[content] evaluation/hard_problems_thinking.jsonl (jsonl)

{"input": [{"role": "user", "content": "Solve the following math problem. Make sure to put the answer (and only answer) inside \\boxed{}.

Find all prime numbers \\( p, q < 2005 \\) such that \\( q \\mid p^2 + 8 \\) and \\( p \\mid q^2 + 8 \\)."}], "output": "<think>
Okay, so I have this problem where I need to find all prime numbers p and q, both less than 2005, such that q divides p² + 8 and p divides q² + 8. Hmm, primes dividing each other's squares plus eight. Let me try to break this down step by step.

First, since both p and q are primes, their divisors are limited. If q divides p² + 8, that means p² ≡ -8 mod q. Similarly, p divides q² + 8, so q² ≡ -8 mod p. So we have these two congruences:

1. p² ≡ -8 (mod q)
2. q² ≡ -8 (mod p)

I need to find primes p and q that satisfy both conditions. Let me think about possible strategies here. Maybe I can use the first congruence to express p² in terms of q and then substitute into the second? Not sure yet. Alternatively, maybe there's a way to relate these congruences to each other.

Since p and q are primes, Fermat's Little Theorem might be useful here. Fermat's tells us that for a prime a, a^(b-1) ≡ 1 mod b when a and b are coprime. But since p and q are primes, they are coprime unless one is equal to the other. But if p = q, then the conditions become p divides p² + 8. Let's check that case first.

If p = q, then we need p divides p² + 8. Which simplifies to p divides 8. The primes that divide 8 are 2. So p = q = 2. Let me check: q=2 divides p² + 8. If p=2, then p² + 8 = 4 + 8 = 12, which is divisible by 2. Similarly, p=2 divides q² + 8 = 4 + 8 = 12. So 2 divides 12. So (2,2) is a solution. But wait, the problem states primes less than 2005, so 2 is allowed. But maybe there are other solutions where p ≠ q.

So let's check the case where p ≠ q.

So assume p ≠ q. Then since p and q are distinct primes, they are coprime. So from the first congruence: p² ≡ -8 mod q. Which means p² + 8 is divisible by q. Similarly, q² + 8 is divisible by p.

Let me try to relate these two congruences. Let's suppose that both p and q are odd primes. Then p and q are both odd, so p² and q² are 1 mod 4. Then p² + 8 would be 1 + 0 = 1 mod 4. Similarly, q² + 8 is 1 + 0 = 1 mod 4. So if p and q are odd, then q must divide a number congruent to 1 mod 4, which is possible. If either p or q is 2, let's check that case first.

Case 1: p = 2. Then we need q divides 2² + 8 = 4 + 8 = 12. So q divides 12. The primes dividing 12 are 2, 3. But since p=2, and we're in the case where p ≠ q (since we already considered p=q=2), so q=3. Now check if p=2 divides q² + 8. q=3, so q² + 8 = 9 + 8 = 17. Does 2 divide 17? No. So that's not valid. Therefore, when p=2, the only possible q is 3, but it doesn't satisfy the second condition. So p=2 only works when q=2. But we saw that (2,2) is a solution.

Case 2: q = 2. Then we need p divides q² + 8 = 4 + 8 = 12. So p divides 12. The primes dividing 12 are 2, 3. Since q=2, if p=2, then that's the same as before. If p=3, check if q=2 divides p² + 8. p=3, p² + 8 = 9 + 8 = 17. 2 divides 17? No. So same problem. So again, the only solution here would be p=2, q=2, which is already covered.

So the only possible case where one of p or q is 2 is (2,2). Now, moving on to the case where both p and q are odd primes.

So both p and q are odd primes greater than 2. Let's consider the first congruence: p² ≡ -8 mod q. Which implies that -8 is a quadratic residue modulo q. Similarly, since q² ≡ -8 mod p, so -8 is a quadratic residue modulo p as well.

Quadratic residues might have some properties here. Let's recall that for a prime modulus, -1 is a quadratic residue modulo q if and only if q ≡ 1 mod 4. Similarly, 2 is a quadratic residue modulo q if and only if q ≡ ±1 mod 8. So maybe we can analyze the possible quadratic residues.

