Find all $(k,n)$, for which $(2^k-1)^n + (2^k+1)^n$ is a perfect square.
I did something like this. $(2^k-1)^n + (2^k+1)^n = d^2$
Since both are odd, $d$ must be even. I considered this as $\mod 3$, $\mod 4$, since $x^2$ congruent to $0$ or $1$ $\mod(3,4)$. But this too did not help me. I now have no idea how to proceed further.
I could not solve this problem but i tried very much.
Check problem 25. You may also note down the other problems as well.
My solution is not complete, but I hope someone can fill in the missing things.
My result: $n$ cannot be even, and also it should be an odd number with $n\equiv 1\pmod 8$.
Let's proceed: Suppose $$d^2=(2^k-1)^n+(2^k+1)^n$$ If $n\geq 2$ is even, then using binomial expansion we have $$d^2=2[(2^k)^n+\binom{n}{2}(2^k)^{n-2}+\cdots+1]$$ Clearly $d^2$ is divisible by $2$ but not $4$, hence every even $n$ doesn't work for sure.
If $n\geq 1$ is odd, we first take $n=1$, and we see $$d^2=2^{k+1}$$ Hence we have $(n,k)=(1,1),(1,3),(1,5), \cdots$.
If $n\geq 3$, then \begin{align*}d^2&=2[(2^k)^n+\binom n2(2^k)^{n-2}+\cdots+\binom{n}{n-1}2^k]\\ &=2^{k+1}\bigg[(2^k)^{n-1}+\binom{n}{2}(2^k)^{n-3}+\cdots+\binom{n}{n-3}(2^k)^2+n\bigg]\end{align*} We see that since $n$ is odd, the big square bracket is odd, hence $2^{k+1}$ must itself be a perfect square, hence $k$ is odd. Then the big square bracket should itself be an odd square, and we see any odd square should have the property $1\pmod 8$ since for any positive integer $m$ we have $$(2m+1)^2\equiv 4m^2+4m+1\equiv 4m(m+1)+1\equiv 1\pmod 8.$$ Hence when $n$ is an odd number with $n\equiv 1\pmod 8$, $k$ is an odd number, but I cannot found any example when $n> 1$, there should be no solution at this stage but I cannot prove it.