q is power of odd prime.How to show that  $-1$ is square in $\mathbb{F_q}$ if $ q=1(mod4)$ i.e order of field divide 4 and -1 is not square if $ q=3(mod4)$.  I know that since q is power of odd prime;  order of $\mathbb{F_q}^*$ is $q-1$ and its cyclic group of even order. I also know only element of order 2 there is -1. Then how to proceed for both this claims?
121335
GF(q) is a finite field of order q, where q is odd. Prove that $a\in GF(q), a\neq0$ has a root in $GF(q)$ iff $a^{(q-1)/2}=1$.  I tried to prove it this way:  Suppose a has a root in $GF(q)$, so there is some $b\in GF(q)$ so that $b^2=a$. There is a primitive element $g$ in $GF(q)$, so that $g^x=a, g^y=b$. Therefore $(g^y)^2=g^x\to 2y=x$. Furthermore $a^{(q-1)/2}=(g^x)^{(q-1)/2}=g^{(x/2)*(q-1)}=g^{y(q-1)}$ and since $y$ is an integer, $g^{y(q-1)}=1^y=1$. Now suppose $a^{(q-1)/2}=1$, then again $(g^x)^{(q-1)/2}=1\to g^{(x/2)*(q-1)}=1$. That means that $x/2$ is an integer (or else $g^{(x/2)*(q-1)}\neq 1$), and therefore $0\le x/2\le q-1$ being an integer, $g^{x/2}$ creates a certain element $b\in GF(q)$ so that $b^2=a$.  I am a bit unsure of this proof - first of all since I don't use the fact that $q$ is odd, which means I am wrong somewhere. Second of all, I feel extremely 'weak' with my argument regarding $x/2$ being an integer, yet I have no idea how to strengthen it. Any hints/directions will be extremely appreicated!
67452
The following question popped up during a conversation with a friend of mine: is there any integer $n$ such that $n$ is not a square in $\mathbb{Z}$ but it is a square in every field $\mathbb{Z}_p$ with $p&gt;n$?  It sounds like an elementary question, and I am almost sure that the answer is no (mostly because of some distant memories of mine about a theorem of Hasse and Minkowsky), but I was not able to find a definitive answer.    I tried to apply some Model Theory to it, but I don't think it can be a fruitful path to follow. Another thing I noticed is that if I can prove that there is an $a$ such that $n=a^2$ in two different fields $\mathbb{Z}_p$ and $\mathbb{Z}_q$, then the the statement must be false: otherwise, writing $$a^2=n+bp$$ and $$a^2=n+cq$$ for the smallest possible $b$ and $c$, we would have $bp=cq$, a contradiction since neither $b$ nor $c$ can be $0$. But again, I got stuck after this consideration. Do you have any ideas or suggestions?
105162
Suppose $n = pq$ with $p$ and $q$ distinct odd primes.   Suppose that $\gcd(a,pq)=1$. Prove that if the equation $x^{2} \equiv a\pmod n$ has any solutions, then it has four solutions.   Proof: Suppose $n = pq$ with $p$ and $q$ distinct odd primes and that gcd($a,pq$) = 1. Let us have the equation $x^{2} \equiv a\pmod n$. Then, $x^{2} \equiv a$ (mod $pq$), $p \not = q$. By definition, we can separate the equation into two equations such that $y^{2} \equiv a \equiv b$ (mod $p$) and $z^{2} \equiv a \equiv c$ (mod $q$). Let $g_{p}$ be a primitive root modulo $p$ and $g_{q}$ be a primitive root modulo $q$. Then, $b$ is equal to some power of $g_{p}$ and $c$ is equal to some power of $g_{q}$. With the fact that $b$ has a square root modulo $p$ (i.e. $r^{2} \equiv b$(mod $\ p)$) and $c$ has a square root modulo $q$ (i.e. $t^{2} \equiv c$(mod $\ q)$), there is an even power of $g_{p}$ and of $g_{q}$ such that $b = g_{p}^{2k_{1}}$(mod $\ p)$ and $c = g_{q}^{2k_{2}}$(mod $\ q)$ for some $k_{1}, k_{2} \in Z$. By computing, we have the following:  $r^{2} \equiv b$(mod $\ p)$       $\equiv b^{(p + 1) / 2}($mod $\ p)$       $\equiv (g_{p}^{2k_{1}})^{(p + 1) / 2}($mod $\ p)$       $\equiv (g_{p}^{p + 1})^{k_{1}}($mod $\ p)$       $\equiv g_{p}^{2k_{1} + (p - 1)k_{1}}($mod $\ p)$       $\equiv b \cdot g_{p}^{(p - 1)k_{1}}($mod $\ p)$       $\equiv b$ and $t^{2} \equiv c$(mod $\ q)$        $\equiv c^{(q + 1) / 2}($mod $\ q)$        $\equiv (g_{q}^{2k_{2}})^{(q + 1) / 2}($mod $\ q)$        $\equiv (g_{q}^{q + 1})^{k_{2}}($mod $\ q)$        $\equiv g_{q}^{2k_{2} + (q - 1)k_{2}}($mod $\ q)$        $\equiv c \cdot g_{q}^{(q - 1)k_{2}}($mod $\ q)$        $\equiv c$  Hence, $r$ is a square root of a modulo $p$ and $t$ is a square root of a modulo $q$, which means there are two solutions for each $p$ and $q$. Since $p \not = q \in \mathbb{Z}_{n}$, we have isomorphism such that $\mathbb{Z}_{n} \simeq \mathbb{Z}_{p} \cdot \mathbb{Z}_{q}$. Therefore, if the equation $x^{2} \equiv a$ (mod $n$) has any solutions, it must have four solutions. $\blacklozenge$  What do you think about the proof I wrote?
