Sorry just a stupid post !

siddigss wrote:

Your solution is obviously wrong because if $a\equiv 2 \mod 3$ then $4a+3\equiv 2\neq 0 \mod 3$. By the way, if you try to find some prime $p$ such that $p\mid x_n$ for some $n$ you will never find such prime letting $a\equiv -1 \mod p$.

I claim that the answer is two. If $x_1 = 2$, then $x_2 = 5$, $x_3 = 11$. $2^2-1 = 3$ and $2^5-1=31$ are primes, but $2^11-1 = 23\cdot 89$ is not a prime. (this achieves $k = 2$) Assume $x_1$ is odd from now on. Lemma: if $2^k-1$ is prime, then $k$ is prime. Proof: If $p\mid k$, then $2^p-1 \mid 2^k-1$. Now, assume to the contrary that $k > 2$, so that $2^{x_1}-1$, $2^{x_2}-1$, and $2^{x_3}-1$ are all prime. Then $x_3 = 4x_1+3$, and since $x_1$ is odd, $x_3 \equiv 7 \pmod{8}$. By the lemma, $x_3$ is prime. Thus $2$ is a quadratic residue mod $x_3$. Let $2 = k^2 \pmod{x_3}$. Then $2^{x_2}-1 \equiv k^{2x_2}-1 = k^{x_3-1}-1 \equiv 0 \pmod{x_3}$ by Fermat's Little Theorem. Thus $y_3 \mid 2^{x_2}-1$, and $2^{x_2}-1 > x_3$ (unless $x_3 = 5$ or $x_3 = 7$, which don't work), a contradiction to the fact that $2^{x_2}-1$ is prime. Thus the minimum $k$ is two.

I do not quite like the fact that quadratic reciprocity is really the only nice way to solve this problem.. there should be a nicer method that does not invoke such mechanics. Does anyone else have a nice more elementary solution to this problem?

Computing $\left ( \frac{2}{p} \right )$ is rather simple and extremely elementary... the full strength of reciprocity laws is not needed.

I will use $M_n$ to denote the $n-th$ Mersenne Number which is $2^n-1$ Now observe that a necessary condition for $M_n$ to be prime is that $n$ should be prime. Now also observe that for primes $p$ and $q=2p+1$ we have $q|M_p$ if $p\equiv 3(mod4)$ So assume $a\equiv 1(mod4)$. Now $x_2=2a+1\equiv 3(mod4)$ and it is a prime for $y_2$ to be a prime. So if $x_3=2(2a+1)+1$ is a prime, $x_3|y_2$, thus the maximum such $k$ is 2 - added that $x_3$, and thus $y_3$ is composite.

Firstly, $x_n = 2^{n-1}(a+1) -1$ Using basic properties of Mersenne primes, if $y_i$s are primes, $x_i$s must also be primes. Assuming $k>2$, $x_1, x_2, x_3$ are all primes. So, $a$ is a prime, $2a+1$ is a prime and also $4a+3$ is a prime. Then by using Euler's criterion, $2^{2a+1}-1 \cong 2^{\frac{(4a+3)-1}{2}}\cong \left( \frac{2}{4a+3} \right)(mod$ $4a+3)$ Suppose now that $a=2$. Then, we see that $y_3= 2^{11} -1= 2047=23.89$, contradiction. So, $a$ must be odd. Therefore $4a+3 \cong 7(mod$ $8)$. So we have $2^{2a+1} \cong \left( \frac{2}{4a+3} \right) \cong 1(mod$ $4a+3)$, since $2$ is a quadratic residue modulo any prime of the form $8m+7$.

