Let  $p$ be prime and  and $b \in \{1,2 \dots p-1\}.$ How do I show that $b^2$ is not divisible by $p,$ i.e. $p\nmid b^2?$ I'm guessing one has to use prime factor decomposition of $b$ first, so say $$b=\prod_{i=1}^{k} p_i^{\alpha_i} \implies b^2=\prod_{i=1}^{k} p_i^{2\alpha_i} $$ Next: if we assume the statement isn't true, i.e. $p \mid b^2=\prod_{i=1}^{k} p_i^{2\alpha_i} ,$ then I think we should end up having one of these $p_i$'s must be equal to $p,$ and we should arrive at a contradiction - this is where I'm a little stuck. Could you please write out a solution, - if possible, as an answer? Secondly, I think the problem can be generalized from prime to any non-perfect square $p$, but how do give the proof for that? Thank you!
59946
Let $A$ be a UFD. Assume that $a,b \in A$ are relatively prime, $c \in A$ and $a | bc$. To prove that $a|c$, is the following approach correct (or do you have to use some type of prime factorization argument)?  By the relatively prime assumption, $\gcd(a,b)$ is a unit. Call that unit $u \in A^{\times}$. Then we can write $u = ax + by$ for some $x,y \in A$. But then, $c = c \cdot u = c(ax+by)=cax+cby$. However, it is clear that $a|cax$ and the fact that $a|bc$ implies $a|cby$. Thus, $a|cax+cby=c$.
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
43536
Can you prove that if $p$ is a prime greater than $5$, then the sum of the squares of the quadratic nonresidues modulo $p$ is divisible by $p$?  Note that I have just proved that the sum of the quadratic residues modulo $p$ is divisible by $p$ for $p$ greater than $3$.
107290
I came to this conclusion: the primes that can't be written as the sum of two squares (lets use$p\prime$ to denote) can be written as ${p}\prime \equiv 3 \;(\bmod\; 4)$.  I need a simple proof to show this and am stuck. I would appreciate guidance!
493
The following is a well known problem   If $a|b^2; b^2|a^3; a^3|b^4; ....$ then $a=b$.  I started with $a$ is a prime $p$ and the conclusion immediately holds.  So I was comparing powers of primes but can't get anything useful.  Please help.
122670
... namely that if $n &gt; 1$ and $(n − 1)!\equiv−1\pmod{n}$, then $n$ is prime.  This is for a number theory class I'm in at Penn State. My idea is to follow accordingly, but I can't get it concretely..  Reason by contradiction, suppose that $n$ is not prime.   Then $n = ab$, where $1 &lt; a, b &lt; n$  In particular $(n-1)! = ai$ for some $i\in\Bbb{Z}$. $\\$ If $(n-1)! = nj - 1$ for some $j\in\Bbb{Z}$ then it follows that a probably doesn't divide $-1$?
105138
if $p&gt;6$ is prime then 1920 divides $p^4-10p^2+9$ Take out the factors of 1920 which are $$1920=2^7(3)(5)$$ and factor the expression that remains as $$p^4-10p^2+9=(p-3)(p-1)(p+1)(p+3)$$ and I assumed as p is prime then p-1 and p + 1 is even then it is divisible by $2^7$ but I think it is not correct, can you help me?
99288
Let $a \in \Bbb Z$ be such that $a = b^2 + c^2,$ where $b,c \in \Bbb Z \setminus \{0\}.$ Then $a$ cannot be written as $:$   $(1)$ $p d^2,$ where $d \in \Bbb Z$ and $p$ is a prime with $p \equiv 1\ \left (\text {mod}\ 4 \right ).$   $(2)$ $p q d^2,$ where $d \in \Bbb Z$ and $p,q$ are distinct primes with $p,q \equiv 3\ \left (\text {mod}\ 4 \right ).$  $(1)$ is false because $2^2 + 1^2 = 5 = 5 \cdot 1^2,$ where $d = 1 \in \Bbb Z$ and $5$ is a prime with $5 \equiv 1\ \left (\text {mod}\ 4 \right ).$ How do I prove or disprove the other option? Any help in this regard will be appreciated. Thanks for your time.
