The problem was proposed by Stanislav Dimitrov from Bulgaria

Is it plus or minus?

The problem is quite similar to APMO 2012 and Turkey EGMO TST 2016.Just looking at order and quite difficult caseworks

sorry forgot the statement.

I think it is plus. Method?

Just take $3p^{q-1}+1=4^k 7^m t$. where $(t,14)=1$ and note that $t\equiv 1 \pmod p$. And it is easy to show that $k\le 2$ and that $m\le 1$ then take both sides $\pmod p$ and finish small cases

why $m \leq 1$ ?

$LTE$ yields that

rightways wrote: Just take $3p^{q-1}+1=4^k 7^m t$. where $(t,14)=1$ and note that $t\equiv 1 \pmod p$. And it is easy to show that $k\le 2$ and that $m\le 1$ then take both sides $\pmod p$ and finish small cases Nice rightways, thanks

Indeed, $(p,q)=(3,3)$ is the only solution. If $p=2$ one can easily check that there are no solutions so suppose $p\ge 3$. Suppose that $q$ is odd; this gives $v_2(3p^{q-1}+1)=2$. Take $r$ an odd prime dividing $3p^{q-1}+1$. Then $r$ divides $11^p+17^p$. As $11$ and $17$ are coprime, we get that $r$ divides $a^p+1$, where $a=11\cdot 17^{-1} (\mathrm{mod}\ r)$. Thus, $r$ divides $a^{2p}-1$ so the order of $a$ mod $r$ has to divide $2p$. If the order is $1$ or $p$), then $r$ divides both $a^p-1$ and $a^p+1$, so $r$ divides $2$, a contradiction. If the order is $2$, as $r$ divides $a^p+1$, we get that $r$ divides $a+1$ so $r$ divides $28$ and hence $r=7$. Otherwise the order is $2p$ so $2p$ divides $r-1$. Thus, if $3p^{q-1}+1=7^t4r_1^{a_1}....r_k^{a_k}$ for some odd distinct primes $r_i$ and some nonnegative integer $t\ge 0$ (which by LTE is less or equal to $2$), by the previous we have $r_i\equiv 1 (\mathrm{mod}\ p)$ so $3p^{q-1}+1\equiv 4,28, 196(\mathrm{mod}\ p)$ and thereby we get $p\in \{3,5,13\}$. A quick check gives $q=3$ the only possibility. If $q=2$, we have $3p+1$ divides $11^p+17^p$. If $3p+1$ is not a power of two, proceeding as in the previous case we get that any odd prime divisor of $3p+1$ must be congruent with $1$ mod $2p$. As $3p+1$ is even, we get that $3p+1\ge 2(2p-1)$ which gives again $p=3$ so $q=3$, a contradiction. Thus $3p+1=2^k$ for some positive integer $k$. Clearly $k$ has to be even, $k=2\ell$ so $3p=(2^{\ell}-1)(2^{\ell}+1)$. As $(2^{\ell}-1,2^{\ell}+1)=1$ and $2^{\ell}-1<2^{\ell}+1$ we get that either $2^{\ell}-1=1,\ 2^{\ell}+1=3p$ or $2^{\ell}-1=3,\ 2^{\ell}+1=p$. We therefore infer that either $p=3$ (in which case again $q=3$ but we supposed $q=2$) or $p=5$. However, $p=5$ and $q=2$ would give $16$ dividing $11^5+17^5$, a contradiction.

Let $r$ be an odd prime dividing $3p^{q-1}+1$, so consequently $r$ divides $11^p+17^p$. ( if $p$ is odd we have that $3p^{q-1}+1 \equiv 4 \pmod 8$, and since this number is clearly bigger than $4$, we know that it must have an odd prime divisor, and if $p$ is $2$, our number is odd, and so our hypothesis is obvious) If $p$ is $2$ we verify by hand. We have that $11^p \equiv -17^p \pmod r$, so $(\frac {11}{17})^{2p} \equiv 1 \pmod r$. Let the order of $(\frac {11}{17})$ be $x$. Thus, $x$ divides both $r-1$ and $2p$ and, since it can't be odd, it has to be either $2$ or $2p$. If it is $2$, we then have that $r$ divides $168$ and, since $r$ is odd and not $3$, it has to be $7$. If $x$ is $2p$, we have that $r \equiv 1 \pmod {2p}$. Taking these into consideration, we get that $3p^{q-1}+1=4^m7^ny$, where $(y,14)=1$ and $y\equiv 1 \pmod p$. By LTE, we have that $n$ is $2$ if $p=7$(we verify by hand), or $n\le 1$. Also, $m=1$. Taking $\pmod p$, we get that $1 \equiv 4,28 \pmod p$, so $0 \equiv 3, 27 \pmod p$. Thus, $p=3$ and, after some casework, $q=3$. Thus, the only solution is $(3,3)$. Edit: Oops, forgot about $q=2$. If there exists a prime $r$ that divides $3p+1$ and $r$ is not $7$, we proceed as previously. if $3p+1$ is a power of $2$, then it's exponent has to be even. Thus, $3p=(2^x-1)(2^x+1)$, and so $x$ is $1$ or $2$, which means that $p$ is either $3$ or $5$. We get a contradiction for both cases. If $7$ divides $3p+1$, then $3p+1=2^k7$. $11^p$ is either $3$ or $1$ $\pmod 8$, and $17^p$ is $1$ $\pmod p$. Thus, $k\le 2$, and we verify the cases by hand.

