MHB How Do You Solve This Floor Function Equation?

[anemone]

Solve the equation $\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor\left\{\dfrac{a^2+ 2a +2}{a^2+ 1}\right\}=\dfrac{2a -a^2}{a^2 + 1}$, where $\{a\}$ denotes the fractional part of $a$.

 

[lfdahl]

My attempt:

Spoiler

\[\left \lfloor \frac{2a+1}{a^2+1} \right \rfloor\left \{ \frac{a^2+2a+2}{a^2+1} \right \}=\frac{2a-a^2}{a^2+1}\\\\ \left \lfloor \frac{2a+1}{a^2+1} \right \rfloor\left \{ \frac{a^2+1+2a+1}{a^2+1} \right \}=\frac{2a+1-(a^2+1)}{a^2+1}\\\\ \left \lfloor \frac{2a+1}{a^2+1} \right \rfloor\left \{ \frac{2a+1}{a^2+1} \right \}=\frac{2a+1}{a^2+1}-1\]

Let $q(a) = \frac{2a+1}{a^2+1}$

The function $q$ has two extrema and two asymptotes: View attachment 4761

From the graph it is obvious, that $q$´s range is included in the open interval $(-1,2)$:

\[q: \mathbb{R}\rightarrow Y \subset (-1;2)\]

If $|q| < 1$ there is no solution, because:

\[\left \lfloor q \right \rfloor\left \{ q \right \}=0\cdot q \neq q-1\;\;\; 0 \le q<1\]

And

\[\left \lfloor q \right \rfloor\left \{ q \right \}=(-1)\cdot q \neq q-1 \;\;\; -1<q < 0\]

If $1 \le q < 2$ (for $0 \le a \le 2$) you get:

\[\left \lfloor q \right \rfloor\left \{ q \right \}=(+1)\cdot (q-1) = q-1\]

So the set, $S$, of solutions is: \[S=\left \{ a \in \mathbb{R}\: \: \: |\: \: \: 0 \le a \le 2 \right \}\]

 

[anemone]

Well done, lfdahl, and thanks for participating!:)

My solution:

Spoiler

$\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor\left\{\dfrac{a^2+ 2a +2}{a^2+ 1}\right\}=\dfrac{2a -a^2}{a^2 + 1}$

$\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor\left\{1+\dfrac{2a +1}{a^2+ 1}\right\}=\dfrac{2a+1 -a^2-1}{a^2 + 1}$

$\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor\left\{\dfrac{2a +1}{a^2+ 1}\right\}=\dfrac{2a+1}{a^2 + 1}-1$

$\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor\left\{\dfrac{2a +1}{a^2+ 1}\right\}=\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor+\left\{\dfrac{2a +1}{a^2+ 1}\right\}-1$

$1-\left\{\dfrac{2a +1}{a^2+ 1}\right\}=\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor-\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor\left\{\dfrac{2a +1}{a^2+ 1}\right\}$

$1-\left\{\dfrac{2a +1}{a^2+ 1}\right\}=\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor\left(1-\left\{\dfrac{2a +1}{a^2+ 1}\right\}\right)$

$1=\left\lfloor{\dfrac{2a + 1}{a^2+ 1}}\right\rfloor$

Solving it for $x$ we get $\{x:0≤ x≤ 2\}$