Let $a,b,r_i\in \mathbb{Z}$. How to show that if $b$ is squarefree and $a \equiv r_i^2 \mod p_i$ for every prime divisor $p_i$ of $b$, then $a$ is also a square modulo $b$.  Approach: I think I have to use CRT in some way, but I don't see how it follows from that. Can someone explain it to me? Thanks!
121016
Suppose you want to prove that $\exists x \in \mathbb{Z}$ with $a \equiv x^2 \mod b$. Write $b = \prod_{i = 1}^{k} p_i^{e_i}$, the prime factorisation of $b$.   Why is the equivalent with finding solutions to $a \equiv x_i^2 \mod p_i$? How does one apply the Chinese Remainder Theorem here?   Thanks in advance!
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
4812
Problem: $(\forall a\in\mathbb{Z^+})(m\in\mathbb{Z^+}\to a^m\equiv a^{m-\phi(m)}\pmod{m})$  My work: Start by letting $m=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$. If $(a,p_i)=1$ for some integer $i$, then we have that  $$ a^{\phi(p_i^{a_i})}\equiv 1\pmod{p_i^{a_i}} $$ by Euler's theorem. Since $\phi(p_i^{a_i}) \mid \phi(m)$, we have that $p_i^{a_i}\mid (a^{\phi(m)}-1)$ if $(a,p_i)=1$.   Beyond this, I am stuck at the moment.
30745
Let $p$ be a prime. Show that for all integers $x$, $x^2\equiv 1 (\text{mod} \ p) $ if and only if $x \equiv 1 (\text{mod} \ p)$ or $ x \equiv p-1 \ (\text{mod} \ p ). $    I know that I have to prove this both forward and backward. I'm not really sure where to begin
90498
I have tried to search similar posts to this, but unfortunately I may not have enough keywords for it. I'm looking for some help so check the following: Given a prime number $p$, let $k, m \in\mathbb N \backslash\{0\}$ such that $k \equiv m(\bmod (p-1))$. Now given $ u, v \in\mathbb Z$ such that $u \equiv v\pmod p$, check that $u^k \equiv v^m\pmod p$. Any tips to show this or where should I start? Thank you!
29843
Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that:     Lemma: If $(D,g)$ is a Euclidean domain such that its invertibles are $\{1,-1\}$, and $x$ is an element of minimal degree among the elements that are not invertible, then $D/(x)$ is isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.   Now the exercise asks:     Prove that $A$ is not a Euclidean Domain.   Everything hints to an argument by contradiction: let $(A, d)$ be a ED and $x$ an element of minimal degree among the non invertibles we'd like to show that $A/(x)$ is not isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.  How do we do that? My problem is that, since I don't know what this degree function looks like, I don't know how to choose this $x$!  I know that the elements of $A/(x)$ are of the form $a+(x)$, with $a$ of degree less than $x$ or zero. By minimality of $x$ this means that $a\in \{0, 1, -1\}$. Now I'm lost: how do we derive a contradiction from this?
14710
There is not any perfect square $k$ such that $k  \equiv 3\ (\textrm{mod}\ 5)$? Why? How can I prove it?
25854
Prove that a perfect square is either a multiple of $4$ or of the form $4q+1$ for some $q\in \mathbb{Z}$.  Any ideas on how to start? Do I use a proof by contraposition? Also what's the definition of a perfect square?
1779
Page 105 of D. Burton's A First Course in Rings and Ideals reads     It is not difficult to show that if (I'll call them $n$ and $m$ instead of $n_1$ and $n_2$) $n$, $m$ are square free integers, then $\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})$ if and only if $n=m$.   Well, it is getting difficult for me, any help or would be appreciated.    My progress so far: Suppose $\phi :\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ is an isomorphism. Then $\phi(u)=u$ for every $u \in \mathbb{Q}$, so that $\phi(\sqrt{n})^2=\phi(n)=n$. If $\phi(\sqrt{n})=a+b\sqrt{m}$ for some $a,b\in\mathbb{Q}$, then $(a+b\sqrt{m})^2=a^2+b^2m + 2ab\sqrt{m}=n$ implies $$a^2+b^2m = n$$ and $$2ab=0.$$  Then $b=0$ cannot happen since that would imply that $\sqrt{n}\in\mathbb{Q}$, where $n$ is a square-free integer. Also $a=b=0$ cannot happen. Hence $a=0$ and we have $$b^2m=n.$$ If $\psi:\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ were an isomorphism then there would exist some $s\in\mathbb{Q}$ such that $$s^2m=n,$$ then $s=b$ or $s=-b$. Therefore the only isomorphisms from $\mathbb{Q}(\sqrt{n})$ to $\mathbb{Q}(\sqrt{n})$ are $\phi(u+v\sqrt{n})=u+vs\sqrt{m}$ and $\psi(u+v\sqrt{n})=u-vs\sqrt{m}$.    I don't know if there's an easier way and I don't know if I'm in the correct way here. Thanks beforehand :)
121966
I am trying to prove that for every integer $n \ge 1$, there exists uniquely determined $a &gt; 0$ and $b &gt; 0$ such that $n = a^2 b$, where $b$ is squarefree.  I am trying to prove this using the properties of divisibility and GCD only. Is it possible?  Let me assume that $n = a^2 b = a'^2b'$ where $a \ne a'$ and $b \ne b$'. Can we show a contradiction now?