Illuzion wrote: why $m \leq 1$ ? Hamel wrote: $LTE$ yields that Can someone please elaborate this ?? I am new to LTE .... I mean how do we bound $\nu _{(7)} 3p^{q-1}+ 1$

Aryan-23 wrote: Illuzion wrote: why $m \leq 1$ ? Hamel wrote: $LTE$ yields that Can someone please elaborate this ?? I am new to LTE .... I mean how do we bound $\nu _{(7)} 3p^{q-1}+ 1$ you dont! you use LTE to show V7 of the right hand side is 1. so the left hand side can not be devisible by 49.

Oh yeah , stupid me , thanks

Solution from OTIS office hours: The answer is $(3,3)$ only which works. We first dispense of several edge cases: One can check $p=2$ has no solutions. One can check $p=3$ has only the solution $(3,3)$. Claim: Every prime dividing $11^p+17^p$ is either $2$, $7$, or $1 \pmod p$. Proof. Consider a prime $r$ which divides $11^p+17^p$. Evidently $r \ne 17$. Apparently, $(-11/17)^p \equiv 1 \pmod r$, so either $-11/17 \equiv 1 \pmod r$ meaning $r \mid 28$, or $-11/17$ has order $p$, as needed. $\blacksquare$ Thus, we find $3p^{q-1}+1$ is the product of $2$, $7$, and $1 \pmod p$ primes. We now consider some new cases: If $p \ne 7$ and $q \ne 2$, then $3p^{q-1}+1 \equiv 4 \pmod 8$, while $\nu_7(11^p+17^p) = 1$. Thus, we must have \[ 3p^{q-1} + 1 = 2^2 \cdot 7^{0\text{ or }1} \cdot \left( \text{$1 \bmod p$ primes} \right). \]This gives that either $1 \equiv 4 \pmod p$ or $1 \equiv 28 \pmod p$, but this gives $p = 3$. If $p = 7$, the same analysis holds except that the exponent of $7$ must be exactly equal to $0$ (despite $\nu_7(11^p+17^p)=2$). If $q=2$, the same argument works except that we need the additional observation that $11^p+17^p \not\equiv 0 \pmod 8$; hence \[ 3p + 1 = 2^{1\text{ or }2} \cdot 7^{0\text{ or }1} \cdot \left( \text{$1 \bmod p$ primes} \right). \]Hence we have one extra case $(p,q) = (13,2)$ to check, which fails. This gives a complete proof that $(p,q) = (3,3)$ is the only solution.

microsoft_office_word wrote: Find all primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$ Proposed by Stanislav Dimitrov,Bulgaria For $p = 2$ there are no solutions. We begin like v_enhance, so if $q \lvert 11^p + 17^p$, then $q = 2, 7$ or $q \equiv 1 \pmod{p}$. Looking at $\pmod{32}$, there are no solutions so $2^4$ is maximum value of $\nu _ 2(3p^{q-1} + 1)$ and similarly $\nu _ 3(3p^{q-1} + 1) \leq 1$. Now, annoying casework gives $(p, q) = \boxed{(3, 3)}$ as the only solutions.

arshiya381 wrote: Aryan-23 wrote: Illuzion wrote: why $m \leq 1$ ? Hamel wrote: $LTE$ yields that Can someone please elaborate this ?? I am new to LTE .... I mean how do we bound $\nu _{(7)} 3p^{q-1}+ 1$ you dont! you use LTE to show V7 of the right hand side is 1. so the left hand side can not be devisible by 49. how do you get $\nu_7(11^p+17^p) = 1$?

