I've been working on this problem for a few hours now, and haven't been able to make much progress answering any of the parts. I'm looking more for a push in the right direction than direct answers. Any help is very much appreciated.    These are my attempts at solutions    a) Let $A=a+b\sqrt d$, $B=x+y\sqrt d$    $N(AB)=N(ax+\sqrt d (ay+bx+by \sqrt d))=(ax+\sqrt d (ay+bx+by \sqrt d))(ax-\sqrt d (ay+bx+by \sqrt d))$    I tried to show that this is equivalent to this, but with no luck: $N(A)N(B)=(a^2+db^2)(x^2+dy^2)=a^2x^2+da^2y^2+db^2x^2+d^2b^2y^2$  bi) I can't completely wrap my head around the concept of units, so I was only able to use brute force to come up with units, and I have no way of knowing if I missed any. I tried using this equation, but didn't get anywhere. $(a+b\sqrt{-5})(x+y\sqrt{-5})=1$  bii) Similar problem to bi. I'm not sure how I might approach generating units.
115397
Find all units in $\mathbb{Z}[\sqrt n] = \{a + b\sqrt n \mid a, b \in \mathbb Z \}$  and $n \in\mathbb N$, $n\ge 2$.  First let $c +d\sqrt n$ be a unit so $$(a +b\sqrt n)(c +d\sqrt n) = 1,$$ $$ac + bdn +(bc + ad)\sqrt n = 1.$$  What next?
122205
I think I have proved that that a cannot be greater than b (ie a &amp; b must be equal) then using this fact I got a quadratic in terms of a and c, trying out some values I got that (3,3,4) is a solution, but how do I prove this is the only solution or if this is not the only solution then what are the other solutions and how do I find them?
73369
I saw this in quora.  What are all the integer solutions to $a^{b^2} = b^a$ with $a, b \ge 2$?  Solutions I have found so far:  $a = 2^4 = 16, b = 2, a^{b^2}  = 2^{4\cdot 4} =2^{16}, b^a = 2^{16} $.  $a = 3^3, b = 3, a^{b^2} = 3^{3\cdot 9} =3^{27}, b^a = 3^{3^3} =3^{27} $.  In the general case, $a$ and $b$ have the same set of prime divisors, so let $a =\prod_P p_i^{a_i}$, $b =\prod_P p_i^{b_i} $ with each $a_i \ge 1, b_i \ge 1$.  $b^a =b^{\prod p_i^{a_i}} =(\prod p_j^{b_j})^{\prod p_i^{a_i}} =\prod p_j^{b_j\prod p_i^{a_i}} $  $a^{b^2} =a^{\prod p_i^{2b_i}} =(\prod p_j^{a_j})^{\prod p_i^{2b_i}} =\prod p_j^{a_j\prod p_i^{2b_i}} $  Therefore, for each $p_j$, $b_j\prod p_i^{a_i} =a_j\prod p_i^{2b_i} $.  I haven't gotten any further than this.  I conjecture that there are no other solutions.
103197
Question : $$\begin{align}a+b+c &amp;= 0\\ a^3 +b^3 +c^3 &amp;= 12\\ a^5 +b^5 +c^5 &amp;= 40\end{align}$$ Then , $a^4 + b^4 + c^4 = ?$  My try : As a common perspective I just went to find any trick related to it and my first step is as usual as a common man will think $3abc = 12$ and got $abc = 4$ After that I am unable to link it with any other equation
16851
I'm asked to show that there are infinitely many units in the ring $\mathbb Z[\sqrt 3]$.   But I don't really see a good approach to this one, so far.   Some thoughts: The inverse of $a+\sqrt3 b$ should be given by $\pm(2 - \sqrt 3 b)$, since the norm   $$N: \mathbb Z[\sqrt{3}] \to \mathbb Z, \qquad N(x+\sqrt{3} y) = x^2 - 3y^2$$  is multiplicative. So if $(a+\sqrt{3}b)^{-1}=x+\sqrt 3 y$, then   $$1 = N(1) = N((a+\sqrt{3}b)(x + \sqrt{3}y)) = (a^2 - 3b^2)(x^2 - 3y^2)$$  Hence we must have $\pm 1 = (a^2 - 3b^2) = (a+\sqrt{3}b)(a-\sqrt{3}b)$.  Therefore I need to show that there are infinitely many $a,b$ such that $a^2 - 3b^2 = \pm 1$.   Here I don't know how to proceed.   Maybe rewriting as $a = \sqrt{3b^2 \pm 1}$, and now trying to prove that there are infinitely many $b$ for which $3b^2 \pm 1$ is a square?  A hint would be appreciated! =) Thanks.
46957
In  in these notes, it is said that  Since $x − a$ has leading coefficient $1$, which is a unit, we may use the Division Algorithm...  Why is this true? I thought that the Division Algorithm is only guaranteed to work in Euclidean domains not any integral domain.
