BdMO Online Forum

$\displaystyle YA.YD=YB.YC \rightarrow \frac {YA}{YB}=\frac{YC}{YD} \rightarrow \frac {AG}{BG}=\frac{CH}{DH}$
Similarly , $\displaystyle \frac {AE}{DE}=\frac{CF}{BF}$

Using sine rule $\displaystyle \frac {YA}{YB}=\frac{XA}{XB}=\frac {sin \angle YAB}{sin \angle YBA} \rightarrow
\frac {AG}{BG}=\frac {AE}{DE} \rightarrow \frac{CH}{DH}=\frac{CF}{BF} \rightarrow GE||BD||HF $

And $\displaystyle \frac {BG}{AG}=\frac{DE}{AE}=\frac{BF}{CF}=\frac{DH}{CH} \rightarrow GF||AC||EH $

We have $(n+1)^{n}-1=2n^{k}+3n\Rightarrow n^{n}+\binom{n}{1}n^{n-1}+...+\binom{n}{n-1}n+1-1=2n^{k}+3n$
Since $n$ is a positive integer,we can divide both sides by $n$. This implies,
$n^{n-1}+\binom{n}{1}n^{n-2}+...+\binom{n}{n-1}=2n^{k-1}+3$. Now in the last equation $n$ divides the left hand side,and also it divides the first term of the right hand side. So we must have $n|3$. Thus the possible values for $n$ are $1,3$. Checking implies that $(n,k)=(3,3)$ is the only solution.