The answer is $(p,q)=(3,3)$ only, which clearly works. We can verify that for $p=2$ no solutions exist, and for $p=3$ we only have $q=3$. Henceforth assume $p>3$ We now begin with the following important claim: Claim: If a prime $n$ divides $11^p+17^p$, then $n=2,7$ or $n \equiv 1 \pmod{2p}$. Proof: Let $n$ be such a prime. Clearly $n \neq 11,17$. Then we have $$11^p+17^p \equiv 0 \pmod{n} \implies \left(\frac{11}{17}\right)^p \equiv -1 \pmod{n} \implies \left(\frac{11}{17}\right)^{2p} \equiv 1.$$For convenience, let $a=\tfrac{11}{17}$. Then we have $\mathrm{ord}_n(a) \mid 2p$ but $\mathrm{ord}_n(a) \nmid p$, so either $\mathrm{ord}_n(a)=2$ or $\mathrm{ord}_n(a)=2p$. If the former is true, then we have $\tfrac{11}{17} \equiv -1 \pmod{n} \implies 28 \equiv 0 \pmod{n}$, which gives $n \in \{2,7\}$. If the latter is true, then since $\mathrm{ord}_n(a) \mid n-1$, we have $p \mid 2p \mid n-1 \implies n \equiv 1 \pmod{p}$, which is the desired conclusion. $\blacksquare$ Since $3p^{q-1}+1$ is a divisor of $11^p+17^p$, it follows that $3p^{q-1}+1$ is the product of $2,7$ and primes $1 \pmod{2p}$. Further, we have $$11^p+17^p \equiv 3^p+1 \equiv 4 \pmod{8}$$as $p$ is odd, so $\nu_2(11^p+17^p)=2$. Now suppose that $q \neq 2$. In this case, $q-1$ is even, so $3p^{q-1}+1 \equiv 3+1 \equiv 0 \pmod{4}$, so we require $ \nu_2(3p^{q-1}+1)=2$. Further, by exponent lifting we have $\nu_7(11^p+17^p)=\nu_7(28)+\nu_7(p)$, so it follows that $$3p^{q-1}+1=4\cdot 7^\epsilon\cdot P,$$where $P$ is the product of primes that are $1 \pmod{2p}$ and $\epsilon \in \{0,1\}$—this is because if $p \neq 7$ we have $\nu_7(11^p+17^p)$ and if $p=7$ we have $7 \nmid 3p^{q-1}+1$ anyways. Note that $P \equiv 1 \pmod{2p}$. Now taking $\pmod{p}$, we have $3p^{q-1}+1 \equiv 4\cdot 7^\epsilon \pmod{p}$, which is either $4$ or $28$. But we also have $3p^{q-1}+1 \equiv 1 \pmod{p}$, so either $p \mid 3$ or $p \mid 27$, giving $p=3$. Since we supposed $p>3$ this case gives no additional solutions If $q=2$, then we need $\nu_2(3p^{q-1}+1) \in \{1,2\}$. Similarly to the previous case, if $p=7$ then $7 \nmid 3p^{q-1}+1$, otherwise we have $\nu_7(3p^{q-1}+1)\leq 7$, so $$3p^{q-1}+1=2\cdot2^{\epsilon_1}\cdot 7^{\epsilon_2}\cdot P,$$where $P$ is the product of primes $1 \pmod{2p}$ and $\epsilon_1,\epsilon_2 \in \{0,1\}$. LIke the previous case, this means that at least one of $\{2,4,14,28\}$ is $1 \pmod{p}$, but these respectively imply that $p \mid 1, p \mid 3, p \mid 13, p \mid 27$. Since $p>3$, the only possibility is therefore $p=13$. But we can manually check that $(p,q)=(13,2)$ fails, so $(3,3)$ is the only solution. $\blacksquare$