78124
Assume that $p$ and $q$ are distinct add primes such that $p-1\mid q-1$. If $\gcd(a,pq)=1$ ,show that: $$a^{q-1} \equiv 1 \pmod{pq}$$   I have tried as follows: $$a^{q-1} \equiv 1 \pmod{q} \quad \text{and} \quad a^{p-1} \equiv 1 \pmod{p}$$ $$\implies a^{(q-1)(p-1)} \equiv 1 \pmod{q} \quad \text{and} \quad a^{(q-1)(p-1)} \equiv 1 \pmod{p}$$ $$\implies a^{(q-1)(p-1)} \equiv 1 \pmod{pq}$$ But then I am stuck - please help.
14947
Find all pairs of prime nummbers $p,q$ such that $p^3 - q^5 = (p+q)^2$.  It's obvious that $p&gt;q$ and $q=2$ doesn't work, then both $p,q$ are odd. Assuming $p = q + 2k$ we conclude, by the equation, that $k|q^3 - q - 4$ because $\gcd(k,q)=1$ (else $p$ is not prime) and $k=1$ has no solution.  I also tried to use some modules, but I couldn't.
98573
I'm trying to show that $Q[\sqrt3]$ is a field. I am particularly struggling with showing that every element is a unit.   So for some element, we want to show that for some element $a+b\sqrt3$, there exists $c+d\sqrt3$ in $Q[\sqrt3]$ such that $(a+b\sqrt3)(c+d\sqrt3) = 1$.  Is that the right direction? I am not really sure of how to come up with that inverse though. Any help would be appreciated.
96611
For which odd positive integer $n$ , is it true that $-1$ is not a positive power of $2$ modulo $n$ i.e. $[-1] \ne [2^k] , \forall k &gt;0$ in $\mathbb Z_n$ ? Is there any ( at least sufficient ) characterization for such odd $n$ ?   ( I have only found $n=7$ is such a number ) . Please help . Thanks in advance
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This is exercise 8 on page 147 of Jacobson's Basic Algebra:     Let $p$ be a prime of the form $4n+1$ and let $q$ be a prime such that $ \left( \frac{q}{p} \right) =-1$. Show that $\mathbb Z[\sqrt{pq}]$ is not factorial.   Here, $ \left( \frac{q}{p} \right) =-1$ means that $\overline{q}$ is not a quadratic number in the field $\mathbb Z / (p)$.  In the special case when $p=5$ and $q=2$, $9 = 3 \cdot 3 = (\sqrt{10} +1 )(\sqrt{10 }-1)$, with $3$ and $\sqrt{10} +1$ or $\sqrt{10} - 1$ nonassociates. But I have no idea what can I do with the general case.  I know that primes of the form $4n+1$ can be written in the form $m^2 + n^2$ with $m,n \in \mathbb Z$. But when $q$ is odd and not of the form $4n+1$, $pq$ is not the sum of two quadratic numbers. How can I use the fact $\left( \frac{q}{p} \right) =-1$?  Thanks~
20982
My attempt :  If $n$ is odd, then the square must be 2  (mod 3), which is not possible.  Hence $n =2m$  $2^{2m}+3^{2m}=(2^m+a)^2$  $a^2+2^{m+1}a=3^{2m}$  $a (a+2^{m+1})=3^{2m} $  By fundamental theorem of arithmetic,   $a=3^x $  $3^x +2^{m+1}=3^y $  $2^{m+1}=3^x (3^{y-x}-1) $  Which is not possible by Fundamental theorem of Arithmetic   Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 $(n) $are used.
