Question

Find n if \[{2^{200}} - {2^{192}} \cdot 31 + {2^n}\] is a perfect square

Answer 1

Hint: As we can see, we need to determine the value of n, if \[{2^{200}} - {2^{192}} \cdot 31 + {2^n}\] is a perfect square. A perfect square or square number is an integer that is the square of an integer, which means the product of some integer with itself.

Complete step-by-step answer:
Given data: The perfect square number is \[{2^{200}} - {2^{192}} \cdot 31 + {2^n}\] , where we need to find the value of n.
We will make the number a little simpler.
So, \[{2^{200}} - {2^{192}} \cdot 31 + {2^n} = {2^{192 + 8}} - {2^{192}} \cdot 31 + {2^{n - 192 + 192}}\]
After, taking \[{2^{192}}\] common from the number, we get
= ${2^{192}}({2^8} - 31 + {2^{n - 192}})$
 = ${2^{192}}(256 - 31 + {2^{n - 192}})$ & $[{2^8} = 256]$
 = ${2^{192}}(225 + {2^{n - 192}})$
Let us take k = n-192, which will give \[{2^{192}}(225 + {2^k})\] .
Now, We will use a trial method to find out the value of n.
Firstly, In \[{2^{192}}(225 + {2^k})\] , you can see, \[{2^{192}}\] is a perfect square because \[{2^{192}}\] can be written as \[{2^{2 \times 192}} = {({2^{96}})^2}\] .
Now, since \[{2^{200}} - {2^{192}} \cdot 31 + {2^n}\] is a perfect square, \[(225 + {2^{n - 192}})\,i.e.:225 + {2^k}\]must be a perfect square too, because we know that the product of two perfect square is always a perfect square.
Now, We will do a trial for the value of k = n-192, so that \[225 + {2^k}\] is a perfect square or not.
When k=1, we get
\[225 + {2^1} = 227\] Not a perfect square
When k =2 , we get
\[225 + {2^2} = 229\] Not a perfect square.
When k =3 , we get
\[225 + {2^3} = 225 + 8 = 233\] Not a perfect square
When k =4 , we get
\[225 + {2^4} = 225 + 16 = 241\] Not a perfect square
When k =5 , we get
\[225 + {2^5} = 225 + 32 = 257\] Not a perfect square
When k =6 , we get
\[225 + {2^6} = 225 + 64 = 289 = 17\] A perfect square.
So, in the 6th trial , when k =6, we get \[225 + {2^k}\] as a perfect square.
i.e.:k = 6
$ \Rightarrow$ n - 192 = 6
$ \Rightarrow$ n = 192 + 6
$ \Rightarrow$ n = 198
Hence, if \[{2^{200}} - {2^{192}} \cdot 31 + {2^n}\] is the value of n is 198.

Note: Students can also solve it without using the trial method, you are given option for values of n, just put the values of n in \[225 + {2^{192 + n}}\] and you will get the value of n if it is a perfect square.