29905
I'm reading a proof in my number theory textbook that all primes of the form $p = 4k+1$ are uniquely the sum of two squares. I'm stuck right at the beginning of the proof, where they say:  To establish the assertion, suppose that $$ p = a^2 + b^2 = c^2 + d^2 $$ where $a,b,c,d$ are all positive integers. Then $$ a^2 d^2 - b^2 c^2 = p(d^2 - b^2). $$  Perhaps I'm just missing something obvious, but I can't figure out how they managed to conclude that $a^2 d^2 - b^2 c^2 = p(d^2 - b^2).$ Please advise.
10320
If $a$ and $b$ are two relatively prime positive integers such that $ab$ is a square, then $a$ and $b$ are squares.   I need to prove this statement, so I would like someone to critique my proof. Thanks  Since $ab$ is a square, the exponent of every prime in the prime factorization of $ab$ must be even.   Since $a$ and $b$ are coprime, they share no prime factors.   Therefore, the exponent of every prime in the factorization of $a$ (and $b$) are even, which means $a$ and $b$ are squares.
1196
If $p$ and $q$ are primes,  which binomial coefficients $\binom{pq}{n}$, $1 \le n &lt; pq$,  are divisible by $pq$?  In particular, if $p$ and $q$ are distinct odd primes, and $n$ is even, does $pq \mid \binom{pq}{n}$?  This question is inspired by my answer here:   According to Kummer's theorem (), "the largest power of $p$ dividing $\binom{m+n}{m}$ equals $p^c$,  where $c$ is the number of carries when $m$ and $n$ are added in base $p$."  If $m+n = pq$, $m = pq-n$. So, if there is at least one carry when $n$ and $pq-n$ are added in base $p$, then $p \mid \binom{pq}{n}$. So, we want to know for which $n$ there is at least one carry in base $p$ and base $q$.  An obvious generalization is to ask, if $p, q, \ldots, r$ are primes,  which of the binomial coefficients $\binom{pq\cdots r}{n}$ are divisible by $pq\cdots r$. But I don't even know how to do this for two primes.  Note: The answers to this question may be useful:
63293
Primes of the form $p=4k+1\;$  have a unique decomposition as sum of squares $p=a^2+b^2$ with $0&lt;a&lt;b\;$, due to .  Is it correct to say that, primes of the form $p=4n+3$, never have a decomposition into $2$ squares, because sum of the quadratic residues $a^2+b^2$ with $a,b\in \Bbb{N}$ $$ a^2 \bmod 4 +b^2 \bmod 4 \le 2? $$  If so, are there alternate ways to prove it?
27206
a, b are integers. p is prime. I want to prove: $(a+b)^{p} \equiv a^p + b^p \pmod p$  I know about Fermat's little theorem, but I still can't get it I know this is valid: $(a+b)^{p} \equiv a+b \pmod p$ but from there I don't know what to do.  Also I thought about $(a+b)^{p} = \sum_{k=0}^{p}\binom{p}{k}a^{k}b^{p-k}=\binom{p}{0}b^{p}+\sum_{k=1}^{p-1} \binom{p}{k} a^{k}b^{p-k}+\binom{p}{p}a^{p}=b^{p}+\sum_{k=1}^{p-1}\binom{p}{k}a^{k}b^{p-k}+a^{p}$ Any ideas?  Thanks!
44204
Let $\pi \in \mathbb{Z}[i]$ irreducible and $\pi$ divides $p$ ($p\in\Bbb Z$ prime). Then show that $N(\pi)=p$.   ($N(\alpha)= a^2 +b^2$ where $\alpha = a+ib$.)
107652
Find a prime number $p$ and an integer $b&lt;p$ such that $p$ divides $b^{p−1}−1$.  First I think of long divisions but it didn't work out. Now I'm stuck..
99272
Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.    My attempt:  $p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3$ So, we have, $a^{3k}\equiv b^{3k}\mod p$ and by Fermat's Theorem we have, $a^{3k+1}\equiv b^{3k+1}\mod p$ as $p$ is of the form $p=3k+2$.    I do not know what to do next. Please help. Thank you.
17993
Let $p$ be a prime number. I could already show, that $\mathbb{Z}[\sqrt{-2}]$ is an euclidean domain and that $$p\text{ reducible }\Leftrightarrow\text{ }p=a^2+2b^2\text{ for some $a,b\in\mathbb{Z}$.}$$  Now I wonder how to prove  $$p\equiv 1,2,3\text{ mod }8 \Rightarrow p=a^2+2b^2\text{ for some $a,b\in\mathbb{Z}$.}$$  Maybe there is a possibility to show  $$p\equiv 1,2,3\text{ mod }8 \Rightarrow p\text{ reducible }?$$