36969
So I am trying to show that $f(x) = x^4-3$ will factor over $\Bbb{F}_p$ when $p\equiv 3\mod 4$, with $p\neq 3$. Specifically, I am trying to show that $f$ will factor into irreducibles such that there is at least one factor of degree 2.  My Work So Far  So if I assume that it can be factored, we would have $$ x^4-3 = (x^2+ax+b)(x^2-ax+d) = x^4 + (b+d-a^2)x^2 + a(d-b)x + bd.$$ Because we want the linear term to be 0, we need either $a=0$ or $d-b = 0$. Now if $a=0$ we get that $b=-d$, so $-b^2 = -3$. That is, $3$ is a square. If $d-b = 0$, then we would have $b^2 = -3$, so $-3$ is a square. Because $-1$ is not a square when $p\equiv 3\mod{4}$, we know that one of $-3$ or $3$ must be a square, but not both, so $a=0$ or $d=b$, but not both.  If $a=0$, that is $3$ is a square, then we have $f(x) = (x^2 +b)(x^2-b)$. Then one of $b$ or $-b$ is a square, but not the other, so $f$ will factor into two linear terms and a quadratic. So this case works.  The trouble I am having is with the second case. If $d=b$, then we have  $$ f(x) = (x^2 + ax + b)(x^2 -ax +b)$$ with $2b = a^2$. This is where I get stuck. How do I know that such an $a$ can exist?  Edit: If there is a better approach to solving this, that would be great to know as well. I haven't been taught any of the relevant number theory to solve this, so I'm kind of out of my element here.
101932
I'd really love your help with proving that for  $q, p$ odd primes such that $p \neq q$, there is not primitive root modulo $pq$.   There's a theorem says that there's primitive root only for $2, 4, q^l, 2q^l$ for prime $q$ and $l \in {Z}$, but I guess I can't use it.  any other suggestions?  Thanks a lot
53376
I am looking at the following question from my undergraduate Number Theory textbook:  Show that if p,q are different odd primes, and if gcd(p,q)=1, then a$\Phi$(pq)/2 $\equiv$  $1$ mod $pq$.  So far, the approach I have taken is trying to split up a$\Phi$(pq)/2 = (a$\Phi$(p)/2)$\Phi$(q) and apply Euler's theorem but I don't think I am really getting anywhere.  Any help would be greatly appreciated, thank you!
72551
Following my  the book then asks      Use this to determine when the field extension  $\mathbb{Q}(\sqrt{a+\sqrt{b}})$ over $\mathbb{Q}$  is biquadratic (where $a,b\in\mathbb{Q}$)   "this" means what proved in the linked post.  My thoughts: First, I don't know what to say when $a^2-b$ is not a square. When $a^2-b$ is a square I think that the extension is of degree $2$ if $a=b$, this leaves the case $a\neq b$ which I am also having problems with and I don't know what to do (I have some cases like if $m,n$ are different primes then the degree of the extension is $4$, but I don't have something that is general)  Can someone please help my do this exercice  ?
38390
Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible.  I am trying to prove it with Eisenstein's criterion but without success: for p=2, it divides -4 and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find $f(x\pm c)$ which is irreducible:  $f(x+1)=x^4+4x^3+2x^2-4x+13$, but 13 has the divisors: 1 and 13, so don't exist a prime number p such that to apply the first condition: $p|a_i, i\ne n$; the same problem for $f(x-1)=x^4+...+13$ For $f(x+2)=x^4+8x^3+20x^2+16x+16$ is the same problem from where we go, if we set p=2, that means $2|8, 2|20, 2|16$, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for x-2  Now I'll verify for $f(x\pm3)$, but I think it will be fall... I think if I verify all constant $f(x\pm c)$ it doesn't work with this method...  so have any idea how we can prove that $f$ is irreducible?
17078
Question :   If $|G|=pq$ where $p$ and $q$ are primes that are not necessarily distinct.   Prove that the order of $Z (G) =1$ or $pq$.   Showing the order is $pq$ is trivial.   I unsure how to start with showing the order is $1$. Hints are appreciated.  Thanks in advance.
17571
If $x$ is odd then prove that $x^2-1$ is divisible by $8$.   I start by writing: $x = 2k+1 $  where $k\in\mathbb{N}$.  Then it follows that: $(2k+1)^2 -1 = 4k^2 +4k + 1 -1 $    Therefore: $$\frac{4k^2 +4k}{8} = \frac{k(k+1)}{2}$$  At the end part I can see that for what $k$ is, the number on top is divisible by 2. I was expecting the end result to be a number, not a fraction. Or is it "divisible" from the definition $\text{even} = 2k$ that completes the proof?
109631
So first of all, I used the following from my lecture notes:   If $f \in \mathbb{Z}[x]$ is primitive (gcd of all the coefficients is 1) - $f$ irreducible in $\mathbb{Z}[x] \Leftrightarrow$ f irreducible in $\mathbb{Q}[x]$.   So now I only have to show that $f_n = x^n + x + 3$ is irreducible in $\mathbb{Z}[x]$ for all $n \geq 2. $  Is there a pattern I should be appealing to in some way?