2021/023) Find n when $2^n + 105$ is a perfect square

 Let $2^n + 105 = p ^2$

Let us work mod 3. for a number to be perfect square it as to  1 or zero mod 3. 

So $2^n + 105 \equiv 0 \pmod 3$ or $2^n + 105 \equiv 0 \pmod 3$

or $2^n \equiv 0 \pmod 3$ or $2^n \equiv 1 \pmod 3$

but $2^n \equiv 0 \pmod 3$ is not possible 

$2^n \equiv 1 \pmod 3$

Now $2^x \equiv 1 \pmod 3$ if x is even and -1 if x is odd

so n is even say 2m

so $2^{2m} + 105 = p^2$

or $p^2- 2^{2m} = 105$

or $(p+2^m)(p - 2^m ) = 105 = 105 * 1 = 35 * 3 = 15 * 7 = 21 * 5$

Now $(p + 2^m) - (p- 2^m) = 2^{m+1}$ a power of 2

from the above the combination we get

$p+2^m = 35$ $p-2^m= 3$ giving p =19 and m = 4 or n = 8 

Or 

$p+2^m = 15$ $p-2^m= 7$ giving p =11 and m = 2 or n = 4

Or $p+2^m = 21$ $p-2^m= 5$ giving p = 13 and m = 3 or n = 6

so n = 4 or 6 or 8