$n, k, m$ is positive integer and $n, k$ satisfies $n=7k-1$. Prove that for every $d$ which is positive divisor of n, $d+\frac{n}{d}\ne3^{3m-2}$.

Question

$n, k, m$ is positive integer and $n, k$ satisfies $n=7k-1$. Prove that for every $d$ which is divisor of n, $d+\frac{n}{d}\ne3^{3m-2}$.

What I've tried:

I've shown that there's no such $n$ that works as a counterexample when $ a<50000000$, but I have no clue to prove the problem strictly.

I've thought of proving it by showing that the quadratic eqation of $d, d^2-3^{3m-2}d+n=0$ doesn't have positive integer solution, but it didn't really turn out helpful in my case.

Answer 1

Can we check it case wise, as follows?

Given $$n\equiv 6 \pmod 7.$$

We need to check if

$$9d^2 +9n = 3^{3m} =27^md.$$

If this thing happens, then we must have

$$2d^2+2n\equiv (-1)^md \pmod 7.$$

That is

$$2d^2+5 \equiv (-1)^m d\pmod 7,$$ (as $2n\equiv 5 \pmod 7.$) Now, discuss the cases:

$\textbf{Case 1:}$ Suppose $d\equiv 1 \pmod 7$. Then, clearly $2d^2+5 \equiv 0\pmod 7$ which is not congruent to $(-1)^m d \equiv (-1)^m \pmod 7$, for whatever the value of $m$ may be.

$\textbf{Case 2:}$ $d \equiv 2 \pmod 7$. Then, $2d^2+5 \equiv 6 \pmod 7$ and $(-1)^md \equiv \pm 2 \pmod 7$. Hence both cannot be congruent modulo $7$.

$\textbf{Case 3:}$ $d \equiv 3 \pmod 7$. Then,$2d^2+5 \equiv 2 \pmod 7$ and $(-1)^md \equiv \pm 3\pmod 7$.

$\textbf{Case 4:}$ $d \equiv 4 \pmod 7$. Then, $2d^2+5 \equiv 2 \pmod 7$ and $(-1)^md \equiv \pm 4\pmod 7$.

$\textbf{Case 5:}$ $d \equiv 5 \pmod 7$. Then, $2d^2+5 \equiv 6 \pmod 7$ and $(-1)^md \equiv \pm 5\pmod 7$.

$\textbf{Case 6:}$ $d \equiv 6 \pmod 7$. Then, $2d^2+5 \equiv 0 \pmod 7$ and $(-1)^md \equiv \pm 6\pmod 7$.

$\textbf{Case 7:}$ $d \equiv 0 \pmod 7$. Then, $2d^2+5 \equiv 5 \pmod 7$ and $(-1)^md \equiv \pm 0\pmod 7$.

Hence, in none of the cases above $$2d^2+5\equiv (-1)^md \pmod 7.$$

Thus, there cannot exist such a $d$ with

$$d+\frac{n}{d} = 3^{3m-2}.$$

Answer 2

$$\begin{array} {r|l} \text{d} & (3^{3m-2}-d)\cdot d + 1 \ne 0 \pmod 7 \\ \hline 0 & 1 \ne 0 \pmod 7 \\ 1 & 3^{3m-2} \ne -1 \cdot 1 + 1 = 0 \pmod 7 \\ 2 & 3^{3m-2} \ne -1 \cdot 4 + 2 = 5 \pmod 7 \\ 3 & 3^{3m-2} \ne -1 \cdot 5 + 3 = 5 \pmod 7 \\ 4 & 3^{3m-2} \ne -1 \cdot 2 + 4 = 2 \pmod 7 \\ 5 & 3^{3m-2} \ne -1 \cdot 3 + 5 = 2 \pmod 7 \\ 6 & 3^{3m-2} \ne -1 \cdot 6 + 6 = 0 \pmod 7 \\ \end{array}$$

$$\begin{array} {rl} 3^{3m-2} \not \in \{0, 2, 5\} & \pmod 7 \\ 3 \cdot 27^j \not \in \{0, 2, 5\} & \pmod 7 \quad j \ge 1 \\ 3 \cdot (-1)^j \not \in \{0, 2, 5\} & \pmod 7 \\ \end{array}$$

Answer 3

Conceptually: $\,d+\frac{n}d = \color{#0af}{3^{3m-2}}\Rightarrow \bmod 7\!:\ d,\ \frac{n}d\,$ are the roots of $\,x^2\color{#0a0}{\pm4}{x}\color{darkorange}{-1}\,$ by Vieta, by root product $\,\frac{n}d\,d\equiv n\equiv \color{darkorange}{-1}\,$ and sum $\,d+\frac{n}d\equiv \color{#0af}{27^m/9}\equiv (-1)^m/2\equiv \color{#0a0}{\pm4},\,$ $\rm\color{#c00}{contra}$ the quadratic has no roots, by discriminant $\,20\equiv -1\,$ is nonsquare by Euler's Criterion, i.e. $(-1)^3\!\equiv\! {-}1$.