Gọi số chính phương cần tìm là n^2, thương là p\,(p\in\mathscr{P}).
Khi đó ta có:
n^2=39p+1
\Leftrightarrow n^2-1=39p
\Leftrightarrow (n-1)(n+1)=39p
Vì 39p có các ước dương là \{1;3;13;39;p;3p;13p;39p\}, ta xét các trường hợp sau:
*) \left\{\begin{array}{l}n-1=1\\n+1=39p\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}n=2\\p=\dfrac{1}{13}\end{array}\right. (loại)
*) \left\{\begin{array}{l}n-1=3\\n+1=13p\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}n=4\\p=\dfrac{5}{13}\end{array}\right. (loại)
*) \left\{\begin{array}{l}n-1=13\\n+1=3p\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}n=14\\p=5\end{array}\right. (thoả mãn)
*) \left\{\begin{array}{l}n-1=39\\n+1=p\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}n=40\\p=41\end{array}\right. (thoả mãn)
*) \left\{\begin{array}{l}n-1=p\\n+1=39\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}n=38\\p=37\end{array}\right. (thoả mãn)
*) \left\{\begin{array}{l}n-1=3p\\n+1=13\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}n=12\\p=\dfrac{11}{3}\end{array}\right. (loại)
*) \left\{\begin{array}{l}n-1=13p\\n+1=3\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}n=2\\p=\dfrac{1}{13}\end{array}\right. (loại)
*) \left\{\begin{array}{l}n-1=39p\\n+1=1\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}n=0\\p=\dfrac{-1}{39}\end{array}\right. (loại)
Vậy các số chính phương thoả mãn là: 196;1444;1600.