Cubes over, now are squares!

You know that $a^{2} + b^{2} + 2ab = (a+b)^{2}$

Now this is in the form of $a^{2} + b^{2} + 2ab$

Noe it is clear that $2^{13}$ can't be $a^{2}$ or $b^{2}$ , because in $a^{2}$ or $b^{2}$ the power is even but in the other it is odd. So it is clear the $2^{13} \neq a^{2} \neq b^{2}$

Now let us take $a^{2} = 2^{10}$ $a= 2^{5}$

Therefore , $2^{13} = 2ab$

or, $2^{13} = 2\times 2^{5}\times b$

or, $2^{13} = 2^{6}\times b$ or $2^ {7} = b$

Now, $b= 2^{7}$

But here as I have mentioned in the first that the given expression is expressed in the expanded form of $(a+b)^{2}$

Therefore, $2^{x} = b^{2}$

or, $2^{x} = 2^{7\times 2}$

or, $2^{x}= 2^{14}$

or, $x=14$

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2^x + 2^13 + 2^10

= 2^10(2^x-10 + 2^3 + 1)

= 2^10(2^x-10 + 9)

(2^x-10 + 9) must be a square number out of this 9 is square number

to get next square number 27 add 16 to 9

so 2^(x-10) = 16 = 2^4

then x – 10 = 4 so x = 14

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Actually. I used a little bit of "trial and error" thingy...... Let us assume x=10 (Case i) X<10 (Case ii) or, x>10 (Case iii). Clearly, x ≠ 10 (Substitute x = 10 and take 2^10 common, we don’t get factors for a perfect square… Case ii: x<10 Given question simplified: 2^10 [1 + 8 + 2^(10-x)] If x<10 and even, Given question would be, 2^x (2^(13-x) + 2^(10-x ) But 13 –x and 10 –x can only result in a combination of odd and even numbers.. So, we can’t have a perfect combination of even numbers. SO, x can’t be even Similarly, if x is odd, 13 –x and 10 –x can only result in a combination of even and odd numbers. So, x can’t be odd either. Therefore, x can be only greater than 10. (Case (iii) is only right) Given simplified question: 2^10 [(9 + 2^(x-10 )] 2^10 is already a perfect square. So, 9 + 2^(x-10) is a perfect square. So, That gives x = 14. (9 +16 = 25, 25 is a perfect square) X =14 is the ONLY POSSIBLE ANSWER