But first, perhaps we can find some relations between p and q. Let's suppose that both congruences hold. Let me write the first equation as p² ≡ -8 mod q. So there exists an integer k such that p² + 8 = kq. Similarly, the second equation: q² + 8 = lp, where l is an integer. Since p and q are primes less than 2005, k and l must be positive integers.

So we can write:

p² = kq - 8

q² = lp - 8

Maybe substitute one into the other? Let's see. Let's take the first equation: p² = kq - 8. Then, we can substitute p² into the second equation. Wait, but the second equation is in terms of q². Alternatively, perhaps use the two equations to relate k and l.

Alternatively, perhaps express k and l in terms of p and q. From the first equation, k = (p² + 8)/q, and l = (q² + 8)/p. Since k and l are integers, we have that q divides p² + 8 and p divides q² + 8.

Let me consider that if p and q are primes, then unless p divides q² + 8 and q divides p² + 8. Since p and q are primes greater than 2, they are odd, so p² and q² are 1 mod 4, hence p² + 8 is 1 + 0 = 1 mod 4, so q must divide a number congruent to 1 mod 4. Similarly for q² + 8. So q must be a prime that divides a number 1 mod 4. Since q is odd, that's possible.

Alternatively, maybe consider combining the two congruences. For example, if we have p² ≡ -8 mod q and q² ≡ -8 mod p, perhaps we can find some relation between p and q.

Let me try substituting one congruence into the other. For instance, since p² ≡ -8 mod q, we can raise both sides to the power of q modulo q. Wait, but since by Fermat's Little Theorem, p^(q-1) ≡ 1 mod q. So p² ≡ -8 mod q, then p^(2k) ≡ (-8)^k mod q. But not sure if that helps.

Alternatively, from p² ≡ -8 mod q, then (p² + 8) is divisible by q. Similarly, q² + 8 is divisible by p. Let's try substituting variables. Let me consider multiplying p and q.

Wait, perhaps if we consider the first equation p² + 8 = kq, and the second equation q² + 8 = lp. Then maybe substitute kq into the second equation.

But not sure. Let me think. Let me write both equations:

p² + 8 = kq --> (1)
q² + 8 = lp --> (2)

From (1): k = (p² + 8)/q

From (2): l = (q² + 8)/p

Since p and q are primes greater than 2, both k and l must be integers. Let me see if we can find a relationship between p and q. Let's suppose that p and q are both primes greater than 2, so they are odd. So p and q are at least 3.

Let me try to use equations (1) and (2) together. Let's express p from equation (2): p = (q² + 8)/l. Substitute this into equation (1):

[( (q² + 8)/l )² + 8] = kq

But this might complicate things further. Alternatively, since both k and l are integers, perhaps we can find bounds on their values.

From equation (1): since p² + 8 ≥ 3² + 8 = 17, and kq = p² + 8, so q ≤ p² + 8. But q is a prime less than 2005, so p² + 8 must be at least q, which is at least 3.

Similarly, from equation (2): q² + 8 = lp, so p ≤ q² + 8. But p is also a prime less than 2005.

Alternatively, maybe if we consider that k = (p² + 8)/q. Since p and q are primes, and q divides p² + 8, q must divide p² + 8, which is equivalent to p² ≡ -8 mod q.

Similarly, since p divides q² + 8, then q² ≡ -8 mod p.

So maybe we can use these two congruences to find some relations.

Let me take the first congruence: p² ≡ -8 mod q. Then, raise both sides to the power of (q-1)/2. By Fermat's Little Theorem, p^(q-1) ≡ 1 mod q. Therefore, (p²)^((q-1)/2) ≡ 1 mod q. But p² ≡ -8 mod q, so (-8)^((q-1)/2) ≡ 1 mod q.

Similarly, (-8)^((q-1)/2) ≡ 1 mod q. Which is equivalent to (-1)^((q-1)/2) * 8^((q-1)/2) ≡ 1 mod q.

