MHB Find $x$ in $\lfloor\frac{x-1}{2}\rfloor+\lfloor\frac{x^2-x}{3}\rfloor=x$

[anemone]

Find the values of integer $x$ such that $\left\lfloor{\dfrac{x-1}{2}}\right\rfloor+\left\lfloor{\dfrac{x^2-x}{3}}\right\rfloor=x$.

 

[kaliprasad]

Find the values of integer $x$ such that $\left\lfloor{\dfrac{x-1}{2}}\right\rfloor+\left\lfloor{\dfrac{x^2-x}{3}}\right\rfloor=x$.

Spoiler

we take 2 separate cases

case 1) x is odd say 2m+1



then we get

$m+\lfloor\dfrac{2m(2m+1)}{3}\rfloor= 2m+ 1$

or $\lfloor\dfrac{2m(2m+1)}{3}\rfloor= m+ 1$

or $m+1\le \dfrac{2m(2m+1)}{3}\lt m+ 2$

or $3m+3\le 2m(2m+1) \lt m+ 6$

or $ 3 \le 4m^2 -m \lt 6$

m = 1 or = -1 giving x = 3 or -1case 2) x is even say 2mthen we get

$m-1 +\lfloor\dfrac{2m(2m-1)}{3}\rfloor= 2m $

or $\lfloor\dfrac{2m(2m-1)}{3}\rfloor= m+1 $

or $m+1\le \dfrac{2m(2m-1)}{3}\lt m+2 $

or $3\le 4m^2-5m \lt 6$
giving no solution

so solution x = -1 or 3

 

[anemone]

Spoiler

we take 2 separate cases

case 1) x is odd say 2m+1



then we get

$m+\lfloor\dfrac{2m(2m+1)}{3}\rfloor= 2m+ 1$

or $\lfloor\dfrac{2m(2m+1)}{3}\rfloor= m+ 1$

or $m+1\le \dfrac{2m(2m+1)}{3}\lt m+ 2$

or $3m+3\le 2m(2m+1) \lt m+ 6$

or $ 3 \le 4m^2 -m \lt 6$

m = 1 or = -1 giving x = 3 or -1case 2) x is even say 2mthen we get

$m-1 +\lfloor\dfrac{2m(2m-1)}{3}\rfloor= 2m $

or $\lfloor\dfrac{2m(2m-1)}{3}\rfloor= m+1 $

or $m+1\le \dfrac{2m(2m-1)}{3}\lt m+2 $

or $3\le 4m^2-5m \lt 6$
giving m = 2 or x = 4so solution x = -1 or 3 or 4

Hi kaliprasad,

Spoiler

Thanks for participating but one of the solutions doesn't satisfy the original equation...(Mmm)

 

[kaliprasad]

Hi kaliprasad,

Spoiler

Thanks for participating but one of the solutions doesn't satisfy the original equation...(Mmm)

my mistake

Spoiler

m =2 and x = 4 is the wrong solution

I shall correct the solution

 

[anemone]

Okay, and thanks, kaliprasad, for participating!

My solution:

Spoiler

The given equality can be rewritten as $\left\lfloor{\dfrac{x-1}{2}}\right\rfloor+\left\lfloor{\dfrac{x(x-1)}{3}}\right\rfloor=x$.

For $x≥1$, we know all the three terms $\left\lfloor{\dfrac{x-1}{2}}\right\rfloor,\,\left\lfloor{\dfrac{x(x-1)}{3}}\right\rfloor$ and $x$ are positive terms therefore $x≥\dfrac{(x² − x)}{3}$ must be correct, solving it for the range of $x$ we get $1≤x≤4$, and $x=3$ is the only possible answer.

A check shows that $x=0$ is clearly out of the question to be a solution to the problem.

Whereas for $x<0$, all the three terms are negative and this leads to $-\left(\dfrac{x-1}{2}\right)≥-x$ must hold, solving it for the range of $x$ we get $x≥-1$, therefore $x=-1$ is the only answer. Thus, the integers $x$ that satisfy the given equality are $-1$ and $3$.

 

[kaliprasad]

Okay, and thanks, kaliprasad, for participating!

My solution:

Spoiler

Whereas for $x≤0$, all the three terms are negative.

Spoiler

for x = 0 the 2nd term is zero

 

[anemone]

Spoiler

for x = 0 the 2nd term is zero

Ops, you're absolutely right...

 

[johng1]

Here's another solution.

Spoiler

 

[anemone]

Here's another solution.

Thanks johng for participating! Your method is good and your solutions are correct.:)