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.