But 8 is 2^3, so 8^((q-1)/2) = 2^(3*(q-1)/2). By Fermat's Little Theorem, 2^(q-1) ≡ 1 mod q, so 2^(3*(q-1)/2) ≡ ±1 mod q, depending on whether 3*(q-1)/2 is even or odd. Wait, maybe this is getting too complicated.

Alternatively, since (-8) is a quadratic residue modulo q, the Legendre symbol (-8|q) = 1. Similarly, (-8|p) = 1.

Using properties of the Legendre symbol: (-8|q) = (-1|q)*(8|q) = (-1)^((q-1)/2) * (8|q). Similarly, (8|q) = (2|q)^3. Since (2|q) is 1 if q ≡ ±1 mod 8, and -1 if q ≡ ±3 mod 8. So:

(-8|q) = (-1)^((q-1)/2) * (2|q)^3

For this to equal 1, the product must be 1. Similarly for the other congruence.

Alternatively, perhaps we can look for primes p and q where p² +8 is divisible by q and vice versa. Let's try small primes to see if there's a pattern.

Let me test p=3. Then p² +8 = 9+8=17. So q must be a prime divisor of 17, which is 17. Then check if p=3 divides q² +8. q=17, q² +8=289 +8=297. 297 divided by 3 is 99. So 3 divides 297. So yes, that works. So (3,17) is a solution.

Similarly, let's check q=17. Now, check if 17 divides p² +8, where p=3, which we already did. Now check other primes. Let's see if there are other primes p where q=17 divides p² +8. For example, p=?

But let's stick to our previous solution. So (3,17) is a solution. Let's check the reverse. If q=3, then p must divide q² +8=9+8=17. So p=17. Then check if q=3 divides p² +8=289+8=297. 297/3=99. So yes, (17,3) is also a solution. So the pairs are (3,17) and (17,3).

Similarly, perhaps check p=17. Then p² +8=289+8=297. So q must divide 297. The prime factors of 297 are 3 (since 297=3*99=3*9*11=3^3*11). So possible q's are 3, 11. Check q=3: already considered. q=11: check if p=17 divides q² +8=121+8=129. 129 divided by 17 is 7.588... not an integer. So q=11 doesn't work. So only q=3 and q=17.

Wait, but p=17: q must divide 297, which factors into 3^3*11. So q could be 3, 11. q=3 works, but q=11 doesn't. So (17,3) is valid, but (17,11) is not. So (17,3) is a solution.

Similarly, check p=5. p² +8=25 +8=33. So q divides 33, primes are 3, 11. Check if p=5 divides q² +8. If q=3: q² +8=9 +8=17. 5 doesn't divide 17. If q=11: q² +8=121 +8=129. 129 divided by 5 is 25.8. Not integer. So neither works. So p=5 gives no solution.

Next, p=7. p² +8=49+8=57. Prime factors of 57 are 3,19. Check q=3: q² +8=17. 7 doesn't divide 17. q=19: check if 7 divides 19² +8=361 +8=369. 369 divided by 7 is 52.714... No. So no solution here.

p=11: p² +8=121 +8=129. Factors: 3*43. So q=3 or 43. Check q=3: q² +8=17, 11 doesn't divide 17. q=43: check if 11 divides 43² +8=1849 +8=1857. 1857 divided by 11 is 168.818... No. Not an integer. So no.

p=13: p² +8=169 +8=177. Factors: 3*59. So q=3 or 59. q=3: 3² +8=17. 13 doesn't divide 17. q=59: check if 13 divides 59² +8=3481 +8=3489. 3489 /13 = 268.384... No.

p=19: p² +8=361 +8=369. Factors: 3*123=3*3*41. Primes q=3, 41. q=3: 3² +8=17. 19 doesn't divide 17. q=41: check if 19 divides 41² +8=1681 +8=1689. 1689 /19=88.894... No.

