Can u solve? Part-5

$2^{200}-(31)2^{192}+2^n = 2^{200}-(32-1)2^{192}+2^n$

$= 2^{200}-(2^5-1)2^{192}+2^n = 2^{200}- 2^{197}+ 2^{192} +2^n$

$= 2^{192}(2^8- 2^5 +1) +2^n = 2^{192}(2^8- 2\dot{}2^4 +1) +2^n$

$= 2^{192}(2^4-1)^2 +2^n = 2^{192}(15^2) +2^n = 2^{192}(15^2 +2^{n-192})$

We note that $2^{200}-(31)2^{192}+2^n$ is perfect square, when $15$ and $\sqrt{2^{n-192}}$ are the smaller numbers of a set of Pythagoras numbers and the set is $\{8,15,17\}$ :

$\Rightarrow 2^{n - 192} = 8^2 = 2^6 \quad \Rightarrow n - 192 = 6 \quad \Rightarrow n = \boxed {198}$ .

The question can be re-writen as $(2^{192})(2^{8} - 31 + 2^{n-192})$ So $256-31+2^{n-192}$ should be a perfect square Which implies $225 + 2^{n-192}$ should be a perfect square. (225 + 64 = 289) Therefore n = 192+6 = 198

$2^{200}-2^{192}*31+2^n = 2^{192}(2^8-31)+2^n$

= $2^{192}*(256-31)+2^n$

= $2^{192}*225+2^n$

Therefore for some $m\epsilon N$

$2^n=m^2-2^{192}*225$

= $m^2-(2^{96}*15)^2$

= $(m-2^{96}*15)(m+2^{96}*15)$

So $m-2^{96}*15=2^a$ and $m+2^{96}*15=2^{a+b}$ for some non-negative integers a,b.

Hence $2^{97}*15=2^{a+b}-2^a=2^a(2^b-1)$

$\Rightarrow 2^a=2^{97}$ and $2^b-1=15 i.e., a=97 and b=4$

$\therefore n=2a+b=198$

This is a question from IIT foundation class 9 book

Supomos $n = 192$ , segue que:

$2^{200} - 2^{192} \cdot 31 + 2^n = \\\\ 2^{192}(2^8 - 31 + 2^0) \\\\ 2^{192} \cdot 226$

Como podemos notar, 226 não é um quadrado perfeito.

Logo, devemos encontrar um número, cuja potência é da base dois, que somado a 225 seja um quadrado perfeito!

O número em questão é o $2^6$ .

Daí,

$2^n = 2^{192} \cdot 2^6 \\\\ 2^n = 2^{198} \\\\ \boxed{n = 198}$