Like the title said:

Is it true that if \(\lceil x\lfloor x\rfloor\rceil+\lfloor x\lceil x\rceil\rfloor=A\) has real solutions of \(x\) with positive range of \(\alpha\le x\le\beta\), then \(A=\alpha\left\lceil\ \sqrt{\frac A2}\ \right\rceil+\beta\left\lfloor\ \sqrt{\frac A2}\ \right\rfloor\) must be true?

Inspired by Raghav Vaidyanathan which was inspired by Pi Han Goh.

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## Comments

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TopNewestHmm , nice inspirations !

@Raghav Vaidyanathan , @Pi Han Goh

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C++ code... Tested for a lot of numbers.. seems that it isn't always true... especially in the vicinity of numbers of form \(2n^2\)

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So you've found a counterexample? Can you show me which values of \(A\) shows that is not true?

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I think \(A=129\) doesn't satisfy. And also, numbers like \(A=128\) have \(\alpha=\beta\).

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So my statement still holds.

If this is a cop out explanation, then you've successfully disproven my statement! (and I can't have that, haha!)

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Okay this is disproven! Thank you for your cooperation. I knew this is too good to be true.

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That means to say that we can't take \(A=137\) as an example because the range of solution of positive \(x\) is an open interval: \( \left(8, \frac{73}9\right)\).

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@Pi Han Goh Sir , does the equality hold on the upper limit side?

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I don't understand your question at all. The claim clearly states that we have to first find \(\alpha\) and \(\beta\).

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