Let $r$ be any prime dividing $3p^{q-1} + 1$. Clearly $r\neq 11,17$. Hence, we can write $a\equiv \frac{17}{11}\pmod{r}$ so that $a^p\equiv -1\pmod{r}$. It follows that $\text{ord}_r(a)\in\{2, 2p\}$, so either $r| 28$ or $r\equiv 1\pmod{2p}$. We have several cases. Case 1: $q = 2$. We can manually verify that $p = 2$ does not work, so $p$ is odd. Moreover, clearly $2p+1$ does not divide $3p+1$, so the only prime divisors of $3p+1$ are $2$ and $7$. If $49 | 3p+1$, then $p\neq 7$ so $7 || 11^p + 17^p$ by LTE, a contradiction. Hence, $v_7(3p + 1) \le 1$. But since $p$ is odd, $v_2(3p+1)\le v_2\left(11^p + 17^p\right) = v_2(11 + 17) = 2$. Combining these results, we obtain $3p + 1 | 28$. However, the only solution to this is $p = 2$, which we have already checked. Case 2: $p = 2$. We can manually verify that there are no solutions in this case. In all remaining cases, $p$ and $q$ are odd and consequently $3p^{q-1} + 1\equiv 4\pmod{8}$. Case 3: $p = 3$. We can manually verify that the only solution in this case is $q = 3$. Case 4: $p = 7$. Then $7$ does not divide $3p^{q-1} + 1$, so $\frac{3\cdot 7^{q-1} + 1}{4}\equiv 1\pmod{7}$, a contradiction. Case 5: $q\neq 2$ and $p\not\in \{2,3,7\}$. By LTE, $v_7\left(11^p + 17^p\right) = v_7(11+17) = 1$. It follows that either $\frac{3p^{q-1} + 1}{4}$ or $\frac{3p^{q-1} + 1}{28}$ is equivalent to $1$ mod $p$, a contradiction as $p\neq 3$. In summary, the only solution is $(p,q) = \boxed{(3,3)}$.

wait is the condition that q is prime necessary (other than, say, parity and divisibility reasons)

Note: All of the $\left(\frac{11}{17}\right)$'s in the proof are the actual fraction, not the value of the Legendre/Jacobi symbol. Lemma 1. Let $r$ be a prime that divides $3p^{q-1}+1$. Then $r\in\{2,7\}$ or $r\equiv 1\pmod{2p}$. Proof. $r=17$ obviously fails, so \[r\mid\left(\frac{11}{17}\right)^p+1.\]If we let $a=\text{ord}_r\left(\frac{11}{17}\right)$, we have $a\mid 2p$. Also, we have that $a\nmid p$ unless $r=2$(which is one of the cases anyway), so $a\in\{2,2p\}$. If $a=2$, then \[r\mid \left(\frac{11}{17}\right)^2-1\rightarrow r\mid 121-289\rightarrow r\mid 168\rightarrow r\in\{2,3,7\}.\]But $r=3$ would imply $a=1$, so $r\in\{2,7\}$. Otherwise, $a=2p$, so $2p\mid r-1$, completing our proof. Lemma 2. If $q\ne 2$, then $p\in\{2,3\}$. Proof. Suppose that $p\ne 2$. Since $q$ is odd, $p^{q-1}$ is a square, so let $b^2=p^{q-1}$ and $b\in\mathbb{Z}^+$. Taking mod $8$, we have \[\nu_2\left(3b^2+1\right)=2.\]Also, \[\nu_7\left(3b^2+1\right)\le \nu_7\left(11^p+17^p\right)=1+\nu_7(p)=\begin{cases}1&p\ne7\\2&p=7\end{cases}.\]Thus $1\equiv 3b^2+1\pmod p$ has to be equivalent to at least one of $2^2$, $2^2\cdot 7$, and $2^2\cdot 7^2$ mod $p$. So $p\mid 3$, $p\mid 27$, or $p\mid 195$. Since $7\nmid 195$, that case doesn't work, so $p=3$, which is what we wanted. Lemma 3. If $q=2$, then $\nu_2(3p+1)\le 3$ and $\nu_7(3p+1)\le 2$. Proof. We have the following by LTE: \[\nu_2(3p+1)\le\nu_2\left(11^p+17^p\right)\le\nu_2(11+17)+\nu_2(p)\le 3\]\[\nu_7(3p+1)\le\nu_7\left(11^p+17^p\right)\le\nu_7(11+17)+\nu_7(p)\le 2.\] It then remains to casework using the cases from lemma 2. Case 1, $p=2$: Then $11^2+17^2=121+289=410$, so \[3\cdot 2^{q-1}+1\in\{1,2,5,10,41,82,205,410\}.\]Then \[2^{q-1}\in\{0,3,27,68\},\]so there are no solutions in this case. Case 2, $p=3$: Then \[11^3+17^3=(11+17)\left(17^2-11\cdot 17+11^2\right)=28\cdot 223.\]Therefore, \[3^q+1\in\{1,2,4,7,14,28,223,446,892,1561,3122,6244\}.\]We can see that \[3^q\in\{0,1,3,6,13,27,222,445,891,1560,3121,6243\}\rightarrow 3^q\in\{3,27\}.\]But $q=1$ is not a prime, so only $(p,q)=(3,3)$ works here. Case 3, $q=2$: Then \[3p+1\mid 11^p+17^p.\]By lemma 1, all prime factors of $3p+1$ are either $2$, $7$, or $1$ mod $2p$. In the third case, $4p+1$ is too big, so $2p+1\mid 3p+1$, absurd. So \[3p+1=2^m 7^n\]for nonnegative integers $m,n$ where $m\le 3$ and $n\le 2$(by lemma 3). Therefore, \[3p+1\in\{1,2,4,8,7,14,28,56,49,98,196,392\}.\]This implies that \[p\in\{0,1,2,9,16,65\},\]all of which are impossible. Combining all cases, our answer is just $(p,q)=\boxed{(3,3)}$, which obviously works. QED.