Hmm, so far, the only solutions other than (2,2) are (3,17) and (17,3). Let me check p=17 and q=3 again to confirm. p=17, q=3: 17 divides 3² +8=17? 3² +8=17, so yes, 17 divides 17. And q=3 divides p² +8=289 +8=297. 297/3=99. So yes, that works. So (17,3) is good.

Wait, another approach: suppose that (p, q) is a solution. Then (q, p) is also a solution. Because if q divides p² +8 and p divides q² +8, then reversing roles, p divides q² +8 and q divides p² +8. So solutions come in pairs unless p=q. So the only symmetric solution is (2,2). So if we find (3,17), then (17,3) must also be a solution. So maybe these are the only ones. Let's check for more.

Next, p=23: p² +8=529 +8=537. Factors: 537 divided by 3 is 179. So primes q=3,179. Check q=3: 3² +8=17. 23 doesn't divide 17. q=179: check if 23 divides 179² +8=32041 +8=32049. 32049 /23: 23*1393=32039, remainder 10. So no.

p=179: p² +8=179² +8=32041 +8=32049. Prime factors: 32049=3*10683=3*3*3561=3*3*3*1187. 1187 is a prime? Let me check. 1187 divided by primes up to sqrt(1187) ~34.4. 1187 is not divisible by 2,3,5,7,11,13,17,19,23,29,31. So 1187 is prime. So q=3, 1187. Check q=3: q² +8=17. 179 doesn't divide 17. q=1187: check if 179 divides (1187² +8). 1187² = (1200 -13)^2 = 1200² -2*1200*13 +13² = 1,440,000 - 31,200 + 169 = 1,408,969. Then +8=1,408,977. Divided by 179. Let's compute 179*7870=179*7000=1,253,000. 179*870=179*(800+70)=179*800=143,200; 179*70=12,530. So 143,200 +12,530=155,730. Then 1,253,000 +155,730=1,408,730. Then 1,408,977 -1,408,730=247. So 179*7870=1,408,730, remainder 247. Not divisible.

So p=179 also doesn't work. Maybe trying higher primes is not helpful. Let's think of another approach.

Suppose that p and q are primes such that q divides p² +8 and p divides q² +8. Let's consider the first equation: p² ≡ -8 mod q. Then, taking this congruence, we can write that (p²) ≡ -8 mod q. Similarly, from the second equation, q² ≡ -8 mod p.

If we can link these two, perhaps substitute one into the other. Let me try substituting. From the first congruence, p² ≡ -8 mod q. Let's raise both sides to the power of (q-1)/2. By Fermat's Little Theorem, p^(q-1) ≡ 1 mod q. So (p²)^((q-1)/2) ≡ (-8)^((q-1)/2) mod q. The left side is p^(q-1) ≡ 1 mod q. So 1 ≡ (-8)^((q-1)/2) mod q. Therefore, (-8)^((q-1)/2) ≡ 1 mod q.

Similarly, for the second congruence: q² ≡ -8 mod p. Raise both sides to the power of (p-1)/2. Then q^(p-1) ≡ (-8)^((p-1)/2) mod p. By Fermat's Little Theorem, q^(p-1) ≡ 1 mod p. So 1 ≡ (-8)^((p-1)/2) mod p. Therefore, (-8)^((p-1)/2) ≡ 1 mod p.

Therefore, both (-8) must be quadratic residues modulo q and p. But this might lead us to conditions on p and q modulo some numbers.

Alternatively, suppose we take both congruences:

From p² ≡ -8 mod q and q² ≡ -8 mod p.

Let me consider substituting p² from the first equation into the second. But how?

Wait, from p² ≡ -8 mod q, so p² = qk -8 for some integer k. Then q² ≡ -8 mod p --> q² +8 is divisible by p, so q² ≡ -8 mod p.

But q² = pl -8 for some integer l. Not sure.

Alternatively, let's consider the first equation p² ≡ -8 mod q. Then, p² +8 is a multiple of q. Let's denote this multiple as kq. Then p² = kq -8.

Similarly, from the second equation, q² = lp -8.

Now, substitute p² = kq -8 into the second equation's expression. Let's see:

From q² = lp -8, and p² = kq -8.

