We devour this with Quadratic Reciprocity and Residues We claim that the only solution is $\boxed{n=2}$ when we obtain $19 \mid 209$. Case 1: Let $n=2k$ be some even number. If the divisibility holds we must have, $$3^n+200 \equiv 0 \ (mod \ 2^n+15) \implies 3^n \equiv -200 \ (mod \ 2^n+15) \implies (3^k)^2 \equiv (-200) \ (mod \ 2^n+15)$$We introduce jacobi symbol and argue that $(\dfrac{-200}{2^n+15})=1$ but notice that by multiplicativity, this becomes $$(\dfrac{-1}{2^n+15})(\dfrac{200}{2^n+15})=1$$. We have $n \ge 2$ so obviously $2^n+15 \equiv 3 \ (mod \ 4)$ and thereby, $(\dfrac{-1}{2^n+15})=(-1)$. So we are forced to have, $(\dfrac{200}{2^n+15})=(-1)$ which again by multiplicativity becomes, $$(\dfrac{2}{2^n+15})^3*(\dfrac{5}{2^n+15})^2=(-1)$$however, we can again check $n=2$ to give us our valid solution pair $\boxed{(19,209)}$. Otherwise we have, $2^n+15 \equiv 7 \ (mod \ 8)$ which leads to $(\dfrac{2}{2^n+15})=1$. Combining these facts we end up with, $(\dfrac{5}{2^n+15})^2=(-1)$ which is impossible. Case 2: Since $n \ge 3$ we obviously can let $n=2k+1$. We end up with $$3^{2k+1} \equiv -200 \ (mod \ 2^n+15) \implies 3^{2k+2} \equiv -600 \ (mod \ 2^n+15) \implies (\dfrac{-600}{2^n+15})=1 \implies (\dfrac{-1}{2^n+15})(\dfrac{600}{2^n+15})=1$$Again we have $2^n+15 \equiv 7 \ (mod \ 8)$ so both $(\dfrac{2}{2^n+15})=1$ and $(\dfrac{-1}{2^n+15})=-1$ hold. By using complete multiplicativity we obtain, $$(\dfrac{600}{2^n+15})(\dfrac{-1}{2^n+15})=1 \implies (\dfrac{2}{2^n+15})^3*(\dfrac{5}{2^n+15})^2*(\dfrac{3}{2^n+15})=(-1) \implies (\dfrac{5}{2^n+15})^2*(\dfrac{3}{2^n+15})=(-1)$$However obviously $gcd(5,2^n+15)=gcd(5,2^n)=1$ and so, $(\dfrac{5}{2^n+15})^2=1$ is forced. This leaves us with $(\dfrac{3}{2^n+15})=(-1)$. Notice that with Quadratic Reciprocity as $3 \equiv 3 \ (mod \ 4)$ and $2^{2k+1}+15 \equiv 3 \ (mod \ 4)$ we get that this is equivalent to, $$(\dfrac{2^n+15}{3})=1 \implies (\dfrac{2^n}{3})=1 \implies (\dfrac{2}{3})^{2k+1}=1 \implies (-1)^{2k+1}=1 \implies (-1)=1$$which is a strict contradiction! Q.E.D $\blacksquare$

jungle_wang wrote: Find all positive integers $n$ such that $$2^n+15|3^n+200$$ quite the same idea as this one https://artofproblemsolving.com/community/c6h3107338p28104291

The solution behind this task uses quadratic residues First case: If n is even number Therefore: (-200) is quadratic residue of mod (2^n+15) (-200/2^n+15)=(-1/2^n+15).(2/2^n+15)^3.(5/2^n+15)^2 Now, we have that 2^n+15 is comparable of (mod 4) to 3, n>=3 Therefore (-1/2^n+15)=-1 Now, 2^n+15 is comparable of (mod 8) to 7, n>=3 Therefore (2/2^n+15)=(-1)^(7^2-1/8)=1 Now (5/2^n+15)^2=1 Therefore we have that (-200/2^n+15)=1=(-1).1.1=-1 Contradiction Now, we are left to case that n is odd Let multiply the whole thing to 3 Therefore (-3.200) is quadratic residue of (mod 2^n+15) We have that (-600/2^n+15)=1 (-200/2^n+15)=-1 We must to have that to (3/2^n+15)=? (3/2^n+15).(2^n+15/3)=(-1)^(1). (-1)^(2^(n-1)+7)=(-1) If n>1 We must to have that (2^n+15/3)=1 (2^n+15/3)=(2^n/3), n is odd, therefore (2^n/3)=2,, which is comparable to (-1), contradiction Now we are left to n<3 For n=1, 203 is prime, for n=2, we have a solution(209, 19), we are ready

Can anyone solve it without Jacobi symbol?

We claim that $n=2$ is the only answer. Clearly when $n=2, 2^2+15=19 \mid 209 = 3^2+200$. We use Jacobi symbols throughout. Since $n=1$ is clearly not a solution, we henceforth assume $n \geq 3$. Assume that $2^n+15 \mid 3^n+200$. Since $gcd(3^n, 2^n+15)=1$, we have $\left( \frac{3^n}{2^n+15} \right) = \left( \frac{-200}{2^n+15} \right)$. Since $\frac{2^n+15-1}{2}=2^{n-1}+7$ is odd and $\frac{(2^n+15)^2-1}{8}=(2^n+14)(2^{n-3}+2)$ is even, we have $\left( \frac{-1}{2^n+15} \right)=-1, \left( \frac{2}{2^n+15} \right)=1$. So, $\left( \frac{-200}{2^n+15} \right) = \left( \frac{-1}{2^n+15} \right) \left( \frac{2}{2^n+15} \right) =-1$. If $n$ is even, then $\left( \frac{3^n}{2^n+15} \right) = \left( \frac{3^\frac{n}{2}}{2^n+15} \right)^2 =1$, contradiction. If $n$ is odd, then since $\frac{(3-1)(2^n+15-1)}{4}=2^{n-1}+7$ is odd, $\left( \frac{3^n}{2^n+15} \right) = \left( \frac{3}{2^n+15} \right) = - \left( \frac{2^n+15}{3} \right) = - \left( \frac{2^n}{3} \right) = - \left( \frac{2}{3} \right) = 1$. So there is no solution for $n \geq 3$. $\square$