Find all values of t such that floor(t) = 3t + 4 - t^2. If you find more than one value, then list the values you find in increasing order, separated by commas.

Answer 1

We begin by simplifying the expression on the right side of the equation. We have $3t + 4 - t^2 = -t^2 + 3t + 4$. Since we are looking for values of $t$ that satisfy the equation $\lfloor t \rfloor = -t^2 + 3t + 4$, we consider the equation $-t^2 + 3t + 4 = t'$ where $t'$ is some integer.

Since $\lfloor t \rfloor$ is an integer, $t$ must also be an integer. Therefore, we let $t = k$ where $k$ is an integer. We then have $-k^2 + 3k + 4 = k'$. Rearranging, we get $-k^2 + 2k + 4 - k' = 0$. We can now find when this quadratic equation has integral roots $k$. Using the quadratic formula, we get
$$k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$where $a = -1$, $b = 2$, and $c = 4-k'$. In order for $k$ to be an integer, the discriminant $\sqrt{b^2 - 4ac}$ must be an integer. In other words, $b^2 - 4ac = m^2$ for some integer $m$. Substituting our values, we have $2^2 - 4(-1)(4-k') = m^2$ $\Longrightarrow 4k' - 12 = m^2$.

Taking the equation modulo $4$, we have $+0 - 0 \equiv m^2 \pmod{4} \Longrightarrow m^2 \equiv 0 \pmod{4}$. Therefore, $m$ must be even, so let $m = 2n$ where $n$ is an integer. Substituting this back into the equation, we get $4k' - 12 = (2n)^2$ $\Longrightarrow 2k' - 6 = n^2$. Taking modulo $2$ now, we obtain $0 \equiv n^2 \pmod{2}$. Therefore, $n$ must be even. Let $n = 2p$ where $p$ is an integer. Substituting this back into the equation, we have $2k' - 6 = (2p)^2 \Longrightarrow k' - 3 = 2p^2$. Rearranging, we find $k' = 2p^2 + 3$.

Thus, we have shown that all values of $k$ that satisfy $-k^2 + 2k + 4 - k' = 0$ for some integer $k'$ will be of the form $k = 2p^2 + 3$. Plugging this back into our original expression, we need $\lfloor 2p^2 + 3 \rfloor = t$. We do casework on $2p^2 + 3 \bmod{1}$:
[list]
[*] If $2p^2 + 3 \equiv 0 \pmod{1}$, then we have $2p^2 + 3 = n$ for some integer $n$. This means that $n = 2p^2 + 3$ from our work above on $k'$. Therefore, our expression for $t$ simplifies to $2p^2 + 3 = t$. From this we obtain one solution, $t=3$. Checking the equation to make sure that it works, we find that $\lfloor 3 \rfloor = 3 = 3(3) + 4 - 3^2$.

[*] If $2p^2 + 3 \equiv 1 \pmod{1}$, then we have $2p^2 + 3 = n + \frac{1}{2}$ for some integer $n$. Rearranging, we find $n = 2p^2 + \frac{5}{2}$. From this we obtain one solution, $t=\frac{15}{4}$. Checking the equation to make sure that it works, we find that $\lfloor \frac{15}{4} \rfloor = 3 = 3\left(\frac{15}{4}\right) + 4 - \left(\frac{15}{4}\right)^2$.

[*] If $2p^2 + 3 \equiv -1 \pmod{1}$, then we have $2p^2 + 3 = n - \frac{1}{2}$ for some integer $n$. Rearranging, we find $n = 2p^2 + \frac{5}{2}$. From this we obtain one solution, $t=-\frac{7}{4}$. Checking the equation to make sure that it works, we find that $\lfloor -\frac{7}{4} \rfloor = -2 = 3\left(-\frac{7}{4}\right) + 4 - \left(-\frac{7}{4}\right)^2$.

[*] If $2p^2 + 3 \equiv \frac{1}{2} \pmod{1}$, then we have $2p^2 + 3 = n + \frac{1}{2}$ for some integer $n$. Rearranging, we find $n = 2p^2 + \frac{5}{2}$. However, there are no integers $p$ such that $2p^2 + \frac{1}{2}$ is an integer.

[*] If $2p^2 + 3 \equiv -\frac{1}{2} \pmod{1}$, then we have $2p^2 + 3 = n - \frac{1}{2}$ for some integer $n$. Rearranging, we find $n = 2p^2 + \frac{5}{2}$. However, there are no integers $p$ such that $2p^2 + \frac{1}{2}$ is an integer.
[/list]Let $S$ be the set of all values of $t$ that satisfy the equation in the problem. We have shown that $S = \{ -\frac{7}{4}, \frac{15}{4}, 3 \}$. Therefore, the answer is $\boxed{-\frac{7}{4}, \frac{15}{4}, 3}$.