But how? Let me try substituting p from one equation into another. Let me solve for p from the second equation: lp = q² +8, so p = (q² +8)/l. Then substitute this into the first equation:

[(q² +8)/l]^2 = kq -8

Multiply both sides by l²:

(q² +8)^2 = l²(kq -8)

But this seems complex. Maybe consider specific values. Let's take the previous solution (3,17). Here, p=3, q=17. Then k=(3² +8)/17=17/17=1, so k=1. l=(17² +8)/3=297/3=99. So l=99.

Similarly, if we take (17,3): p=17, q=3. Then k=(17² +8)/3=297/3=99, l=(3² +8)/17=17/17=1. So l=1.

So in the first case, k=1, l=99. In the second case, k=99, l=1. So maybe there's a relationship here. Let's see.

Suppose that k=1. Then from p² +8 = q. So q = p² +8. Then q must be a prime. Then check if p divides q² +8. q= p² +8. Then q² +8 = (p² +8)^2 +8. Let's see if p divides that. (p² +8)^2 +8 = p^4 +16p² +72. Let's compute this mod p. Since p divides p^4 and 16p², so (p^4 +16p² +72) ≡ 0 +0 +72 mod p. So 72 ≡ 0 mod p. Therefore, p divides 72. The primes dividing 72 are 2,3. So p must be 2 or 3.

Check p=2: Then q=2² +8=12, which is not prime. So invalid.

Check p=3: q=9 +8=17, which is prime. Then check if p=3 divides q² +8=289 +8=297. 297/3=99. Yes. So this works. So (3,17) is a solution when k=1.

Similarly, if l=1. Then from q² +8 = p. So p = q² +8. Then p must be a prime. Check if q divides p² +8. p= q² +8. Then p² +8 = (q² +8)^2 +8. Similarly, compute mod q. (q² +8)^2 +8 ≡ 0 + 64 +8=72 mod q. So 72 ≡0 mod q. Therefore q divides 72. Primes q=2,3.

Check q=2: p=4+8=12, not prime.

Check q=3: p=9 +8=17, which is prime. Then check if q=3 divides p² +8=289 +8=297. 297/3=99. Yes. So (17,3) is a solution.

So these solutions come from when k=1 or l=1, leading to p and q such that q = p² +8 or p = q² +8, and primes dividing 72. Which only worked for p=3 and q=17.

Therefore, the only solutions in this case are (3,17) and (17,3).

Are there any other solutions where k and l are greater than 1?

Suppose we consider k and l greater than 1. Let's try with k=2. Then p² +8 =2q. So q=(p² +8)/2. Since q must be a prime, then (p² +8) must be even. Since p is a prime greater than 2, it's odd, so p² is odd, so odd +8 is odd + even = odd. So (p² +8) is odd, which would make q=(odd)/2, which is not an integer. Therefore, k=2 is impossible for odd p. If p=2, then k=(4 +8)/q=12/q. For k=2, q=6, which is not prime. So no solutions for k=2.

Similarly, k=3. Then q=(p² +8)/3. q must be a prime. Let's check possible primes p.

Take p=3: q=(9 +8)/3=17/3, not integer. p=5: (25 +8)/3=33/3=11, which is prime. Then check if p=5 divides q² +8. q=11: q² +8=121 +8=129. 129 divided by 5=25.8, no. So invalid.

p=7: (49 +8)/3=57/3=19, prime. Check if 7 divides 19² +8=361 +8=369. 369/7≈52.714. Not divisible. No.

p=11: (121 +8)/3=129/3=43, prime. Check if 11 divides 43² +8=1857. 1857 /11=168.818... No. Doesn't divide.

p=13: (169 +8)/3=177/3=59, prime. Check 13 divides 59² +8=3489. 3489 /13=268.384... No.

Same as before. So even though q is prime, p doesn't divide q² +8. Therefore, k=3 doesn't yield solutions.