This is such an incredibly boring but also slightly difficult problem. Let $p_1$ be a prime that divides $3p^{q-1} + 1$. Then as $p_1 \mid 11^p + 17^p$, either $p_1 \equiv 1 \pmod p$ or $p_1 \mid 28$. The rest of the problem is finding constraints on $\nu_2(3p^{q-1} + 1)$. Notice that if $p, q$ are both odd primes, then $\nu_2(3p^{q-1} + 1) = 2$ precisely. On the other hand, $2$ is not a quadratic residue mod $7$, so this means that $7 \nmid 3p^{q-1} + 1$. But this means now that $4 \equiv 1 \pmod p$ as $$3p^{q-1} + 1 \equiv 1 \pmod p,$$so $p=3$, and correspondingly $q = 3$ too. Now if $q = 2$, then note that $\nu_2(3p+1) \leq 2$ and $\nu_7(3p+1) \leq 1$, which upon a finite case check yields no solutions. Thus $(3, 3)$ is the only solution.

HamstPan38825 wrote: Notice that if $p, q$ are both odd primes, then $\nu_2(3p^{q-1} + 1) = 2$ precisely. No. Try $p=3$, $q=7$ for example

Very hard. What. Assume that $p$ and $q$ are odd primes. Note that $\nu_2(3p^{q-1} + 1) = 2$ and hence due to size reasons we must have an odd prime factor $r$. Now consider a prime factor of $11^p + 7^p$, say $t$. Then we have, \begin{align*} \left(\frac{-11}{7}\right)^{2p} &\equiv 1 \pmod{t} \end{align*}Then we have $\text{ord}_t(-11 \cdot 7^{-1}) \mid 2p$. Then either $-11 \equiv 7 \pmod{t}$ and hence $t \mid 28$, or we have $2p \mid t - 1 \iff t \equiv 1 \pmod{2p}$. Then all prime factors of $11^p + 7^p$ are $1$, $2$ or $7$ modulo $2p$. As a result all prime factors of $3p^{q-1} + 1$ are equal to $2$ or $7$ or are congruent to $1$ modulo $2p$. Now note that $\nu_7(3p^{q-1} + 1) \leq 1$ as if $p \neq 7$ it is easy to see that $\nu_7(11^p + 17^p) = 1$ else if $p = 7$, then $\nu_7(3 \cdot 7^{q-1} + 1) = 0$. Assume $7 \mid 3p^{q-1} + 1$. However then, \begin{align*} 3p^{q-1} + 1 \equiv 0 \pmod{7}\\ p^{q-1} \equiv 2 \pmod{7} \end{align*}However as $2$ is a NQR modulo $7$ we cannot have $7 \mid 3p^{q-1} + 1$. Thus $3p^{q-1} + 1 = 4 \cdot P$ where $P$ is the product of primes congruent to $1$ modulo $2p$. However then obviously we require $3p^{q-1} + 1 \equiv 4 \pmod{p}$ or $3 \equiv 0 \pmod{p}$ and hence $p = 3$. Now we then have, \begin{align*} 3^q + 1 \mid 11^3 + 17^3 \end{align*}from which we find $q = 3$. Now if $q = 2$ we find, \begin{align*} 3p + 1 \mid 11^p + 17^p \end{align*}Note that from our claim the primes dividing $3p+1$ are $2$, $7$ or are $1$ modulo $2p$. Clearly $2p + 1 \nmid 3p + 1$ and we require $\nu_p(3p + 1) \leq 7$ and $\nu_p(3p + 1) = 2$. Hence $3p + 1 = 2^x7^y$ and a finite case check leads to no solutions.