108016
$p \gt 2$ is a prime, then  there are infinite primes $q$  such that $q$  is a quadratic residue modulo $p$.  With , the problem is  easy! How about  Without the use of the theorem? Can one find  an elementary proof?    p.s.  there are  several  elementary ways(without Dirichlet) to show that there are infinite primes $q$  such that $q$  is a quadratic non-residue modulo $p$  Proof 1  The least $q&gt;0$ quadratic non-residue modulo $p$  is a prime . Consider   $$a_0=q, a_1=q+p,a_2=q+pa_1,a_3=q+pa_1a_2,\dotsc, a_n=q+pa_1a_2\dotsb a_{n-1},\dotsc $$ then $(a_i,a_j)=1$  Proof 2  Only $\dfrac{p-1}2$ quadratic non-residues modulo $p$, so there is a integer $a$ such that $(\dfrac{a^2+1}p)=-1$.   In not, only finite $q_1,q_2,\dotsc, q_n$. then there exists $k\in\Bbb Z$ such that $kq_1q_2\dotsb q_n \equiv a \pmod p$. Consider   $$ (kq_1q_2\dotsb q_n)^2 +1$$
90444
I've stucked at the following problem:  Let $p$ be an odd prime number. Show that for all $a\in\mathbb{Z}$  $$a^p \equiv a\ (\mathrm{mod}\ 2p)$$  How can I prove it? I know within Fermat's theorem that $a^p \equiv a\ \mathrm{mod}\ p$.
85930
An integer $n$ is a perfect square if and only if it can be expressed as the square of some other integer i.e. there exists some integer $a$ such that $n=a^2$.    Prove that if $n$ is a perfect square, then either $n \equiv 1 ~(\mod 4)$, or $n \equiv 0 ~(\mod 4)$.  How can I prove this?
41686
This seems to be one of those tricky examples. I only know one proof which is quite complicated and follows by localizing $\mathbb{Z}[\sqrt[3]{2}]$ at different primes and then showing it's a DVR. Does anyone know any simple quick proof?
72954
There is a hint to show that if there were points in $\mathbb{Q}$, then there would exist $r,s, t \in \mathbb{Z}$ with $\gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$ , and then show that any prime dividing $s$ is a quadratic residue modulo $17$.  I'm stuck on the first bit - if you suppose $\left(\frac{a}{b}, \frac{c}{d}\right)$ is a point on the curve in $\mathbb{Q}$ (so $a, b, c, d \in \mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?
90462
Possible Duplicate:          I am trying to prove that for every $n \ge 1$ there exist uniquely determined integers $a \gt 0$ and $b \gt 0$ such that $n = a^2b$ where $b$ is square-free.  The fact that such $a$ and $b$ exist is easy to prove.  From the fundamental theorem of arithmetic, $n$ can be uniquely represented as $p_1^{a_1} p_2^{a_2} \cdots p_s^{a_s}$ where $s$ is a positive integer. Thus  \begin{align*} n &amp; = \prod_{i=1}^s p_i^{a_i} \\\\   &amp; = \prod_{i=1}^s p_i^{\left(2 \left\lfloor \frac{a_i}{2} \right\rfloor + a_i \bmod{2}\right)} \\\\   &amp; = \prod_{i=1}^s p_i^{\left(2 \left\lfloor \frac{a_i}{2} \right\rfloor\right)} \cdot \prod_{i=1}^s p_i^{a_i \bmod{2}} \\\\   &amp; = \left(\prod_{i=1}^s p_i^{\left\lfloor \frac{a_i}{2} \right\rfloor}\right)^2 \cdot \prod_{i=1}^s p_i^{a_i \bmod{2}}. \end{align*}  Clearly, $\left(\prod_{i=1}^s p_i^{\left\lfloor \frac{a_i}{2} \right\rfloor}\right)^2$ is a perfect square and $\prod_{i=1}^s p_i^{a_i \bmod{2}}$ is square free. Hence, we have shown that such $a$ and $b$ exist.  Now, how do we show that such a pair of $a$ and $b$ is unique?  I know how to start proving such a theorem. Let us assume that $n = a^2b = a'^2b'$ such that $a' \ne a$ and $b' \ne b$. Now since this is not possible this should lead us to some contradiction. But, I'm unable to reach a contradiction from this assumption. Could you please help me?
4708
I am trying to prove the following: Let $R=\mathbb{Z}[i]$, $a, b \in \mathbb{Z}, ab \neq 0$. Show that $a+bi$ is an irreducible element in $R$ if and only if $a^2 + b^2 ∈ \mathbb{Z}$ is prime. Since $R$ is a Euclidean domain, irreducible elements are prime. I think this is relevant, but am stuck on where to go from here.