easy olympiad problem

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kjhgyuio

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kjhgyuio

#1 h Apr 17, 2025, 2:00 PM

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Find all positive integer values of $x$ such that
$\sqrt{x - 2011} + \sqrt{2011 - x} + 10$ is an integer.

This post has been edited 2 times. Last edited by kjhgyuio, Apr 17, 2025, 2:01 PM
Reason: nil

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Mathdreams

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Mathdreams

#2 h Apr 17, 2025, 2:14 PM

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Solution

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Charizard_637

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Charizard_637

#3 h Apr 17, 2025, 7:44 PM

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$x-2011$ and $2011-x$ are each other's negatives, and you can't take the square root of a negative number while having an integer solution. Therefore the only solution that would work is if they were both non-negative, and based on this they must be zero because one positive number will lead to one negative number. Since 0 is neither negative or positive it's what's under both square roots. Therefore, $\sqrt{x-2011} = 0$ . Squaring both sides gives $x-2011 = 0$ , hence $x = 2011$ . This is the only solution.

Edit: I know it's verbose but it's an "olympiad problem"

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vincentwant

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vincentwant

#4 h Apr 18, 2025, 1:25 AM

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Charizard_637 wrote:

$x-2011$ and $2011-x$ are each other's negatives, and you can't take the square root of a negative number while having an integer solution. Therefore the only solution that would work is if they were both non-negative, and based on this they must be zero because one positive number will lead to one negative number. Since 0 is neither negative or positive it's what's under both square roots. Therefore, $\sqrt{x-2011} = 0$ . Squaring both sides gives $x-2011 = 0$ , hence $x = 2011$ . This is the only solution.

Edit: I know it's verbose but it's an olympiad problem

you dont have to do this, this is enough

Notice that for the expression to be real, $x-2011\geq0$ and $2011-x\geq 0$ , otherwise the imaginary part of the expression would be positive. Thus no solutions other than $x=2011$ exist, and inspection gives that $x=2011$ works. Thus the answer is $x=2011$ .

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