations Find all such natural $n$ such that 
$
\dfrac{12n-21143}{25} 
$
and 
$
\dfrac{2n-3403}{25} 
$
both are squares of prime numbers.
So far I have found by computer only   solution:  $ 2014.$
How to prove taht there are  no more solutions?
I was tried  first  to solve the system 
\begin{cases}
12n=21143 \mod 25,\\
2n=3403 \mod 25,
\end{cases}
and find that $n=14 \mod 25$.  Then I substitute $n=14+25 \cdot k$ into the system 
\begin{cases}
12(14+25 \cdot k)-21143= 25 p^2,\\
2(14+25 \cdot k)-3403= 25 q^2,
\end{cases}
and get 
\begin{cases}
12k= 839+ p^2,\\
2k= 135+ q^2,
\end{cases}
or $10k=704 +p^2-q^2.$ It follows that $p^2-q^2=6 \mod 10$.  But how to find now all $k?.$
Edit. If we rewrite it in the way $p^2-q^2=16=4^2 \mod 10$ then we get the $p=5 \mod 10$ and $q=3 \mod 10.$ 
Sorry, one  solutions.

Your equations are
\begin{cases}
12k= 839+ p^2\\
2k= 135+ q^2
\end{cases}
and we observe that $$p^2-q^2 = 10k - 704 = 5(2k-141)+1$$
so $\,p^2 - q^2 = 1 \pmod{5}$ which tells us that $p^2 = 1 \mod 5$ and $q = 0 \mod 5$, or $p=0 \mod 5$ and $q^2 = -1 \mod 5$.  We consider the first case here, and the second later. Since $q$ is prime and divisible by $5$, $q=5$.  Then 
$$
2k =  135+25 = 160 \implies k=80
$$
and 
$$
12k = 960 =- 839+p^2\\
P^2 = 121\\
p=11
$$
Finally,
$$
12 n - 21143 = 25p^2= 3025\\
12n = 24168\\
n=2014
$$
or 
$$
2 n - 3403= 25q^2= 625\\
2n = 4028\\
n=2014
$$
So that is the one solution.
Now consider the second case: $p=0 \mod 5$ and $q^2 = -1 \mod 5$.
$$12k=839+p^2 = 144 \\
k=72$$
And the other equation becomes
$$2\cdot 72 = 135 + q^2\\
q=3$$
So that will give a second solution, with
$$
n = 14 + 25*k = 14 + 25\cdot 72=1814 $$
Kudos to Leox for noting the second case, which I had overlooked in my original answer.