The answer is $\boxed{(p,q)=(3,3)}$, which clearly works. Claim 1: Let $r$ be a prime that divides $3p^{q-1}+1$. Either $r=2$, $r=7$, or $r \equiv 1 \pmod{2p}$. Proof: $r$ obviously cannot be $17$, and $r=2$ clearly works, so assume $r \neq 2, 17$ for the remainder of this proof. We write \[r \mid \left(\frac{11}{17}\right)+1,\] meaning the order of $\left(\frac{11}{17}\right)$ modulo $r$ divides $2p$; denote this order as $o$ Furthermore, $o$ does not divide $p$, so $o$ can be $2$ or $2p$. If $o=2$, \[\left(\frac{11}{17}\right)^2 \equiv 1 \pmod{r} \implies -168 \equiv 0 \pmod{r} \implies r \in \{3,7\}.\] Manually checking each case gives a contradiction for $r=3$, so $r=7$ is the only extra case here. If $o=2p$, we have $2p \mid r-1$, which gives the last solution set. $\square$ Claim 2: If $p, q \neq 2$, then $p=3$. Proof: Suppose that $p \neq 7$. Let $a$ be the positive integer such that $a^2 = p^{q-1}$. Notice that \[3a^2+1 \equiv 4 \pmod{8}.\] Also, \[\nu_7(3a^2+1) \le \nu_7(11^p+17^p) = 1+ \nu_7(p) =1.\] Some quick analysis yields \[3a^2+1 \equiv 2^2, 2^2 \cdot 7 \pmod{p}\]\[\implies 1 \equiv 4 \pmod{p} \ \textbf{ or } \ 1 \equiv 28 \pmod{p}.\] This implies that $p=3$. If $p=7$, we must analogously have \[1 \equiv 196 \pmod{p},\] a contradiction. $\square$ Now, we break this problem into cases: If $p=2$, we have \[3 \cdot 2^{q-1}+1 \mid 410\]\[\implies 2^{q-1} \in \{0,3,27,68\},\] which is impossible. If $p=3$, we have \[3^q+1 \mid 6244\]\[\implies 3^q \in \{0,1,3,6,13,27,222,445,891,1560,3121,6243\}.\] This makes the only solution $q=3$. If $q=2$, we have \[3p+1 \mid 11^p+17^p.\] Claim 1 gives us that the only prime factors of $3p+1$ are $2$, $7$, or $2kp+1$ for some positive integer $k$. Clearly, the latter case is out of the question, so we have that \[3p+1 = 2^m7^n.\] Notice that \[\nu_2(3p+1) \le \nu_2(11^p+17^p) \le 3\]\[\nu_2(3p+1) \le \nu_2(11^p+17^p) \le 2.\] At this point, there are a finite amount of values for $3p+1$, which yield \[p\in\{0,1,2,9,16,65\},\] after some computation. This is a contradiction, so there are no solutions in this case either. Adding up all the solutions from the three cases gives the desired answer.

We will deal with the cases \(q = 2\), \(p = 2\), and \(p = 7\) later, so assume none of these are true. \textbf{Claim 1:} If \(r\) is a prime such that \(r \mid 3p^{q - 1} + 1\), then \(r \equiv 1 \mod{p}\) or \(r \mid 28\). \[ r \mid 3p^{q - 1} + 1 \implies 11^p + 17^p \equiv 0 \mod{r} \implies \left(\frac{-11}{17}\right)^p \equiv 1 \mod{r} \] Let \(k\) be the order of \(\frac{-11}{17} \mod r\). We have \(k \mid (p, r - 1) = 1, p\). If \(k = 1\), then \[ r \mid \frac{-11}{17} - 1 \implies r \mid 28 \implies r = 2, 7 \] If \((p, r - 1) = p\), we have \(p \mid r - 1 \implies r \equiv 1 \mod{p}\). \textbf{Claim 2:} \(v_2(3p^{q - 1} + 1) = 2\) Since \(q - 1\) is even and \(p\) is odd, \[ p^{q - 1} \equiv 1 \mod{8} \implies 3p^{q - 1} + 1 \equiv 4 \mod{8} \] So let \[ 3p^{q - 1} + 1 = 2^2 \cdot 7^\alpha \cdot p_1^{\alpha_1} \cdots p_k^{\alpha_k} \] where \(p_1 \equiv p_2 \equiv \cdots \equiv p_k \equiv 1 \mod{p}\). \textbf{Claim 3:} \(p = 3\), \(q = 3\) From LTE, we have \[ v_7(11^p + 17^p) = v_7(28) + v_7(p) = 1 \] Therefore \(\alpha = 0, 1\), and \[ 2^2 \cdot 7^\alpha \cdot p_1^{\alpha_1} \cdots p_k^{\alpha_k} \equiv 28, 4 \equiv 1 \mod{p} \] So \(p \mid 27, 3 \implies p = 3\). So, we have \[ 3^q + 1 \mid 11^3 + 17^3 = 6244 = 7 \cdot 4 \cdot 223 \] which quickly yields \(q = 3\). If \(p = 2\), we have \[ 3 \cdot 2^q + 1 \mid 11^2 + 17^2 = 410 \] which yields no solutions. If \(q = 2\), we have \[ 3p + 1 \mid 11^p + 17^p \] Here, \(v_2(3p + 1) = 1\), which means \(14 \equiv 1 \mod{p}\), or \(p = 13\), and \(q = 2\), which doesn't work. If \(p = 7\), \[ v_7(11^p + 17^p) = 2 \] so the power of 7 in \(3p^{q - 1} + 1\) can be \(0, 1, 2\), where \(0, 1\) yield the same cases as earlier and \(2\) yields \[ 196 \equiv 1 \mod{p} \implies p \mid 195 \] which is impossible as \(7 \nmid 195\). So, our only solution is \((3, 3)\).

