Giả sử: $a+71=m^2;4a-31=n^2$ với $m,n\in\mathbb{N}$.
Ta có: $4m^2-n^2=315$
$\Leftrightarrow (2m-n)(2m+n)=315$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}2m-n=1\\2m+n=315\end{array}\right.\\\left\{\begin{array}{l}2m-n=3\\2m+n=105\end{array}\right.\\\left\{\begin{array}{l}2m-n=5\\2m+n=63\end{array}\right.\\\left\{\begin{array}{l}2m-n=7\\2m+n=45\end{array}\right.\\\left\{\begin{array}{l}2m-n=9\\2m+n=35\end{array}\right.\\\left\{\begin{array}{l}2m-n=15\\2m+n=21\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}m=79\\n=157\end{array}\right.\Leftrightarrow a=6170\\\left\{\begin{array}{l}m=27\\n=51\end{array}\right.\Leftrightarrow a=658\\\left\{\begin{array}{l}m=17\\n=29\end{array}\right.\Leftrightarrow a=218\\\left\{\begin{array}{l}m=13\\n=19\end{array}\right.\Leftrightarrow a=98\\\left\{\begin{array}{l}m=11\\n=13\end{array}\right.\Leftrightarrow a=50\\\left\{\begin{array}{l}m=9\\n=3\end{array}\right.\Leftrightarrow a=10\\\end{array}\right.$
Vậy số tự nhiên $a$ lớn nhất thỏa mãn là $6170$.