22896
I have the equation $ \mathbf{x}^T A \mathbf{x} = b $, where $b$ is a scalar, $\mathbf{x}$ a vector of size $M$, and A a matrix of size $M\times M$. $b$ and $\mathbf{x}$ are given.   How many such equations do I need to find the (common) $A$, and how do I do it?
80535
Show that $4x^2+6x+3$ is a unit in $\mathbb{Z}_8[x]$.  Once you have found the inverse like , the verification is trivial. But how do you come up with such an inverse. Do I just try with general polynomials of all degrees and see what restrictions RHS = $1$ imposes on the coefficients until I get lucky? Also is there a general method to show an element in a ring is a unit?
15167
The terms $a,b,c$ of quadratic equation $ax^{2}+bx+c=0$ are in A.P. and positive. Let this equation have integral root $\alpha,\ \beta$. Then find the value of $\alpha+ \beta + \alpha \cdot \beta$ ?   please point where I'm wrong:  Let common difference be $d$  $\implies \alpha+ \beta + \alpha \cdot \beta=\dfrac{c-b}{a}=\dfrac{d}{a} \implies a|d \ \ \ \ \longrightarrow  \ \ \ \ \              \because (b=a+d$, $c=a+2d)$  Also , $ax^{2}+(a+d)x+(a+2d)=0$.   $\implies$ $\alpha,\ \beta=\dfrac{-(a+d) \pm \sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}}{2a}$.   For this to be integer $\sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}$ must be perfect square.  $\implies$ ${(a+d)^{2}-4\cdot a \cdot (a+2d)}=p^{2}$ for some $p$.  $\implies -3a^{2}+d^{2}-6ad=p^{2}$  $\implies -3a^{2}+a^{2}q^{2}-6a^{2}q=p^{2}$ $\because$ $a|d \implies aq=d$ for some $q$.  $\implies a^{2}(-3+q^{2}-6q)=p^{2}$   $\implies -3+q^{2}-6q\ $  has to be perfect square. By trial $q=7$  But I need to get this without trial, please help.
29338
Overview I have tried finding a solution to this problem myself and I have flailed. Its just a challenge for me. could you please tell me how far am I in solving this question? My approach for quadratic equation : I am still a student, and when I first saw the quadratic formula I wanted to derive it myself and I did it. This was my logic: the equation $x^2 + bx + c=0$ is obviously supposed to contain 2 roots (real or imaginary). So any 2 points in an axes can be represented as: $n+k$ or $n-k$. so (1) : putting $x = n+k$, (2): putting $x=n-k$ we get two equations; on subtracting $(1) - (2)$ we get an equation in $n$ and $k$ as, $4nk+2kb=0$ which gives $n=-b/2$.  Now substituting this value of $n$ in $(1)$ and $(2)$ and adding them we get $k= (+or-) \sqrt{b^2 -4c}/2 $. My approaches for trying to solve cubic equations: Let $x^3+bx^2+cx+d = 0$ be the equation. My first trials were focused on extracting the roots by various substations. Let me explain some of them. Originally, when I derived the quadratic equation, I wanted to literary &quot;represent the solution&quot; by $n+k$ and $n-k$. most methods that I have thought of for cubic equation similar to that of the quadratic (like assigning 3 unambiguous  variables to uniquely represent 3 roots- $n$ and $k$ represents unique values), by considering 3 unique values, but they end up being ambiguous. Then I was suddenly struck with the idea that I could define certain random operators to &quot;transform&quot; the equation in some way changing the co-coefficients of $x^2, x, x^0$. I have defined the first transformation as: $x^3+bx^2+cx+d=0 \to x^3+ b'x^2 + c'x+d' =0   $ then $b' = b$, $c' =b^2+c$, $d' = bc-d$. Applying this operation changes the modified equation's zeros as follows: $\alpha'=(\alpha+\beta)/2, \beta' = (\beta+\gamma)/2, \gamma' = (\alpha+\gamma)/2 $ if the initial zeros are $\alpha, \beta, \gamma$. My latest assumption is that creating $n$ number of transformations and applying them can help us obtain the roots in some combination. The other operation I have defined is &quot;increment roots by by $\theta$ operation&quot;. in this case, $x^3+ bx^2+ cx+ d \to x^3+b'x^2+ c'x+d=0$. here, $b' = b + 3\theta, c' = 3\theta^2+ 2\theta b+ c, d' = \theta^3 +b\theta^2 + c\theta + d$. This modification transforms the original roots $\alpha, \beta, \gamma$ to $\alpha' = \alpha + \theta, \beta' = \beta + \theta, \gamma' = \gamma + \theta$ Now for my questions: How far am I from solving the cubic polynomial problem myself without any external help? More over, I give up. How to find the roots of a cubic equation? Could my latest idea for solving cubic equations actually be applied (the transformation method)? Is it possible to find the roots with my transformation method? (my second transformation definition was successful in converting the coefficient b or c to zero by transforming the roots linearly)
87766
Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then $$ $$ (i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$) $$ $$ (ii) if $a$ | $b$ then $a$|$bc$ for all integers $c$; $$ $$ (iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.  Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column and the reason for the step in the second column.  My book is really vague. I'm not really sure what to do..  First I was thinking something like this: $a\mid b\Rightarrow b=as$ and  $b\mid c\Rightarrow c=bt$ and $a\mid c\Rightarrow c = au$ $$ $$ $b + c + c = b + 2c$  $$ $$ $=as + 2(bt+au)$  Yea then I just got lost.