Let $r$ be a prime factor of $3p^{q-1}+1$. Then we have that either $r = 7$ or $r\equiv 1 \pmod{p}$ Now observe that if $r_1,r_2, \dots r_i$ be all the prime factors of $3p^{q-1}+1$ we have that all $r_i \equiv 1 \pmod{p}$. But then as $3p^{q-1}+1 = r_1^{\alpha_1}r_2^{\alpha_2}\dots r_i^{\alpha_i} = (pk_1 + 1)^{\alpha_1}(pk_2)^{\alpha_2}\dots (pk_i)^{\alpha_i}$ Here if prime factors is greater than $3$ then we have that the term with $p$ in the product on RHS is even while on LHS it is odd contradiction. Hence only $2,3,7$ can be the prime factors of $3p^{q-1}+1$, then a case bash gives $(3,3)$ as the only solution.

Take a prime $r$ such that it divides $3p^{q-1}+1$ so $\text{ord}_{r}\left(\frac{11}{17}\right)=2$ or $2p$, so $r=2,7$ or $r\equiv 1 \pmod{2p}$. Note that if $7\mid3p^{q-1}+1$ then we get $p^{q-1}\equiv 2\pmod{7}$ which is not possible. Now, $\nu_7(11^p+17^p)=\nu_7(28)+\nu_7(p)=1$. We check for $p=2$ which gives us $3\cdot 2^{q-1}+1\mid 410$ which is not possible. For $p=3$ we have $3^q+1\mid 6244$, after case-bash we get $q=3$. If $p\neq7$ then we have $\nu_7(11^p+17^p)=1$, if $p=7,$ then $\nu_7(3\cdot 7^{p-1}+1)=0$. So $3p^{q-1}+1\equiv 2^2\cdot 7, 2^2\pmod{p}$, this gives us $(p,q)=(3,3)$ to be a solution. For $q=2$ we have $3p+1\mid 11^p+17^p$, since $3p+1$ cannot be written as $1\pmod{2p}$, so it is of the form $2^i7^j$, also we can write it as $3p+1\mid 11^p+17^p$ and because $\nu_2(3p+1)=1$ we have $p=13\implies q=2$, which does not work. So our only answer is $\boxed{(p,q)=(3,3)}$.

Only solution is (3,3) Let $r$ be a prime dividing $3p^{q-1}+1$, then we get by taking the inverse $\frac{11^p}{17^p}$ congruent to 1, giving either $r$ being $2,7$ or 2p divides r-1, for the second case, $3p^{q-1}+1$ is congruent to $1$ modulo 2p, which is impossible for p odd prime. Using LTE on $11^p+17^p$ for $7$, we get that, the most $7$ power is 1 and $8$ can never divide $11^p+17^p$, this means that $3p^{q-1}+1 = 7.2^n$ where $n \leq 2$, check the remaining cases.

