I need to prove that for every integer $a&gt;0$ there is a unique representation $a=r*s^2$ where $r$ is not dividable by any square:  there is no $d&gt;1$ such that $d^2|r$  What I tried is to show a as a unique multiplication of primes and then show the case that a is odd or not, but didn't get anywhere.. any help will be appriciated
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?
19737
I'm trying to teach myself some number theory. In my textbook, this proof is given:  Example (2.3.1) Show that an integer is divisible by 3 if and only if the sum of its digits is a multiple of 3. Let $n=a_0a_1\ldots a_k$ be the decimal representation of an integer $n$. Thus $n=a_k+a_{k-1}10+a_{k-2}10^2+\cdots+a_010^k$ where $0\le a_i&lt;10$ The key observation is that $10\equiv1\pmod3$, i.e., $[10]=[1]$. Hence $[10^i]=[10]^i=[1]$, i.e., $10^i\equiv1\pmod 3$. The assertion is an immediate consequence of this congruence.  I don't understand the last statement. Why does it follow from that congruence?
91504
In Herstein’s Topics in Algebra, p.147, there is a sentence  If $\pi$ in $R$ is a prime element of $R$ and $a \in R$, then either $\pi \mid a$ or $(\pi,a) = 1$, for, in particular, $(\pi,a)$ is a divisor of $\pi$ so it must be $\pi$ or $1$ (or any unit).  I’m having a bit of difficulty seeing why the last part of this sentence is true (“for, in particular…”). Note:  Herstein’s definition of a prime element in a Euclidean ring $R$ is a nonunit $\pi$ such that whenever $\pi = ab$, where $a, b$ are in $R$, then one of $a$ or $b$ is a unit in $R$. This result comes before the Unique Factorization Theorem, so that can’t be used. However, we have proved that every element of a Euclidean ring $R$ is either a unit or can be written as a product of finitely many primes. (I don’t know if this is useful, but it’s what I tried to use, to no avail.)  My attempt: I tried supposing to the contrary, so that $(\pi,a)$ is neither $\pi$ nor $1$ nor a unit. Since $(\pi,a)$ divides $\pi$ we can write $\pi = (\pi,a)k$ for some $k \in R$. Then by definition of $\pi$ being prime, $k$ must be a unit, since $(\pi,a)$ isn’t. So $\pi$ and $(\pi,a)$ are associates. This is where I tried a few other things and got stuck each 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.
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!
53920
I am self-studying real analysis for fun and came across the following question (from A First Course in Real Analysis by Berberian).  (i) Prove the following: If $r$ is a real number such that $0 \leq r \leq 1$, then there exists a unique real number $s$ such that $0 \leq s \leq 1$ and $s^2 = r$. (ii) Generalize (i) to the following: If $a \in \mathbb{R}$, $a \geq 0$, then there exists a unique $b \in \mathbb{R}$, $b \geq 0$ such that $b^2 = a$.  Put $y = 1-r$ and $x = 1-s$. Then we want to find a real number $x \in [0,1]$ such that $(1-x)^2 = 1-y$. I am trying to find an increasing sequence $(x_n)$ such that $0 \leq x_n \leq 1$ (defined recursively). For (ii) how could I use (i) to get the result?
16306
Let $\pi$ denote a prime element in $\mathbb Z[i], \pi \notin \mathbb Z, i \mathbb Z$. Prove that $N(\pi)=2$ or $N(\pi)=p$, $p \equiv 1 \pmod 4, p$ is a prime.   I know that $\pi$ is prime in $\mathbb Z[i]$ implies it is irreducible. Also if $\pi$ has a prime norm, that is $N(\pi) = p$ then $\pi$ is irreducible.  For any $z \in \mathbb Z[i]$ we have $N(z) = a^2 + b^2$. Since $\pi \notin \mathbb Z, i \mathbb Z$ both $a$ and $b$ must be non-zero in the case of $\pi$. By Fermat's Two Square theorem every prime number $\equiv 1 \pmod 4$ can be written as a unique sum of two squares. I have some idea I should utilize this theorem here, but no success so far.  However I've come to a dead end. I don't know how to go on from here. Could someone help me out ?  Thanks.
