Solve the equations $(n+1)^n=2n^k+3n+1$

Question

Solve the equations in positive integers $$(n+1)^n=2n^k+3n+1$$

My attempts:

$$(n+1)^n=2n^k+3n+1$$ $$(n+1)^n-1=n(2n^{k-1}+3)$$ $$(n+1)^{n-1}+(n+1)^{n-2}+(n+1)^{n-3}+...+n+1=2n^{k-1}+3$$ $$(n+1)^{n-1}+(n+1)^{n-2}+(n+1)^{n-3}+...+n=2(n^{k-1}+1)$$

Answer 1

$$(n+1)^n=^nC_0+^nC_1n+...+^nC_nn^n$$

Comparing,

$$^nC_1n=3n$$ $$\implies n=3$$ since $n>0$

So, substituting $n$

$$27=3^k$$ $$\implies k=3$$

Answer 2

[Note:I'm assuming for large $n$ that $k \le n$.]

If $n=1$ we get $2=6$ which is false. If $n=2$ we have $9=2\cdot 2^k+7$, which is true for $k=0$. Otherwise there are at least three terms in the binomial expression for $(n+1)^n\ge n^n+n^n+\frac {n-1}2n^{n-1}$

So we need $$\frac {n-1}2n^{n-1}\le 3n+1$$ or $$\frac n4\cdot n^{n-1}\lt 4n$$ (using crude estimates on both sides $\frac n4\lt \frac {n-1}2$ and $3n+1\lt 4n$) or $$n^{n-1}\lt 16$$

The left-hand side is increasing with $n$ and $4^3=64\gt 16$ so that $n\le 3$

If $n=3$ we have $64=2\cdot 3^k+10$, which gives $k=3$ .

[For $k\le n$, suppose for contradiction $k\ge n+1$ then divide through by $n^n$ to get $(1+\frac 1n)^n=2n{k-n}+\frac 1{n^n}$ The left-hand side is $\lt e$ and for $n\ge 2$ the right-hand side is $\gt 4$]