Take any prime $r \geq 3$ which divides $3p^{q-1}+1$, so $r \mid 11^p + 17^p$ and thus $\text{ord}_{r}\left(11 \cdot 17^{-1}\right)=2$ or $2p$, giving $r=2,7$ or $r\equiv 1 \pmod{2p}$. Suppose $p\geq 3$. Then $\nu_2(11^p + 17^p) = 2$ (by mod $8$) and $\nu_7(11^p + 17^p) = \nu_7(28) + \nu_7(p)$ (by Lifting the Exponent) which is $1$ if $p\neq 7$ and $2$ if $p=7$. So $\nu_7(3p^{q-1} + 1) \leq 1$ (for $p=7$ actually it is $0$) and $\nu_2(3p^{q-1} + 1) \leq 2$ Hence \[ 3p^{q-1} + 1 = 2^{\leq 2}7^{\leq 1} \ \cdot \ (\mbox{primes } \equiv 1 \pmod{2p}). \]Taking the latter mod $p$ yields $p = 2, 3, 13$. Since $11^2 + 17^2 = 2 \cdot 5 \cdot 41$, checking $3 \cdot 2^{q-1} + 1 \mid 5 \cdot 41$ gives no solution. For $11^3 + 17^3 = 28 \cdot (17^2 - 17 \cdot 11 + 11^2) = 2^2 \cdot 7 \cdot 223$ checking $\displaystyle \frac{3^{q}+1}{2} \mid 2\cdot 7 \cdot 223$ gives $q=3$, so $(3,3)$ is a solution. If $p=13 = 2\cdot 7 - 1$, then the right-hand side above must be divisible by $7$, but then mod $7$ yields $3\cdot (-1)^{q-1} + 1 \equiv 0$, contradiction for any positive integer $q$. Remark. We did not use anywhere that $q$ is prime!

Solved a while ago but forgot to post The only solution is $\boxed{(p,q) = (3,3)}$, which works. Claim: Any prime divisor $r\mid 11^p + 17^p$ satisfies $r\in \{2,7\}$ or $r\equiv 1\pmod{2p}$. Proof: Let $r$ be a prime divisor of $11^p + 17^p$ and $r\ne 2,7$. We have \[11^p + 17^p \equiv 0\pmod r\] This implies that \[\left(\frac{11}{17}\right)^{p}\equiv -1\pmod r,\]so \[\left(\frac{11}{17}\right)^{2p}\equiv 1\pmod r\] Thus $\mathrm{ord}_r \left(\frac{11}{17}\right)$ divides $2p$ but not $p$, so it must be either $2$ or $2p$. If $\mathrm{ord}_r \left(\frac{11}{17}\right) = 2$, then we have \[r\mid 17^ 2- 11^2 = 168\]Since $r$ is odd, $\frac{11}{17} \not\equiv 1\pmod r$, so $r\ne 3$. This implies $r=7$, which we assumed to not be true. If $\mathrm{ord}_r\left(\frac{11}{17}\right) = 2p$, then $2p\mid r-1$, $r\equiv 1\pmod{2p}$, as desired. $\square$ Claim: $p\ne 2,7$ Proof: If $p=2$, then \[3\cdot 2^{q-1} + 1 \mid 410\]It's easy to check this is not possible. If $p=7$, then \[3\cdot 7^{q-1} + 1 \mid 11^7 + 17^7\]Since $22\nmid 11^7 + 17^7$, $q=2$ doesn't work. Now assume $q$ is odd. We have $\nu_2(3\cdot 7^{q-1} + 1) = 2$. So any prime factor of \[\frac{3\cdot 7^{q-1} + 1}{4}\]is $1\pmod{14}$, which implies $3\cdot 7^{q-1} + 1\equiv 4\pmod{14}$, absurd. $\square$ If $p=3$, then \[3^q + 1 \mid 11^3 + 17^3 =6244\]One can manually check that this is not the case since $3^q < 6244\implies q\le 7$. From now on, assume $p\not\in \{2,3,7\}$. Case 1: $q=2$ Then we have \[3p+1 \mid 11^p + 17^p\]Note that no prime that if $1\pmod{2p}$ divides $3p+1$, so the only prime factors of $3p+1$ are $2$ and $7$. However \[1\le \nu_2(3p+1) \le \nu_p(11^p + 17^p) = 2\]and \[\nu_7(3p+1)\le \nu_7(11^p + 17^p) = 1\]by LTE. So \[3p+1\in \{2,4,14,28\},\]all of which don't work. Case 2: $q>2$ Then $p^{q-1}\equiv 1\pmod 4$, so $4\mid 3\cdot p^{q-1} + 1$. Thus by Claim 15.1, \[3\cdot p^{q-1} + 1\equiv \{4,28\}\pmod{2p}\]However, \[3\cdot p^{q-1} + 1\equiv p+1\pmod{2p},\]so $p=3$, contradiction.