Similarly, let's try k=4. Then q=(p² +8)/4. q must be a prime. So p² +8 must be divisible by 4. Since p is an odd prime, p² ≡1 mod 4, so p² +8 ≡1 +0=1 mod 4. So 1 ≡0 mod 4? No. Contradiction. So q=(p² +8)/4 is not integer. Thus, no solution for k=4.

k=5: q=(p² +8)/5. Let's check p such that p² +8 is divisible by 5. So p² ≡ -8 mod5 → p² ≡ 2 mod5. Quadratic residues modulo 5 are 0,1,4. So 2 is not a quadratic residue modulo5. Therefore, no solution for k=5.

Similarly, k=6: q=(p² +8)/6. Need p² ≡ -8 mod6 → p² ≡4 mod6. p is prime >3, so p≡1 or 5 mod6. Then p²≡1 mod6. So 1≡4 mod6? No. So no solution.

k=7: q=(p² +8)/7. So p² ≡ -8 mod7. -8 mod7 is -1, so p² ≡6 mod7. Quadratic residues modulo7 are 1,2,4. So 6 is not a quadratic residue mod7. No solution.

k=8: q=(p² +8)/8. Need p² ≡ -8 mod8. p is odd prime. p²≡1 mod8. So 1≡0 mod8. No.

k=9: q=(p² +8)/9. p² ≡-8 mod9. -8 mod9=1. So p²≡1 mod9. Which holds for p≡±1 mod3. But primes are 3, which is 0 mod3, but p=3: q=(9+8)/9=17/9, not integer. Other primes congruent to 1 or 2 mod3. So p²≡1 mod3, so p²≡1 mod9 as well. So p²=9m +1. So q=(9m +1 +8)/9=(9m +9)/9=m+1. So q=m+1. But q must be prime. However, unless m+1 is prime. But we have p²=9m +1. Let me take p=5: p²=25. 25=9m +1→m=(24)/9=8/3, not integer. p=7: p²=49=9m+1→m=48/9=16/3, nope. p=11:121=9m +1→m=120/9=40/3. No. Not working. So no solutions here.

Similarly, seems like for k>=1, the only possible solutions are when k=1 and l=99, leading to (3,17) and (17,3).

Additionally, the pair (2,2) works. So in total, solutions are (2,2), (3,17), and (17,3). Wait, but need to confirm whether 3 and 17 are less than 2005, which they are. Also, 2 is less than 2005. So these are valid.

But wait, when we checked p=3 and q=17, then p=3 and q=17, both primes less than 2005, and they satisfy the conditions. Similarly, (17,3). And (2,2).

Wait, but wait, the problem says \"Find all prime numbers p, q < 2005\". So primes p and q each less than 2005, correct. So 2,3,17 are all less than 2005. So these three pairs should be the solutions.

But let me check if there are any other possibilities for k and l. For example, take k=10. Then q=(p² +8)/10. p² ≡-8 mod10 → p²≡2 mod10. But quadratic residues mod10 are 0,1,4,5,6,9. So 2 is not a quadratic residue mod10. So no solutions.

Alternatively, maybe if we take l=2. Then from the second equation, q² +8=2p. So p=(q² +8)/2. p must be prime. q is a prime. If q is odd, q² is odd, so q² +8 is odd + even = odd. So p=(odd)/2, not integer. So q must be 2. Then p=(4 +8)/2=6, not prime. So no solution.

Similarly, l=3: q² +8=3p. So p=(q² +8)/3. For q=2: p=(4+8)/3=4, not prime. q=3: (9+8)/3=17/3, nope. q=5:25+8=33/3=11, prime. Then check if q=5 divides p² +8. p=11. p² +8=121+8=129. 129 divided by5=25.8. No. So q=5, p=11: doesn't satisfy. q=7: (49 +8)/3=57/3=19, prime. Check if 7 divides19² +8=369. 369/7≈52.714. No. Similarly q=11: (121 +8)/3=129/3=43, prime. Check if 11 divides 43² +8=1857. 1857/11=168.818. No. So same issue.

So l=3 gives no solutions.

