I have a new proof method that im not sure if it works and proves the result, its like proof by contradiction, but rather then obtaining a contradiction and disproving the result, your asumption would satisfy the equation and hence prove the assumption e.g
If you know for a fact and have proven that
$$\begin{align}P(k)&=g(1)(k)p(k+1)-g(2)(k)p(k+2)+…+g(n-k-2)(k)p(k+(n-k-2))\\&-g(n-k-1)(k)p(k+(n-k-1))+g(n-k)(k)\end{align}$$
Where $g(n-k-m)(k)$ and $p(k+m)$ are some functions, if we assume that $p(k+m)=g(n-k-m)(k)$ then…
$g(1)p(k+1)=g(n-k-1)p(k+(n-k-1))$
$g(2)p(k+2)=g(n-k-2)p(k+(n-k-2))$
Etc..
Then in our equation everything cancels out and we are left with $p(k)=g(n-k)(k)$ which satisfies our assumption for $m=0.$
But we know our originial eqaution is true so does that mean the assumption is true because it satisfies thr equation that is true, much like how many if $x+2=8$ and we assume $x=6$ and it satisfies the equation then it prove $x=6$ is a solution?
If you make the substitution and it works, that means it's one solution to the equation you found. But substituting $x=6$ into $x+2=8$ is not the only kind of example.
Imagine you substituted $x=6$ into $x^2-5x=6$. You would get a true statement, but you would be missing another solution: $x=-1$.
Or imagine you had a functional equation: $f(x+y) = f(x) f(y)$. You could substitute $f(x) = 0$ and get a true statement. However, $f(x) = 2^x$ is another solution, and there's many others.
Depending on your situation, getting a single solution might be enough. But you have not proved that your guess is the only possible thing that could be true.