RockmanEX3 wrote: Determine all positive integers $k$ and $n$ satisfying the equation $$k^2 - 2016 = 3^n$$ (Stephan Wagner) First of all, $3 | k$, hence $k=3k_1$ and $k_1^2-224=3^{n-2}$. As $(-1)^{n-2}\equiv 3^{n-2}= k_1^2-224\equiv 0, 1$ (mod 4), then $n-2=2m$. We get the equation $k_1^2-224=3^{2m}$, or $(k_1-3^m)(k_1+3^m)=224$. Then listing all cases we get the answer. I think the only one is $k_1=15$ and $m=0$, so $k=45$ and $n=2$.

$$k^2-2016=3^n \implies 3 ~ | ~k^2 \implies 9 ~ | ~ k^2 \implies n \geq 2 \implies k=3m$$$$k^2-2016=3^n \implies 9m^2=3^n+2016 \implies m^2=3^{n-2}+224$$For $n \geq 3$ $\implies$ $m^2 \equiv 2 \pmod 3$ , hence no solution! For $n=2$ $\implies$ $k=45$

Given \( 2016 = 2^5 \cdot 3^2 \cdot 7 \): 1. If \( 3^n \) is even, then \( K \) is odd. 2. Dividing both sides by 3: \[ \frac{3^n}{3}, \quad \frac{2016}{3} \implies \frac{K}{3} \]3. This leads to: \[ 9m^2 - 2016 = 3^n \]4. Simplifying further: \[ m^2 - 224 = 3^{n-2} \]5. If \( n-2 \) is even, then \( n-2 = 2t \). Substituting this: \[ m^2 - 3^{2t} = 2^5 \cdot 7 \]6. Factoring: \[ (m - 3^t)(m + 3^t) = 2^5 \cdot 7 \]7. Solving yields: \[ m = 15, \quad K = 45, \quad n = 2 \]

$n = 1$: $k^2 = 2019$, there is no solution. $n = 2$: $k^2 = 2025$, $(k,n) = (45,2)$. $n \geq 3$: In modulo $3$, we have $k = 3m$, $m \in Z^{+} \implies 3^2 \cdot m^2 - 2016 = 3^n \implies m^2 - 224 = 3^{n-2}$. Again, in modulo $3$; $m^2 \equiv 2$ (mod 3). There is no solution. There is only 1 solution: $(k,n) = (45,2)$.