For \(n>3\), if \(n\) is odd, then this means \(a^n+b^n\) is a perfect \(n^{th}\) power. By Fermat's Last Theorem,
no positive integer solutions exist.

If \(n\) is even, suppose that \(n=2k\). Then, let \(x=a^2\) and \(y=b^2\). The equation becomes:
\((x+y-d)^k=x^k+y^k\). By Fermat's Last Theorem, no positive integer solutions exist as well.

We now need to check \(k=1,2\). For the first, note that \(a^2+b^2<(a+b)^2\), so no solutions.
For the second,
\(a^2+b^2-d=a^2+b^2\) implies \(d=0\)
So any \(a,b\) work with \(d=0\). However, if the range of \(d\) is exclusive, there are no solutions.

The solution is Provided by our teaching assistant Tarit Goswami.