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Jul
8
asked Bound on the size of the solution to $u_{t} - \Delta u - u \leq 0$
May
23
accepted $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
May
23
comment $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
@user26857: Implicitly throughout, we're viewing $f$ and $g$ now as functions over $\mathbb{F}_{p}$. So we have $f'(x)g(x) = f(x)g'(x)$ for all $x \in \mathbb{F}_{p}$. If one really wanted to be pedantic, I suppose one could add the phrase "over $\mathbb{F}_{p}$" everywhere, but I definitely agree that Matt's solution is correct.
May
23
comment $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
@barto: It turns out that that's not how I define $\deg f$. I define it as $\max(\deg g, \deg h)$, sorry for the confusion.
May
23
comment $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
@user26857: It turns out I was misinterpreting what I was reading, I have changed to question to better reflect what I want to ask.
May
23
revised $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
rewrote the whole question to be more clear
May
23
revised $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
added 58 characters in body
May
23
revised $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
added 58 characters in body
May
23
asked $f'(x) \equiv 0 \pmod{p}$ with $\deg f < p$ implies $f(x) \equiv c \pmod{p}$
May
3
accepted Does $cN \leq \left|\sum_{n= 1}^{N}f(n)\right|$ imply $|f(n)| \geq c/2$ for many $n$?
May
3
comment Does $cN \leq \left|\sum_{n= 1}^{N}f(n)\right|$ imply $|f(n)| \geq c/2$ for many $n$?
$N$ is fixed and large.
May
3
revised Does $cN \leq \left|\sum_{n= 1}^{N}f(n)\right|$ imply $|f(n)| \geq c/2$ for many $n$?
clarified role of n
May
3
asked Does $cN \leq \left|\sum_{n= 1}^{N}f(n)\right|$ imply $|f(n)| \geq c/2$ for many $n$?
Dec
21
awarded  Popular Question
Oct
22
comment A contravariant functor taking colimits to limits is representable.
@ZhenLin: Not OP, but it should be that if $L$ is the left adjoint of $F$, then the representing object is $L(\{\ast\})$ where $\{\ast\}$ is the set with 1 element?
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
11
awarded  Yearling
Mar
8
awarded  Popular Question
Sep
2
asked All possible subsequences converging to same function $f$