Hmm, I feel this is a bit too hard for a #1, though not sure. The key lemma took me a while to find. The answer is $k=2$. Suppose we had a chain of length $3$. Clearly, if $y_n$ is prime, then so is $x_n$ (because $2^a-1\mid 2^b-1$ if $a\mid b$). We have the following miraculous lemma: Lemma: All the $x_i$ for $i\le k$ except potentially $x_1$ are not $1$ or $7$ mod $8$. Proof of Lemma: Note that if $x_{n+1}\equiv 1,7\pmod{8}$, then $2$ is a QR mod $x_{n+1}$. Therefore, $2^{(x_{n+1}-1)/2}\equiv 1\pmod{x_{n+1}}$, so $x_{n+1}\mid 2^{x_n}-1=y_n$. But $y_n$ is prime, so we must have $x_{n+1}=y_n$, so $2x+1=2^x-1$ for $x=x_n$. It is easy to see that $2x+1=2^x-1$ has no solutions for positive integers $x$. Therefore, we have the desired contradiction. $\blacksquare$ Now, note that the action of $2x+1$ mod $8$ is given by \begin{align*} 0&\mapsto 1\\ 1&\mapsto 3\\ 2&\mapsto 5\\ 3&\mapsto 7\\ 4&\mapsto 1\\ 5&\mapsto 3\\ 6&\mapsto 5\\ 7&\mapsto 7.\\ \end{align*}Since our chain of length $3$ can't have anything that's $1$ or $7$ except the start, the last element can only be $1,2,5,6$, so only $5$. Therefore, the chain is $2,5,3$. Now, if $x_1\equiv 2\pmod{8}$ then $x_1=2$ (because its has to be prime), which we can verify doesn't work. $\blacksquare$

Nice problem. Although it's easily doable if one knows what are Mersenne Primes. Anyway, here's my solution: Let $M_s=2^s-1$ denote the $s^{th}$ Mersenne Number. It is well known that if $M_s$ is a prime then $s$ must also be a prime. We claim that $k=2$ is the largest possible value of $k$. To see this, take $x_1=a=3$. Then $x_2=7$ and $x_3=15$, which works as $y_1=M_3=7$ and $y_2=M_7=127$ are primes, while $y_3=M_{15}=32767=7 \times 4681$ is not. Now, we'll show that $k=3$ doesn't work. To see this, assume to the contrary that for some value of $a$, $k=3$ works. Then $x_1=a,x_2=2a+1,x_3=4a+3$ must all be primes. We use the following well-known lemma to supplement our proof- LEMMA If $p \equiv 3 \pmod{4}$ and $q=2p+1$ are both primes, then $q \mid M_p$. PROOF: By FLT, we get that \begin{align*} 2^{q-1} \equiv 1 \pmod{q} \Rightarrow 2^{2p} \equiv 1 \pmod{q} &\Rightarrow 2^p \equiv +1 \pmod{q} \\ &\text{OR } 2^p \equiv -1 \pmod{q} \end{align*}Suppose the latter holds true. Then, using the fact that $p=\frac{q-1}{2}$, and from Euler's Criterion, we get that $2$ is a quadratic non-residue modulo $q$. But, as $q=2p+1 \equiv 7 \pmod{8}$ , using the Second Supplement to the Quadratic Reciprocity Theorem, we get that $2$ is in fact a quadratic residue modulo $q$, which is a contradiction! Thus, the former result must be true, i.e. $2^p-1 \equiv 0 \pmod{q} \Rightarrow q \mid M_p$ $\Box$ Return to the problem at hand. It's easy to see that $a=2$ doesn't work (as $M_{4a+3}=M_{11}=2047=23 \times 89$ is not a prime). So from now on we assume that $a>2$. As $a$ is a prime number, it must be of the form $2t+1$. This gives $x_2=4t+3$. By our Lemma, we get that $x_3=2x_2+1$ (which is a prime) divides $M_{x_2}$, as $x_2 \equiv 3 \pmod{4}$ is a prime, which contradicts the fact that $y_2$ is a prime number. Hence, done. $\blacksquare$

We claim the answer is $k = 2$, which is achieved by $a = 3$. Lemma 1: If $2^m - 1$ is prime for some positive integer $m$, $m$ must itself be prime. Proof of Lemma 1: Suppose not, and write $m = ab$ for some $a, b > 1$. But $2^a - 1 \mid 2^m - 1$, contradicting the fact that $2^m - 1$ is prime. $\blacksquare$ Lemma 2: Suppose $p, 2p + 1$ are both odd primes. If $2^p - 1$ and $2^{2p + 1} - 1$ are both also prime, $p \equiv 1 \pmod{4}$. Proof of Lemma 2: Consider \begin{align*} 2^p \pmod{2p + 1}. \end{align*}This is the Jacobi symbol $\left(\frac 2{2p + 1}\right)$. If $\left(\frac 2{2p + 1}\right) = 1$, we have \begin{align*} 2p + 1 &\mid 2^p - 1, \end{align*}contradicting the fact that $2^p - 1$ is prime. Hence, we must have $\left(\frac 2{2p + 1}\right) = -1$. This is equivalent to \begin{align*} (-1)^{\frac{(2p + 1)^2 - 1}{8}} &= -1 \\ \iff (-1)^{\frac{2p(2p + 2)}{8}} &= -1 \\ \iff (-1)^{\frac{p + 1}{2}} &= -1. \end{align*}Thus, we must have $p \equiv 1 \pmod{4}$, proving the lemma. $\blacksquare$ We now return to the original problem. Suppose $y_1, y_2, y_3$ are all prime. By Lemma 1, $x_1, x_2, x_3$ are also all prime, so by definition $a, 2a + 1, 4a + 3$ are all prime. Now, by Lemma 2, $a \equiv 1 \pmod{4}$, and also $2a + 1 \equiv 1 \pmod{4}$, which is clearly impossible. Therefore, $k \leq 2$, with equality demonstrated by $a = 3$. This completes the proof. $\Box$

