Solving equation of floor function

[anemone]

Solve the equation

$\left\lfloor{\sqrt{x+10}}\right\rfloor-1= \dfrac{x}{2}$

 

[kaliprasad]

Solve the equation

$\left\lfloor{\sqrt{x+10}}\right\rfloor-1= \dfrac{x}{2}$

Spoiler

x has to be even integer else x/2 shall not be integer and 1+ x/2 >=0 so x >=-2
so let x = 2y
$\left\lfloor{\sqrt{2y+10}}\right\rfloor= y+ 1$
or $y+1 \le \left\lfloor{\sqrt{2y+10}}\right\rfloor\lt y+ 2$
or $(y+1)^2 \le 2y + 10 \lt (y+ 2)^2$
or $y^2+2y+ 1 \le 2y + 10 \lt (y^2+4y + 4)$
so $y^2 \le 9 $ and $y^2 +2y \ge 6$ so y is positive
and $ y \le 3 $ and $y^2 + 2y +1 \ge 7$ so y
so $y \le 3$ and $(y+1)^2 \ge 7$
so $y \le 3$ and $y+1 \gt \sqrt{7}$
y = 2 or 3 hence x = 4 or 6

 

[anemone]

Spoiler

x has to be even integer else x/2 shall not be integer and 1+ x/2 >=0 so x >=-2
so let x = 2y
$\left\lfloor{\sqrt{2y+10}}\right\rfloor= y+ 1$
or $y+1 \le \left\lfloor{\sqrt{2y+10}}\right\rfloor\lt y+ 2$
or $(y+1)^2 \le 2y + 10 \lt (y+ 2)^2$
or $y^2+2y+ 1 \le 2y + 10 \lt (y^2+4y + 4)$
so $y^2 \le 9 $ and $y^2 +2y \ge 6$ so y is positive
and $ y \le 3 $ and $y^2 + 2y +1 \ge 7$ so y
so $y \le 3$ and $(y+1)^2 \ge 7$
so $y \le 3$ and $y+1 \gt \sqrt{7}$
y = 2 or 3 hence x = 4 or 6

Well done, kaliprasad!

My solution:

Spoiler

First, note that $x$ has to be an even number and $x\ge -2$. Second, we have $\sqrt{x+10}\ge \left\lfloor{\sqrt{x+10}}\right\rfloor$ and so we get $\sqrt{x+10}\ge \dfrac{x}{2}+1$, solving it for the range of $x$ yields $-6\le x\le 6$.

Checking out for each case ($x=-2,\,0,\,2,\,4,\,6$) in that range gives us the solutions where $x=4,\,6$.