Similarly, l=4: q² +8=4p. So p=(q² +8)/4. q must be even prime, so q=2. p=(4+8)/4=12/4=3, prime. Check if q=2 divides p² +8=9+8=17. 2 doesn't divide17. So no. So (p=3, q=2). But already checked, doesn't work.

Hmm, perhaps this is another way. If we consider that the only solutions where k=1 or l=1 give the pairs (3,17), (17,3), and (2,2). Let me check if there are any more solutions with higher k or l.

Alternatively, let's think about the initial equations:

p² ≡ -8 mod q and q² ≡ -8 mod p.

Let me consider substituting one into the other. For example, since p² ≡ -8 mod q, then p² can be replaced by -8 in any expression modulo q. So, for example, consider q² ≡ -8 mod p. Let's write q² +8 ≡0 mod p. But since p divides q² +8, which is similar to the original condition.

Alternatively, maybe take the equations:

From p² ≡ -8 mod q and q² ≡ -8 mod p.

Let me consider multiplying them:

(p² +8)(q² +8) ≡0 mod pq.

But expanding that: p²q² +8p² +8q² +64 ≡0 mod pq. Which simplifies to (p²q²) +8(p² + q²) +64 ≡0 mod pq.

But since p²q² ≡0 mod pq, the equation simplifies to 8(p² + q²) +64 ≡0 mod pq.

So 8(p² + q² +8) ≡0 mod pq. Which implies pq divides 8(p² + q² +8).

Since p and q are primes, and assuming p≠q, then pq is coprime to 8 unless one of them is 2. Let's split into cases:

Case 1: p=2. Then we already checked, only possible q=2.

Case 2: q=2. Similarly, only possible p=2.

Case 3: Both p and q odd primes.

In that case, pq is coprime to 8. Therefore, pq divides (p² + q² +8). Therefore, p² + q² +8 ≡0 mod pq.

So p² + q² +8 ≡0 mod p and mod q.

Check mod p: p² + q² +8 ≡0 + q² +8 ≡0 mod p. So q² +8 ≡0 mod p. Which is our original condition.

Similarly mod q: p² +8 ≡0 mod q, which is also the original condition. So no new information here.

But the equation p² + q² +8 ≡0 mod pq. Since p and q are primes, and pq divides p² + q² +8, then p² + q² +8 = mpq for some integer m.

Given that p and q are primes less than 2005, m would have to be a positive integer. Let's consider possible values of m.

Since p and q are primes greater than or equal to 2, the left side p² + q² +8 is at least 4 +4 +8=16. The right side is mpq. For p and q at least 2, the minimum mpq is m*2*2=4m. So 4m ≤16 ⇒ m≤4.

But since p and q are primes greater than 2, let's see:

If m=1: p² + q² +8 = pq. Let's rearrange: p² -pq + q² +8=0. Considering p and q are positive integers greater than 2, this equation seems difficult to satisfy. Let's check for small primes.

If p=3: 9 -3q + q² +8 = q² -3q +17 =0. Discriminant: 9 -68= -59. No real solutions.

p=5: 25 -5q +q² +8 = q² -5q +33=0. Discriminant 25 -132= -107. No.

Similarly, no solutions.

m=2: p² + q² +8 =2pq. Rearranged as p² -2pq +q² +8=0 → (p - q)^2 +8=0. Which is impossible since squares are non-negative.

m=3: p² + q² +8 =3pq. Let's rearrange: p² -3pq +q² +8=0. Let's see if this can be possible.

This resembles a quadratic in p: p² -3q p + (q² +8)=0. The discriminant is 9q² -4(q² +8)=5q² -32. To have integer solutions, discriminant must be a perfect square. So 5q² -32 =k².

Looking for primes q such that 5q² -32 is a perfect square. Let me check small primes.

q=3: 5*9 -32=45-32=13, not square.

q=5:5*25 -32=125-32=93, no.

q=7:5*49=245-32=213, no.

q=11:5*121=605-32=573, no.

q=17:5*289=1445-32=1413, no.

Not promising. Maybe m=4: p² + q² +8=4pq.