I claim $k=2$. Construction: $x_1=2$. Lemma 1: If $2^t -1$ is prime, then $t$ must be prime. Proof: Clearly, it $t=uv$, then both $2^u -1$ and $2^v -1$ divide $t$. Lemma 2: For any prime $p \equiv 7 \mod 8$, $ \ \ 2^{\frac{p-1}{2}} \equiv 1 \mod p$ Proof: We have $\left( \frac{2}{p} \right) = (-1)^{\frac{p^2-1}{2}} = 1$. Hence $1\equiv \left( \frac{2}{p} \right) \equiv 2^{\frac{p-1}{2}} \mod p$. Now back to the problem. By Lemma 1, $x_i$'s must be prime for $y_i$'s to be prime. AFSOC, we have $y_1, y_2, y_3$ are all prime. Let $x_1 =a$, where $a$ is an odd prime. Then clearly $x_3 \equiv 7 \mod 8$. Applying Lemma 2, $2^{x_2}-1 \equiv 0 \mod x_3$. But $a \geq 3 \longrightarrow 2^{2a +1}-1 > 4a+3 $ because $2^{2a-1} > 2^a +1 > a+1 $. So $y_2 =2^{x_2}-1$ is composite. Contradiction. Lastly, if $x_1=2$ then $x_2 =5, x_3 =11, x_4 =23$. Again by Lemma 2, $23 \mid 2^{11}-1$ and $2^{11}-1>23$, so it's composite. Therefore, $k=2$.

The key observation is the following lemma:

We can finish by considering the possible chain of $x_n$ modulo 8, and the answer is 2, which could be achieved by $a=3$.

The maximum value of $k$ is $k=2$. Construction is easy; take $a=2$, for example. The following proof of maximality is due to awang11. Suppose that $k=3$ was attainable. Note that $a,2a+1,4a+3$ must all be primes. If $a=2$, we get a contradiction because $2^{4a+3}-1=2^{11}-1=2047=23\cdot 89$. Thus, we can assume $a\equiv 1\pmod{2}$. Then, note that \[2^{(4a+3-1)/2} \equiv 1\pmod{4a+3}\]because $2$ is a quadratic residue modulo $4a+3\equiv -1\pmod{8}$. This implies \[4a+3\mid 2^{2a+1}-1=y_2.\]It remains to deal with the case \[4a+3=2^{2a+1}-1.\]Then, we can rewrite as \[a+1=2^{2a-1}.\]Note that equality holds at $a=1$, then note that increasing $a$ by $1$ increases the RHS by $1$ and the LHS by more than $1$, so equality can never hold for prime $a$.

We claim that that the maximum is $k=2$. We can achieve this with $a=2$. For the sake of contradiction, assume there is a sequence of length at least $3$. First, note that $2^x - 1$ is prime only if $x$ is also prime. Therefore, we have $a$, $2a+1$, and $4a+3$ are prime. Note that when $a$ is even, $y_1$ is prime if and only if $a=2$. We already know the $a=2$ case yields $k=2$, so we can assume $a$ is odd or $a = 2k+1$. Therefore, we have $2k+1$, $4k+3$, and $8k+7$ are prime. Note that for a prime of the form $8k+7$, the number $2$ is a quadratic residue because \[ \left(\frac{2}{p}\right) = (-1)^{\frac18 (p^2-1)} = 1 .\]Let $m^2 = 2 \pmod{8k+7}$. We assumed $y_2$ is prime, but \[ 2^{4k+3} -1 \equiv m^{8k+6} - 1 \equiv 0 \pmod{8k+7} \]by FLT. Therefore, we must have $2^{4k+3}-1 = 8k+7$, which does not provide any solutions. Therefore, we have a contradiction, and we are done.

