Mathematical Enigma - XV

This solution was written by @Chan Lye Lee . I'm posting it just for the sake of my fellow Brilliantians who were unable to do it.

---

Let $m^2=2^{200}-2^{192}× 31 +2^n=2^{192}(225+2^{n-192})$

Then $225+2^{n-192}=a^2$ for some positive integer $a$ .

Let $k_1+k_2=k=n-192$ .

Now $2^k=a^2-225=(a+15)(a-15)$ .

We can express it in $\cases{2^{k_1}=a+15 \\2^{k_2}=a-15 }$ , where $k_1>k_2$ .

The difference of the two equations yields $30=2^{k_1}-2^{k_2}$ , which implies that $2\times 15 = 2^{k_2}\left(2^{k_1-k_2}-1\right)$ .

Clearly, $\cases{k_2=1\\2^{k_1-k_2}-1=15}$ and hence $\cases{k_2=1\\ k_1=5}$ .

Finally, $k=6=n-192$ and thus $n=198$ .