Prove (or disprove): 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 $A=\alpha\lceil\ \sqrt{A/2}\ \rceil+\beta\lfloor\ \sqrt{A/2}\ \rfloor$.

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.

#Algebra

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Comments

Hmm , nice inspirations !

@Raghav Vaidyanathan , @Pi Han Goh

A Former Brilliant Member - 6 years, 1 month ago

#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 $2n^2$

Raghav Vaidyanathan - 6 years, 1 month ago

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