Để \(A\) là số chính phương \(\Rightarrow26n+17=t^2\left(t\in N\right)\)

\(\Rightarrow26n+13=t^2-4\)

\(\Rightarrow13\left(2n+1\right)=\left(t-2\right)\left(t+2\right)\left(1\right)\)

\(\Rightarrow\left(t-2\right)\left(t+2\right)⋮13\)\(\Rightarrow\left[{}\begin{matrix}t-2⋮13\\t+2⋮13\end{matrix}\right.\)

*)Xét \(t+2⋮13\Rightarrow t+2=13m\left(m\in N\right)\)\(\Rightarrow t=13m-2\)

Thay vào \(\left(1\right)\)\(\Rightarrow13\left(2n+1\right)=13m\left(13m-4\right)\)

\(\Rightarrow2n+1=m\left(13m-4\right)\Rightarrow n=\dfrac{13m^2-4m-1}{2}\)

*)Xét \(t-2⋮13\Rightarrow t-2=13m\left(m\in N\right)\)\(\Rightarrow t=13m+2\)

Thay vào \(\left(1\right)\)\(\Rightarrow13\left(2n+1\right)=13m\left(13m+4\right)\)

\(\Rightarrow2n+1=m\left(13m+4\right)\)\(\Rightarrow n=\dfrac{13m^2+4m-1}{2}\)

Vậy-.