This is a very simple problem if p≡0(mod3) p+6>6 and p+6≡0(mod3) So it's a contradiction if p≡2(mod3) p+10>10 and p+10≡0(mod3) So it's a contradiction therefore p≡1(mod3) p+q+1>3 therefore p+q+1≡1or2(mod3) p+q+1≡q+2(mod3) If you do the same as before q≡0(mod3) q is prime therefore q=3 know p+4,p+6,p+10 has to be prime. p+4 and p+6 are twin primes(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)are all twin primes less than 100 by p≡1(mod3),p is prime,p+10 is prime now we can know p can be 7,13,37 thefore (p,q)=(7,3),(13,3),(37,3)

Dear David_Kim_0202, It would be nice if you would use $\LaTeX$, which is a markup language. You can learn it here, in case you haven't. Thanks! Regards, Zeta.

thanks! I've always wondered how to use the markup languages

You missed (97,3)

If $ p + 6 $ is a prime number, $ p $ can't divisible by $3$. If $ p + 10 $ is a prime number, $ p $ can't be equivalent $2$$\pmod{3}$. So, $ p \equiv 1 \pmod{3}$. If $ q + 4 $ is a prime number, $ q $ can't be equivalent $2$$\pmod{3}$. If $p+q+1$ is a prime number, q can't be equivalent $1$$\pmod{3}$. Hence, $q=3$. In conclusion, we have four prime numbers $ p, p + 4, p + 6, p + 10$. With some case work, everyone can see that $ p \equiv \{3,7\} \pmod{10}$. If you investigate the cases, the solutions are : $(13,3)$, $(7,3)$, $(37,3)$, $(97,3)$. $\square$