The answer is $k=2$; this can be achieved, for instance, when $a=3$. Now we show this is the best possible. Assume otherwise that we could achieve three consecutive primes $y_1, y_2, y_3$; for obvious reasons, $x_1=a, x_2=2a+1, x_3=4a+3$ must also be prime. However, by quadratic reciprocity we have $$\left( \frac{2}{4a+3} \right) = (-1)^{(4a+4)(4a+2)/8} = (-1)^{(a+1)(2a+1)} = 1$$and so by Fermat's little theorem we have $2^{2a+1} \equiv 1 \pmod{4a+3}$. This contradicts the assumption that $y_2=2^{2a+1}-1$ is prime, because $2^{2a+1}-1>4a+3$ when $a \ge 2$.

Ultra nice problem and a great exercise for quadratic residues $\color{black}\rule{25cm}{1pt}$ The maximal possible $k$ is $k=2$. Because of the primality condition on $y$ we must have that all of the $x$'s must be prime as well. Now we easily find a formula for $x$, by subtracting the $(n+1)^{th}$ equation from the $(n+2)^{nd}$ equation we have that the characteristic polynomial is $t^2-3t+2=0$, this implies that $x_n = A+B.2^n$. Plugging in the what we got for $x$ we get that $A=-1$, and setting $n=1$ we have that $B=\frac{a+1}{2}$, thus we have that $x_n=2^{n-1}(a+1)-1$. Thus we must have that the numbers $a,2a+1,4a+3,8a+7$ are all prime numbers. But notice the following, we have that $2^{2a+1}=2^{\frac{4a+3-1}{2}}$, since $4a+3$ is prime from Euler's criterion we must have that $2^{2a+1} \equiv \left(\frac{2}{4a+3}\right)\pmod{4a+3}$. We have that $\left(\frac{2}{4a+3}\right) \equiv (-1)^{\left\lfloor \frac{4a+3+1}{4} \right\rfloor} \equiv (-1)^{a+1} \pmod{4a+3}$. Now if $a > 2$ we have that $y_3$ isn't prime, so we have that $a=2$. But now we have that $y_3=2^{11}-1=23.89$, thus we have that $k=2$.

why are we all doing heavynt Our answer is $k = 2$, which can be achieved when $a = 2$. Note that in general, $y_n$ prime implies that $x_n$ is prime; if $y_n$ is prime but $x_n$ is not, then $y_n = 2^{x_n} - 1$ is divisible by $2^m - 1$ for some $m \neq 1, x_n$ satisfying $m \mid x_n$, which is impossible. We will prove that the maximum $k$ for which $x_1, \ldots , x_k$ are all prime is $2$, which finishes the problem. Suppose $(x_1, x_2, x_3) = (a, 2a+1, 4a+3)$ are all prime. We know $a = 2$ yields $k = 2$ so assume $a$ is an odd prime. Write\[2^{2a+1} \equiv 2^{\tfrac12(4a + 3 - 1)} = \left(\tfrac{2}{4a+3}\right) = (-1)^{\tfrac{1}{8}((4a + 3)^2 - 1)} \pmod{4a + 3}\]Since $a$ is odd write $a = 2b+1$, so\[\tfrac18((4a + 3)^2 - 1) = \tfrac18((8b + 7)^2 - 1) = (8b+6)(b+1)\]hence it follows that $2^{2a+1} \equiv 1 \pmod{4a+3}$. Clearly $4a + 3 < 2^{2a+1} - 1$ so $4a+3 \mid 2^{2a+1} - 1$ thus it follows that $2^{x_2} - 1$ has a prime factor, a contradiction. Hence, we can have $k = 2$ at maximum, as desired. $\blacksquare$

