Question

# Find the natural number n, such that ${2^8} + {2^{10}} + {2^n}$ is a perfect square.

Hint: In this question first try to convert ${2^8} + {2^{10}} + {2^n}$ in ${\left( {a + b} \right)^2}$. As we know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ as in the equation their is also three term hence by comparing and some analysis we find that $a = {2^4}, b = {2^5}$ or ${({2^4} + {2^5})^2}$ = ${2^8} + {2^{10}} + {2^n}$ by comparing we will find n.

As in the given question ${2^8} + {2^{10}} + {2^n}$ we have to find the value of n for which it is a perfect square.
For this we have to try to make it like ${a^2} + {b^2} + 2ab$
Hence, ${\left( {{2^4}} \right)^2} + {\left( {{2^5}} \right)^2} + {2^n}$
As for we have change it as ${2^n} = 2ab$
So that is equal to
${\left( {{2^4}} \right)^2} + {\left( {{2^5}} \right)^2} + {2.2^{n - 1}}$
So form this we know that as $a = {2^4},b = {2^5}$ ,
hence
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
${({2^4} + {2^5})^2}$ = ${2^8} + {2^{10}} + {2.2^{4 + 5}}$
Hence by comparing with real equation
${2^8} + {2^{10}} + {2.2^{4 + 5}}$ = ${\left( {{2^4}} \right)^2} + {\left( {{2^5}} \right)^2} + {2^n}$
From this ${2^8},{2^{10}}$ is cancel out from both side the remaining term is
${2^{10}} = {2^n}$
By comparing the power we get $n = 10$
Hence $n = 10$ will be the answer for making this as perfect square

Note: Sometimes the question will frame like that ${3^8} + {3^{10}} + {2.3^n}$ this question is same as the above question solve it , as in the above the question will solve.
Whenever we have to give question like this just try to make it as ${\left( {a + b} \right)^2}$ as in the question that the power of $2$ is $8, 10$ it is multiply of $2$ or written as ${({2^4})^2}, {({2^5})^2}$ and n is unknown.