21368
You learn in algebra that  $$\sqrt{ab}=\sqrt{a} \sqrt{b}$$ and that  $$\sqrt{\frac ab}=\frac {\sqrt a}{\sqrt b}$$ You also learn to never make the fatal mistake of thinking $$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$  However, I am wondering if there is a rule for $\sqrt{a+b}$. I would think it would be pretty complex, if it exists at all.  $$\sqrt{a+b}=\text{?}$$
80460
Have to admit, I forgot how to do simple 9th grade algebra problems. I need to know how to do this stuff for an exam tomorrow.  I am doing recurrences and the only thing I need to do is find $a, b$, and $c$ from the following  $$\begin{align*} a + b + c &amp;= 3\\ 4a + 2b+ c&amp;= 6\\ 9a + 3b + c&amp; = 13 \end{align*}$$  Can someone refresh my memory on how to solve for  $a, b$ and $c$ ?
97063
Consider the equations:   $$\begin{cases} ab+c+d=3\\ bc+d+a=5\\ cd+a+b=2\\ da+c+b=6 \end{cases}$$   Where $a,b,c,d \in \mathbb{R}$. How can we find $a,b,c,d$?   The furthest I've got is by adding the first two equations and the last two equations together to get: $ab+bc+c+d+d+a=(b+1)(a+c)+2d=8$ and $(d+1)(a+c)+2b=8$ We can rearrange these (assuming $b,d \neq -1$) to get $$\frac{8-2d}{b+1}=\frac{8-2b}{d+1} \implies -d^{2}+3d+4=-b^{2}+3b+4$$ Which leads us further to $$b^{2}-d^{2}-3(b-d)=0 \implies (b-d)(b+d-3)=0$$ Now since $b=d$ gives us an absurdity (equations 1 and 4 reduce to 3=6), we must have $b+d=3$. This is as far as I can get unfortunately, since trying the same approach with the other pairs doesn't work in the same way.  Help would be much appreciated; the problem is from the BMO1 2004 paper.
93879
The problem I am given is finding a solution to the following pair of equations:  $3x + 4y$ $\equiv$ $5$ $mod$ $13$  $2x + 5y$ $\equiv$ $7$ $mod$ $13$  By reading the methodology here:   Since the modulos in my problem are also the same, I am able to use substitution, but this leaves fractions in my expressions.  I then used matrix multiplication as follows:  $\begin{bmatrix} 3 &amp; 4\\ 2 &amp; 5\end{bmatrix}$ $\begin{bmatrix}\ x\\ y\end{bmatrix}$ = $\begin{bmatrix}\ 5\\ 7\end{bmatrix}$   And solving this also gives me a matrix with fractions, so I am a bit lost on how to solve this problem.  Any help is appreciated. Thank you very much.
119649
Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$   What I have tried so far is using CBS:  $(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$  $(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$  $(a^2 + b^2)(c^2 + d^2) \geq (ac + bd)^2$  Then, we have:  $(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 9(a + b)^2(c + d)^2$.   Thus, we have to prove that $9(a + b)^2(c + d)^2 \geq 256$.  Then, I used the following substitution:  $c + d = t$ and $a + b = 4 - t$.  We assume wlog that $a \leq b \leq c \leq d$.  Then, $4 = a + b + c + d \leq 2(c + d) = 2t$. Thus, $t \geq 2$.  Then, what we have to prove is:  $9t^2(4 - t)^2 \geq 256$.  We can rewrite this as:  $(3t(4 - t) - 16)(3t(4 - t) + 16) \geq 0$, or $(3t^2 + 2t + 16)(3t^2 - 12t - 16) \geq 0$,  at which point I got stuck.
74091
Problem: Show that if $(b,c)=1$, then for any integer $a$ we have $(a,bc)=(a,b)(a,c)$. [Hint: Prove that each member of the alleged equation divides the other.]  This is the way I proved it (without the hint): $(a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)$  But how should I prove it using that hint or at least a similar proof.