<s>[INST] Solve the equations $(n+1)^n=2n^k+3n+1$ Solve the equations in positive integers
$$(n+1)^n=2n^k+3n+1$$
My attempts:
$$(n+1)^n=2n^k+3n+1$$
$$(n+1)^n-1=n(2n^{k-1}+3)$$
$$(n+1)^{n-1}+(n+1)^{n-2}+(n+1)^{n-3}+...+n+1=2n^{k-1}+3$$
$$(n+1)^{n-1}+(n+1)^{n-2}+(n+1)^{n-3}+...+n=2(n^{k-1}+1)$$
 [/INST] [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 $\gt 4$]
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