Rearrange: p² -4pq + q² +8=0. Which can be written as (p -2q)^2 -3q² +8=0. Let's compute:

Let me rearrange: p² -4pq +q² = -8.

This can be written as (p -2q)^2 -3q² = -8.

Let x = p -2q. Then x² -3q² = -8. So x² =3q² -8.

We need x² =3q² -8. Let's check small primes q.

q=3: 3*9 -8=27-8=19, not a square.

q=5:3*25 -8=75-8=67, no.

q=7:3*49 -8=147-8=139, no.

q=11:3*121=363-8=355, no.

q=17:3*289=867-8=859, no.

Similarly, no solutions. Thus, m=4 is impossible.

Therefore, the only possible solutions are when m=1, which gives no solution, or when we don't have p and q both greater than 2, leading us back to (2,2), (3,17), and (17,3).

Thus, the solutions are (2,2), (3,17), and (17,3). However, wait, but earlier when we considered p=2, q=2, it works. But let's verify if (2,2) satisfies the original conditions.

p=2, q=2: q divides p² +8 → 2 divides 4 +8=12. Yes. And p divides q² +8 →2 divides 4 +8=12. Yes. So (2,2) is valid. But is the problem asking for ordered pairs or sets? Since p and q are both primes less than 2005, and they must satisfy both divisibility conditions, so (2,2), (3,17), and (17,3) are the solutions.

But let me check if there are any other solutions when p and q are not equal but still satisfy the conditions. For example, let's take p=3, q=17 and p=17, q=3. We found those.

Wait, but let's try p=13 and q=7. Wait, p=13, p² +8=177. 177 factors into 3*59. So q=3 or 59. Check q=59. Then check if 13 divides 59² +8=3481 +8=3489. 3489 divided by13: 13*268=3484, remainder 5. Not divisible. So no.

Alternatively, p=7, q=19. p=7: p² +8=57. q=3 or 19. q=19: check if 7 divides 19² +8=369. 369 divided by7≈52.714. No.

Another approach: suppose that p and q are primes such that they form a \"symbiotic\" pair where each divides the other's square plus eight. From our trials, only (2,2), (3,17), and (17,3) work.

Additionally, given that when we check higher primes, like p=5,7,11, etc., we don't find any solutions. So perhaps these are the only solutions.

But let's check p=3, q=17 again. p=3 divides q² +8=289 +8=297. 297/3=99, yes. q=17 divides p² +8=9 +8=17, which is exactly q=17. So yes, perfect. So that's a valid solution.

Similarly, p=17, q=3: p=17 divides q² +8=9 +8=17, which is exactly 17. And q=3 divides 17² +8=297. 297/3=99. So yes.

So unless there are larger primes that I haven't tested, but given the pattern, it's likely these are the only solutions.

Therefore, the prime numbers p and q less than 2005 satisfying the conditions are (2,2), (3,17), and (17,3).

But wait, the problem says \"Find all prime numbers p, q < 2005\". So we need to list all such pairs. However, sometimes in such problems, ordered pairs are considered, but sometimes unordered. Let me check if the problem considers ordered pairs or unordered pairs.

The problem states \"Find all prime numbers p, q < 2005 such that q divides p² + 8 and p divides q² + 8\". So ordered pairs where p and q are primes less than 2005 and satisfy both divisibility conditions. Therefore, (3,17) and (17,3) are distinct solutions. And (2,2) is another.

Therefore, the final answer should be the three pairs: (2,2), (3,17), and (17,3).

But let me just confirm if (2,2) is the only solution where p=q, and whether there are other primes p where p divides p² +8. Suppose p divides p² +8. Then p divides 8. Since p is prime, p=2. Because 8's prime divisors are 2. So yes, only p=2. Therefore, (2,2) is the only solution with p=q.

So in conclusion, the solutions are:

- (2,2)
- (3,17)
- (17,3)

All prime numbers p, q <2005 meeting the conditions.
</think>

The prime numbers \\( p \\) and \\( q \\) less than 2005 that satisfy the conditions \\( q \\mid p^2 + 8 \\) and \\( p \\mid q^2 + 8 \\) are:

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