Quidditch wrote: Suppose that $m, n, k$ are positive integers satisfying $$3mk=3(m+3)^n+1.$$Prove that $k$ is odd. What? From the equation, we have $1 \equiv 0 \pmod{3}$.

Bulgaria 1998

$(m + 3)^n \equiv -1$ mod $3 \Rightarrow m \equiv -1$ mod $3$ and $n \equiv 1$ mod $2$. gcd$(3, m) = 1$ so if $(m + 3)^n + 1 \equiv 0$ mod $m$ then $3m | (m + 3)^n + 1$ so it suffices to show $\frac{(m + 3)^n + 1}{m}$ is odd when it is an integer and $m \equiv -1$ mod $3$. If $m$ is odd then $v_2$ of the denominator and numerator are both $0$. If $m \equiv 2$ mod $4$ then numerator is $2$ mod $4$ and so is denominator so $v_2$ of denominator and numerator are both $1$. If $m \equiv 0$ mod $8$ then numerator is $4$ mod $8$ so the the expression isn't an integer. If $m \equiv 4$ mod $8$ then $v_2$ of denominator is $2$ and $v_2$ of numerator is at least $3$ so we prove it is impossible for such an $m$ to make the expression an integer. Let $m = 4t$ where $t$ is odd and $t \equiv -1$ mod $3$. FTSOC suppose we have $4t | 3^n + 1$ where $n$ is odd. $v_2(3^n + 1) = 2$ and so we need to show $3^n + 1 \equiv 0$ mod $t$ is impossible. Note that $t$ must have a prime divisor which is $-1$ mod $3$ and let this be $p \Rightarrow 3^n \equiv -1$ mod $p$. Note that $\left(\frac{p}{3}\right) = \left(\frac{2}{3}\right) = -1$ $\textbf{Case 1:}$ $p \equiv 1$ mod $4$. $\left(\frac{3}{p}\right)\left(\frac{p}{3}\right) = (-1)^{\frac{2(p - 1)}{4}} = 1 \Rightarrow \left(\frac{3}{p}\right) = -1$ If $g$ is then a primitive root mod $p$ then $3 \equiv g^{j}$ mod $p$ where $j$ is odd so $3^n \equiv g^{jn} \equiv g^{\frac{p - 1}{2}}$ mod $p \Rightarrow jn \equiv \frac{p - 1}{2}$ mod $p - 1$ but $2|p-1$ and since $jn$ is odd and $\frac{p - 1}{2}$ is even we have a contradiction. $\textbf{Case 2:}$ $p \equiv 3$ mod $4$. $\left(\frac{3}{p}\right)\left(\frac{p}{3}\right) = (-1)^{\frac{2(p - 1)}{4}} = -1 \Rightarrow \left(\frac{3}{p}\right) = 1$ If $g$ is then a primitive root mod $p$ then $3 \equiv g^{j}$ mod $p$ where $j$ is even so $3^n \equiv g^{jn} \equiv g^{\frac{p - 1}{2}}$ mod $p \Rightarrow jn \equiv \frac{p - 1}{2}$ mod $p - 1$ but $2|p-1$ and since $jn$ is even and $\frac{p - 1}{2}$ is odd we have a contradiction.