Question 1208295
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I'll assume x and y are nonnegative integers.
There might be a more clever and elegant solution, but I'm going to check possible values of x.


If x = 0, then,
2^13 + 2^10 + 2^x 
= 2^13 + 2^10 + 2^0
= 9217
And,
sqrt(9217) = 96.0052 approximately
Since the square root result isn't a whole number, this shows 9217 isn't a perfect square.
y^2 = 9217 doesn't have an integer solution.
x = 0 won't pair with an integer y value.
We can rule out x = 0.


Let's try x = 1.
2^13 + 2^10 + 2^x 
= 2^13 + 2^10 + 2^1
= 9218
And,
sqrt(9218) = 96.0104 approximately
Same idea as the previous paragraph. 
We can rule out x = 1.


Let's try x = 2.
2^13 + 2^10 + 2^x 
= 2^13 + 2^10 + 2^2
= 9220
And,
sqrt(9220) = 96.0208 approximately
We can rule out x = 2 for similar reasoning as the previous paragraphs.


And so on.
Keep this process going until reaching x = 14.
Use of a spreadsheet is strongly recommended to make this process very quick.
Alternatively you can use a programming language such as Python to write up a quick script.
2^13 + 2^10 + 2^x 
= 2^13 + 2^10 + 2^14
= 25600
And,
sqrt(25600) = 160 exactly
We finally get an integer result. 
This proves that <font color=red>x = 14 and y = 160</font> is one ordered pair solution.
2^13 + 2^10 + 2^<font color=red>14</font> = <font color=red>160</font>^2 


There might be other solutions.
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