We claim $k=2$ is the maximum. Firstly, $k=2$ works since taking $a=2$ gives $2^2-1=3$ and $2^5-1=31$, both primes. Now we show $k=3$ fails. Let $S=\{a,2a+1,4a+3\}$ and suppose $2^s-1$ is prime for all $s\in S$. If $s$ is composite, then so is $2^s-1$ due to the well-known fact that $x\mid y \iff 2^x-1\mid 2^y-1$. Hence all elements of $S$ are prime. Claim: We have $4a+3\mid 2^{2a+1}-1$. Proof: Since $a$ is prime, $a\equiv 1\pmod2$, so $4a+3\equiv 7\pmod8$. Also, $4a+3$ is prime, so \[ \left(\frac{2}{4a+3}\right)=1 \implies 2^{2a+1}\equiv 1 \pmod{4a+3},\]as claimed. $\blacksquare$ However, the claim is a contradiction since $2^{2a+1}-1$ is prime.

dr_Civot wrote:

dr_Civot wrote:

Why a>2 then a is odd?

Cool problem. We claim that the largest $k$ is $2$ which is possible if $x_1 = 2, x_2 = 5$. $\textbf{Claim:}$ $k < 2$ $\textbf{Proof)}$ Assume that $2^{x_1}-1, 2^{x_2}-1, 2^{x_3}-1$ are all primes from which we get that $x_1, x_2, x_3$ are all primes as well because if we let $x_i = a \cdot b$ for some $i \in \{1,2,3\}$, then $$2^{a}-1 \mid 2^{ab}-1$$We therefore have that $p = x_1, 2p+1, 4p+3$ are all primes such that $2^p-1, 2^{2p+1}-1, 2^{4p+3}-1$ are all primes, as well with $p \neq 2$ as $2^11-1$ is not prime. Notice that $$2^p-1 \equiv 2^{\frac{(2p+1)-1}{2}} - 1 \equiv 0 \pmod{2p+1}$$if $2$ is a quadratic residue $\pmod{2p+1}$ which is not possible, consequently, $2$ has to be a quadratic non-residue $\pmod{2p+1}$ and $\pmod{4p+3}$. Then $2p+1 \equiv 3,5 \pmod{8}$, yet if $2p+1 \equiv 5 \pmod{8}$, then $p$ has to be even meaning that $2p+1 \equiv 3 \pmod{8}$. Then $$4p+3 \equiv 2(2p+1)+1 \equiv 2\cdot 3 +1 \equiv 7 \pmod{8} $$meaning that $2$ is a quadratic residue $\pmod{4p+3}$ which is a contradiction. $\blacksquare$

The answer is $k = 2.$ Remark that since $2^{m}-1 \mid 2^{n}-1$ when $m \mid n,$ it follows that all of $\{x_1, \dots, x_k \}$ must be prime. Now, assume we had three primes $a, 2a+1, 4a+3$ such that $2^{a}-1, 2^{2a+1}-1$, and $2^{4a+3}-1$ are all prime. Claim: $4a+3 \mid 2^{2a+1}-1.$ We have \[ \left( \frac{2}{4a+3} \right) = (-1)^{1/8 ((4a+3)^2-1)} = (-1)^{(2a+1)(a+1)} = 1 \]if $a$ is an odd prime. If $a = 2,$ then $y_1 = 2^2-1 = 3, y_2 = 2^5-1 = 31$ and $y_3 = 2^11-1 = 2047,$ which is not prime. Hence, as we also have \[ \left( \frac{2}{4a+3} \right) = 2^{2a+1} \equiv 1 \pmod{4a+3} \]under the assumption that $4a+3$ is prime, we thus have $4a+3 \mid 2^{2a+1}-1.$ Hence, it is impossible to have three primes $a, 2a+1,$ and $4a+3$ such that all of $2^{a}-1, 2^{2a+1}-1,$ and $2^{4a+3}-1$ are prime, meaning we can have at most $k=2,$ which holds by taking $a=2.$

The answer is $k=2$. For construction, take $a=2$. Now suppose $k \geq 3$. I will show one of the first three terms is composite. Assume for the sake of contradiction that $a, 2a+1, 4a+3$ are all primes. Assume that $a$ is odd; $a=2$ can be quickly checked. Now note that $4a+3 \equiv -1 \pmod 8$, thus $\left(\frac 2{4a+3}\right) = 1$ and thus $4a+3 \mid 2^{x_2} - 1 = 4a+3$ as by assumption $2^{x_2} - 1$ is prime. But this obviously impossible for size reasons.

