Solve in positive integers $\sqrt{n+1982^n}+\sqrt n=(\sqrt{1983}+1)^k$

Question

Solve in positive integers $$\sqrt{n+1982^n}+\sqrt n=(\sqrt{1983}+1)^k.$$

If $n=1, k=1$ then $$\sqrt{1983}+1=\sqrt{1983}+1.$$ Let $n>1, k>1$ $$\sqrt{n+1982^n}+\sqrt n=(\sqrt{1983}+1)^k$$ $$1982^n=(\sqrt{1983}+1)^k(\sqrt{n+1982^n}-\sqrt n).$$

Answer 1

Let $$\mathcal K = \{a+b\sqrt{1983},\quad a,b\in\mathcal Z,\quad b\not=0\},$$ then $$RHS, LHS, LHS^2 \in\mathcal K,$$ $$2\sqrt{n(n+1982^n)}\in\mathcal K,$$ $$n(n+1982^n) = 1983 m^2.\qquad(1)$$

Let us consider the next cases.

If $n=1$ then $(1)$ and OP are satisfied for $k=1.$
If $n=2$ then $(1)$ can not be satisfied for any $m.$
If $n=2i+1\ge3,$ then one can get contradict after $$n\equiv1\pmod2,\quad m\equiv1\pmod2,$$ $$\quad n^2\equiv3m^2\pmod4.$$ If $$n=2i\ge4,\quad\max_d\ 2^d\, |\, n = p,$$ then $$\max_d\ 2^d\, |\, m = p,\quad \frac n {2^p}\equiv\frac n {2^p}\equiv1\pmod2,\quad n\ge p+2,$$ and one get contradict after $$\frac n {2^p}\left(\frac n {2^p}+1982^{n-p}\right) = 1983\frac m {2^p},$$ $$\quad \left(\frac n {2^p}\right)^2\equiv3\left(\frac m {2^p}\right)^2\pmod4.$$ So OP equality has the only solution $$n=1,\quad k=1.$$

Answer 2

(Note: This is only a draft; Too long for a comment)

Notice $$1982^n=(\sqrt{1983}+1)^k(\sqrt{n+1982^n}-\sqrt n)\tag1$$ $$1982^k=(\sqrt{1983}-1)^k(\sqrt{n+1982^n}+\sqrt n)\tag2$$

Let $\alpha=\sqrt{1983}+1,\beta=\sqrt{1983}-1$. Then $(1),(2)$ becomes $$\alpha^{n-k}\beta^n=\sqrt{n+(\alpha\beta)^n}-\sqrt n\tag3$$ $$\alpha^k=\sqrt{n+(\alpha\beta)^n}+\sqrt n\tag4$$

$(4)-(3)$: $$\alpha^{n-k}(\alpha^{2k-n}+\beta^k)=2\sqrt n=(\alpha-\beta)\sqrt n\tag5$$