Prove (or disprove): If xx+xx=A\lceil x\lfloor x\rfloor\rceil+\lfloor x\lceil x\rceil\rfloor=A has real solutions of xx with positive range of αxβ\alpha\le x\le\beta, then A=α A/2 +β A/2 A=\alpha\lceil\ \sqrt{A/2}\ \rceil+\beta\lfloor\ \sqrt{A/2}\ \rfloor.

Like the title said:

Is it true that if xx+xx=A\lceil x\lfloor x\rfloor\rceil+\lfloor x\lceil x\rceil\rfloor=A has real solutions of xx with positive range of αxβ\alpha\le x\le\beta, then A=α A2 +β A2 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.

Note by Pi Han Goh
4 years, 3 months ago

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Hmm , nice inspirations !

@Raghav Vaidyanathan , @Pi Han Goh

A Brilliant Member - 4 years, 3 months ago

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#include<iostream.h>
#include<conio.h>
#include<stdlib.h>
#include<math.h>
#include<iomanip.h>

double fccf(double x)
    {
    double d=ceil(x*floor(x))+floor(x*ceil(x));
    return d;
    }

int main()
    {
    clrscr();
    double x,l[2],g[2]={0,0},n;
    int r=0;
    cin>>n;
    l[0]=floor(sqrt(n/2));l[1]=ceil(sqrt(n/2));
    for(x=l[0];x<=l[1];x+=0.0001)
        {
        double k=fccf(x);
        if(k==n&&r==0)
            {
            g[0]=x;
            r=1;
            }
        else if(k>n&&r==1)
            {
            g[1]=x;
            break;
            }
        }
    cout<<g[0]<<" "<<g[1]<<endl<<setprecision(5)<<(g[0]*l[1]+l[0]*g[1]);
    getch();
    return 0;
    }

C++ code... Tested for a lot of numbers.. seems that it isn't always true... especially in the vicinity of numbers of form 2n22n^2

Raghav Vaidyanathan - 4 years, 3 months ago

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So you've found a counterexample? Can you show me which values of AA shows that is not true?

Pi Han Goh - 4 years, 3 months ago

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

Raghav Vaidyanathan - 4 years, 3 months ago

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@Raghav Vaidyanathan There's no solution for xx when A=129A=129 so it doesn't satisfy the condition. Well technically, if AA is in the form of 2B22B^2 for some integer BB, then α=β\alpha = \beta , so by squeeze theorem, x=α=βx = \alpha = \beta which doesn't contradict the statement. In other words, it's trivial to pointless to disprove this statement when xx is an integer.

Pi Han Goh - 4 years, 3 months ago

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@Pi Han Goh I agree with you on numbers of the form 2n22n^2 and also numbers which don't yield a solution. Now try A=137A=137...

Raghav Vaidyanathan - 4 years, 3 months ago

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@Raghav Vaidyanathan Because 8<x<7398 < x < \frac{73}9 is the range for xx when x=137x=137, by applying the inequality, we can have α=8+ϵ,β=739ϵ \alpha =8 + \epsilon, \beta = \frac{73}{9} - \epsilon for an extremely small positive value ϵ\epsilon. In other words α8,β739\alpha \gtrsim 8, \beta \lesssim \frac{73}9.

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 - 4 years, 3 months ago

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@Pi Han Goh The values of α,β\alpha, \beta are one thing, but you may notice that: 9α+8β1379\alpha+8\beta \ne 137. Counterexample.

Raghav Vaidyanathan - 4 years, 3 months ago

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@Raghav Vaidyanathan Oh I failed my math. D=

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

Pi Han Goh - 4 years, 3 months ago

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@Raghav Vaidyanathan Follow up (or modified) question: Is it true that if xx+xx=A\lceil x\lfloor x\rfloor\rceil+\lfloor x\lceil x\rceil\rfloor=A has real solutions of xx with positive range in a closed interval [α,β] [\alpha, \beta ] for αβ\alpha \ne \beta, then A=α A2 +β A2 A=\alpha\left\lceil\ \sqrt{\frac A2}\ \right\rceil+\beta\left\lfloor\ \sqrt{\frac A2}\ \right\rfloor must be true?

That means to say that we can't take A=137A=137 as an example because the range of solution of positive xx is an open interval: (8,739) \left(8, \frac{73}9\right).

Pi Han Goh - 4 years, 3 months ago

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

Ankit Kumar Jain - 1 year ago

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

Pi Han Goh - 1 year ago

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