Let $F$ be a field of order $32$. Then find the number of non-zero solutions $(a,b)\in F\times F$ of the equation $x^2+xy+y^2=0$.  As , $|F|=32$ , so $(F\setminus\{0\},.)$ forms a group of order $31$, which is prime. So , $F\setminus \{0\}\simeq\mathbb Z_{31}$.  Then how I proceed ?
122708
Let $F$ be a field of order 32. Find number of solutions $(x,y)\in F\times F$ for $x^2 +y^2 +xy =0$.   I have figured out that non zero elements of this field forms a cyclic multiplicative group of order 31 and charecteristic of $ F $ must be 2. Have no idea how to proceed further.
75735
What is the total number of solutions of an equation of the form $x_1 + x_2 + \cdots + x_r = m$ such that $1 \le x_1 &lt; x_2 &lt; \cdots &lt; x_r &lt; N$ where $N$ is some natural number and $x_1, x_2, \cdots, x_r, m$ are integers?  Also, what would be the solution in the relaxed case where $x_1, x_2, x_3, \dots, x_r$ could be equal?
11436
Let $a \in \mathbb{R}, a&gt;1,$ and consider the equation $$ze^{a-z}=1$$ How many solutions does this equation have in the unit disk, and how many of those solutions are real?   My attempt to solve was to use the Argument Principle on the function $f(z)=ze^{a-z}-1$, in hopes of recovering the number of zeroes that way. The integral I attempted to compute for this was $$\int_{|z|=1} \frac{e^{a-z}-ze^{a-z}}{ze^{a-z}-1}dz=\int_{|z|=1}\frac{1-z}{z-e^{z-a}}dz$$  I do not know how to proceed from here, as the $e^{z-a}$ in the denominator seems problematic. Thanks for your help!
113034
How to show that a root of the equation $$x (x+1)(x+2) ....... (x+2009) = c $$ can have multiplicity at most 2 , and to find the value of $ c $   for which this is possible.  I proceeded by using the derivative method, but the number of terms involved is huge.I have been trying to find some common property for all these terms, but I've come up with nothing so far.
98209
The equation $X^{n}  + Y^{n}  = Z^{n}$    , where $n   \geq  3$     is a natural number, has no solutions at all where $X,Y,Z$     are integers.   My solution:  False. Because if we let $X=0  ,Y=0$,  then $0^{3}+0^{3}=0^{3}$  . Thus there is a solution for the equation $X^{n}  + Y^{n}  = Z^{n}$,   where $X,Y,Z$   are integers.  Can anyone please give feedback on my answer and state whether it is correct on not.
124300
I am reading a proof that narrows down to the following statement:      It is easy to see that the number of solutions of $x^6+x=a$ in $\mathbb{F}_{2^m}$, where $a\in \mathbb{F}_{2^m}$ and $m\geq 3$ is odd, is same as number of solutions of $x^2+ax+1=0$ in $\mathbb{F}_{2^m}$.   Why is this true? They say that $x^7=1$ for $x \neq 0$, which I also don't see. Thank you!
100707
Prove that any element of $GL_{2}(\mathbb{Z})$ of finite order has order $1,2,3,4,6$ using Field Theory. My idea is to reduce such a finite order matrix say $A$ with order $n$ to modulo a prime $p$. $\det A=\pm1$ so $A$ will land inside $G=\{M\in GL_{2}(\mathbb{F}_{p}):\det M=\pm 1\}$. $|G|=2p(p^{2}-1)$. So $n\mid 2p(p^{2}-1)$ for all prime $p$. I don't know how to proceed after this.
