Gọi số đó là a (a > 12)
Ta có: \(\left\{{}\begin{matrix}a+12=x^2\\a-12=y^2\end{matrix}\right.\) \(\Rightarrow x^2-y^2=24\) \(\Leftrightarrow\left(x-y\right)\left(x+y\right)=24\)
\(\Rightarrow x-y\in\left\{\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24\right\}\)
\(\Rightarrow x+y\in\left\{\pm24,\pm12,\pm8,\pm6,\pm4,\pm3,\pm2,\pm1\right\}\)
Giải ra tìm x,y \(\Rightarrow\) Được a = 13, 37