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.

## Comments

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TopNewestHmm , nice inspirations !

@Raghav Vaidyanathan , @Pi Han Goh – Azhaghu Roopesh M · 2 years ago

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– Pi Han Goh · 2 years ago

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

– Raghav Vaidyanathan · 2 years ago

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

– Pi Han Goh · 2 years ago

There's no solution for \(x\) when \(A=129\) so it doesn't satisfy the condition. Well technically, if \(A \) is in the form of \(2B^2\) for some integer \(B\), then \(\alpha = \beta \), so by squeeze theorem, \(x = \alpha = \beta \) which doesn't contradict the statement. In other words, it's trivial to pointless to disprove this statement when \(x\) is an integer.Log in to reply

– Raghav Vaidyanathan · 2 years ago

I agree with you on numbers of the form \(2n^2\) and also numbers which don't yield a solution. Now try \(A=137\)...Log in to reply

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!) – Pi Han Goh · 2 years ago

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– Raghav Vaidyanathan · 2 years ago

The values of \(\alpha, \beta\) are one thing, but you may notice that: \(9\alpha+8\beta \ne 137\). Counterexample.Log in to reply

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: \( \left(8, \frac{73}9\right)\). – Pi Han Goh · 2 years ago

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Okay this is disproven! Thank you for your cooperation. I knew this is too good to be true. – Pi Han Goh · 2 years ago

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