42449
In   the equation $x^2-1=2^x$ is considered, this is a generalization:  Let $P_n(x)$ a polynomial of degree $n &gt; 0$. It is well know that the equation $P_n(x)=0\;$ has at most $n$ real solutions. But consider the inhomogeneous equations $P_n(x)=2^x$ or even $$P_n(x)=e^{ax},\quad a\ne 0$$  Is it correct to say that the number of solutions is $\le n+1?$   Proof could be via induction on $n$ with $P_{n+1}'(x)/a=e^{ax},\;$ but at the moment I see no obvious way to get back to $P_{n+1}(x)=e^{ax}.$
72465
My question is, can we prove that the number of solutions to $x^n=e$ in group $G$ is divisible by $n$, for $n | |G|$   Apparently the answer is it is indeed true, infact for the equation $x^n = a$ for any $a \in G $ also, however I have had many ideas each of which having a small flaw with them.  For example I attempted induction on order of $G$ as follows:  Assume true for all groups of order $ \leq k$. Consider $|G| = k+1$.  If $G$ is cyclic, we show fairly easily the result holds. (In fact if I recall correctly there are $n$ solutions for each divisor $n$).  If $G$ is not cyclic, we can look at the subgroups of $G$. It'd be nice to say the number of solutions in each group is divisible by $n$ by the induction hypothesis and get the desired answer by inclusion-exlusion for solutions in all the subgroups, however we could potentially have a subgroup of order  less than $n$ and still contain a solution.   So if we were able to show it works for all the subgroups of orders $ &lt; n$ collectively we would be done. I have no idea whether this can work though.  I did see this question  however I couldn't find much help on it. From the hint given I managed to show $no. solutions = \sum_{ d|n} \lambda_{i} \phi(d)$ for some $\lambda_{i}$s, however they could be different so I can't just use the well know summing of euler-phi formula.  Any solution/ideas would be very much appreciated.
107461
I have to prove the following theorem :     Let $p$ be a prime number and let $n \ge 1$,be any integer, then there exists a field of order $p^n$.   My attempt  I started off by considering the polynomial $f(x)$=$x^{p^n}-x \in \Bbb Z_p[x]$.  I took $F$ to be the splitting field of $f(x)$ over $\Bbb Z_p$.  Since $F$ is the splitting field of $f(x)$ over $\Bbb Z_p$ therefore $f(x)$ has exactly $p^n$ zeros in $F$ counting multiplicity.  Also since $f'(x)$ = $p^nx^{p^n-1}-1$ , therefore $f(x)$ will not have any multiple zeros and hence it will have $p^n$ distinct zeros in $F$.  Then I took $K =\{k \in F | k^{p^n} = k\}$.  I was trying to prove that $K$ is a sub-field of $F$.  $1 \in K$ thus $K$ is non-empty.  For $a,b \in K$ $a^{p^n}=a$ and  $b^{p^n}=b$  So $(a+b)^{p^n}$ (should be) = $a^{p^n}$+$b^{p^n}$, this is where I am currently stuck!  I know that this is true when  $p$ is the characteristic of a field, but why would this expansion be true here?
91247
Let $p$ be a prime number, and let $\mathbb{F}_p$ be the field with $p$ elements. How many elements of $\mathbb{F}_p$ have cube roots in $\mathbb{F}_p$?   I had this question on an exam and after reviewing I am still not sure. Any help would be appreciated.
70167
Does $a^2+b^2=1$ have infinitely many solutions for $a,b\in\mathbb{Q}$?  I'm fairly certain it does, but I'm hoping to see a rigorous proof of this statement. Thanks.    Here is my motivation. I'm working on a strange little problem. I'm working in a geometry over an ordered field. Suppose I have a circle $\Gamma$ with center $A$ passing through a point $B$. I want to prove that there are infinitely many points on $\Gamma$. Up to a change of variable, I'm considering the unit circle centered on the origin over $\mathbb{Q}$. To show there are an infinite number of points on $\Gamma$, it suffices to show there are an infinite number of solutions to $a^2+b^2=1$ for $a,b\in\mathbb{Q}$. I could then extend this to showing there are infinite number of solutions to $a^2+b^2=r^2$ for some $r$, which proves that any circle over $\mathbb{Q}$ has an infinite number of points. Then since any ordered field has a subfield isomorphic to $\mathbb{Q}$, I would be finished.
86256
How can I count all integer solutions to the equation $$x^2+dy^2=n$$  given $n$ and $d$, where $n$ is composite (or prime)?  By counting I mean any algorithm faster than brute force enumeration.
45015
Let $f \in \mathbb{Z}[x]$ be non constant. How to prove that exists infinitely many primes such $f$ have root in $\mathbb{Z/_{(p)}}$. I spent much time, but with no benefits.
2216
Let $F$ be a finite field. How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?
121414
Let $a, b, c \in \mathbb{N}_0$. If $a+b+c=12$, how many solutions $(a,b,c)$ satisfy the equation?   Is the answer:      $729$
33668
I need to prove that the equation $x^2 + y^2  = z^3$ has infinitely many solutions for positive $x, y$ and $z$. I got to as far as $4^3 = 8^2$ but that seems to be of no help.  Can some one help me with it?