2
$\begingroup$

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.

  • 0
    $-1814$ is not a natural number; that usually means positive integers (sometimes people include zero). – 2017-01-24
  • 0
    @MarkFischler I'm fairly certain that $n=1814$ is what was meant, and that it's just a hyphen posing as a dash. A negative value couldn't be a square. – 2017-01-24
  • 0
    no, I mean 1814, of cource,, I'll correct – 2017-01-24
  • 0
    Oh...yes, you are right – 2017-01-24
  • 0
    @Leox Hint: for the vast majority of primes $p$, $p^2 \equiv 1,4 \pmod 5$. There's only one exception. – 2017-01-24

2 Answers 2

2

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.

  • 1
    what about the case $p^2=0 \mod 5$, $q^2=-1 \mod 5?$ – 2017-01-27
  • 0
    Right, that does yield another solution. I have corrected the answer. – 2017-01-29
1

You've already done the hard part. Now that you know $p^2 - q^2 \equiv 1 \pmod 5$. But the only non-zero quadratic residues mod $5$ are $\pm 1$, so there are no solutions unless one of $p$ or $q$ is divisible by $5$. There are not many options for primes divisible by $5$.

  • 0
    I have found $4$ solutions $\left\{ p=0,q=2 \right\} , \left\{ p=0,q=3 \right\} , \left\{ p=1,q=0 \right\} , \left\{ p=4,q=0 \right\} $. – 2017-01-24
  • 0
    @Leox Think about it: how many prime numbers satisfy $p \equiv 0 \pmod 5$? "Not many options" refers to the number of primes, not the number of modular roots. – 2017-01-24
  • 0
    yes, I see that $p=5$ or $q=5.$ – 2017-01-24
  • 0
    @Leox It seems that your question has been addressed by at least two different answers and that you understand at least one of them. It would be appropriate for you to accept one of the existing answers. – 2017-01-26