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.
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 $2p^2 + 3 \bmod{1}$:
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[*] 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 $\boxed{-\frac{7}{4}, \frac{15}{4}, 3}$.