Easy to see if $a=1,2$ $k=2$. Let $a\ge3$ Obvious that all of $x_i$ are prime. Let $x_1=p$. From quadratic residue we get $(\frac{-2}{4p+3})=1=(\frac{-1}{4p+3}).(\frac{2}{4p+3})$ so $(\frac{2}{4p+3})=-1$ and $p$ must be even. Contradiction. $\blacksquare$

We know that if $2^t-1$ is a prime, then $t$ itself must be a prime, because if $t'\mid t$ then $2^{t'}-1\mid 2^t-1$. Hence, $x_1, x_2, \ldots, x_k$ must be prime. Note that if $2^{\text{odd}}-1$ is a prime $p$, then $2$ is Quadratic Residue $\pmod{p}$ and hence $p\equiv \pm1\pmod{8}$. Note that $x_3=4x_1+3$ and $x_3\equiv 7\pmod{8}$. Note that $2^{\frac{x_3-1}2}\equiv1\pmod{x_3}$ as $x_3\equiv-1\pmod{8}$. This means that $x_3\mid y_2$ and hence $x_3=y_2$ which is possible only if $x_3=7$ but then $7\mid y_4=2^{15}-1$. So, largest possible $k$ is $2$ because $2^1-1,2^3-1,2^7-1$ are not all prime because first term is 1. $\blacksquare$

Notice we must have $x_1, x_2, \ldots$ prime. Our main claim states that if $x_i \equiv 3 \pmod 4$ and $x_{i+1} = 2x_i+1 \equiv 3 \pmod 4$ are prime, we have $x_{i+1} \mid 2^{x_i}-1 = y_i$. This holds, as \[2^{x_i} \equiv 2^{\frac{x_{i+1}-1}{2}} \equiv \left(\frac{2}{x_{i+1}}\right) \equiv 1 \pmod{x_{i+1}}.\] Thus we can consider cases using this claim, and denote $K$ as the maximum possible value of $k$ in that case. $a=2$: $K=2$. $a \equiv 3 \pmod 4$: Then $x_1 \equiv x_2 \equiv 3 \pmod 4$. If $x_2$ is prime, then $x_1$ is composite, so $K=0$. If $x_2$ is composite, then $K=1$. $a \equiv 1 \pmod 4$: Then $x_2 \equiv x_3 \equiv 3 \pmod 4$, so we can find $K \leq 2$ in a similar fashion. Thus our answer is $k = \boxed{2}$, which can be achieved using $a=2$. $\blacksquare$

Let a = 2, $x_1 = 2$, $y_1 = 3$, $y_2 = 31$, $y_3 = 2047 = 23.89$ $\Rightarrow$ this is an example for k = 2. We will prove that $k < 3$. For this, assume that $y_1$, $y_2$ and $y_3$ are prime. If $b \mid a$ then $2^b - 1 \mid 2^a - 1$ - impossible $\Rightarrow$ a is prime $\Rightarrow$ all $x_i$ are prime. Now let $x_1$ be an odd prime since we checked the case for a = 2. Since $x_1$ is odd we get that $x_3 \equiv 7 \pmod 8$. We will show that for any prime $p \equiv 7 \pmod 8$, $2^{\frac{p-1}{2}} \equiv 1 \pmod p$. Thats true because $\left( \frac{2}{p} \right) = (-1)^{\frac{p^2-1}{2}} = 1$ and $1 \equiv \left( \frac{2}{p} \right) \equiv 2^{\frac{p-1}{2}} \pmod p$. Using this we get $2^{x_2}-1 \equiv 0 \pmod {x_3}$ $\Rightarrow$ $x_3 \mid y_2$ $\Rightarrow$ $x_3 = y_2$, but for $a \geq 3$ we have that $2^{2a +1}-1 > 4a+3$ since $2^{2a-1} > 2^a +1 > a+1$. So $y_2 = 2^{x_2}-1$ is composite, which is impossible and we get our contradiction $\Rightarrow$ k = 2.

The answer is $k=2$. Check $a=1,2$. Next suppose $k=3$ is possible. We may assume $a=p$ is an odd prime and so are $2p+1,4p+3$. If $p\equiv 1\pmod 2$ then $4p+3\equiv 7\pmod 8$ so $2$ is a quadratic residue $\pmod{4p+3}$. Now consider the order of $2 \pmod{4p+3}$. It divides $4p+2$ but must be odd, so it is $2p+1$ and $4p+3\mid 2^{2p+1}-1$ so it is not prime, contradiction.