25520
We represent an element $(x_1,y_1,...,x_n,y_n,x_{n+1},y_{n+1})\in S^{2n+1}$ as an element $(z_1,...,z_{n+1})\in \mathbb C^{n+1}$ where $z_k = x_k+iy_k$. Now I'm considering the action of $S^1$ on $S^{2n+1}$ as $e^{iθ}·(z_1,...,z_{n+1}) = (e^{iθ}z_1,...,e^{iθ}z_{n+1})$. Under this action we identify any two elements as $(z_1,...,z_{n+1}) ∼ e^{iθ} ·(z_1,...,z_{n+1})\forall (z_1,...,z_{n+1})\in S^{2n+1}\subseteq \mathbb C^{n+1}$ and $e^{iθ}\in S^1$. For the resulting quotient space $S^{2n+1}/S^1$, I'm trying to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$. How do I prove this? Any solution will be highly appreciated.
117510
From Ireland,Rosen's Number Theory book, ex.8.12.  Let $p\equiv1\mod{4}$, then we can write $p=a^{2}+b^{2}$, where $a,b\in\mathbb{Z}$. Now, let $a$ be odd and $b$ be even and $a,b&gt;0$. How can we show that $a$ and $b$ are uniquely determined? The hint says to use the fact that $\mathbb{Z[i]}$ is a UFD and $p=a^{2}+b^{2}\implies a+bi$ is prime in $\mathbb{Z[i]}$
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
56314
Q: Is every element of $\mathbb Z/p\mathbb Z$ represented by a prime number? More generally, let $m, n \in \mathbb Z$ be coprime. Is there a prime number congruent to $m$ modulo $n$?   the affirmative answer is given by .     A: Yes, in fact there are infinitely many such primes.   I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?
87485
Is it true that all numbers of the form $\sqrt[b_1]{a_1}+\sqrt[b_2]{a_2}+\cdots+\sqrt[b_k]{a_k}$ are not integers, where $a_i,b_i$ are positive integers such that none of the $k$ terms is an integer? I think this must be a theorem somewhere, probably a fundamental theorem in number theory, but not sure what it would be called.
97895
Prove that every positive integer $n$ has a unique expression of the form: $2^{r}m$ where $r\ge 0$ and $m$ is an odd positive integer  if $n$ is odd then $n=2^{0}n$, but I dont know what to do when $n$ is even  and to prove that this expression is unique is it a good choice to  use the fundamental theorem of arithmetic?  I would really appreciate your help
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?
115780
How can we prove that for every pair $N \in \mathbb{N}$, and natural number $\beta\in [2, \infty)$ there exists a unique set of integers $x_i \in [0, \beta -1]$, $k\in [0,\infty)$ such that:  $$N = \sum_{i=0}^{k}\beta^ix_{i}, \quad $$  It's a question if every positive integer has its unique representation in every base $\beta$. Is the validity of above statement somehow related to Horner's rule?
73198
Show that any non-negative rational integer may be written as the sum of two squares if it is of the form $$2^ip_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}q_1^{2j_1}q_2^{2j_2}\cdots q_n^{2j_n}$$ where the $p_1,p_2,\dots,p_m$ are all primes congruent to $1\bmod 4$, the $q_1,q_2,\dots,q_n$ are all primes congruent to $3\bmod 4$ and $i,k_1,k_2,\dots,k_m,j_1,j_2,\dots,j_n$ are all non-negative integers.   I am completely blank on how to form a solution, so help would be much appreciated.
49227
Prove that for each $n$, there exist $n$ consecutive integers, each of which is divisible by a perfect square larger than $1$. Can any one help with this? I was stuck because I know how to solve for a particular $n$ value, but I was helpless in the generalized form.