Given the following module operation: $$8x + 51\pmod{10}$$ Details:  $x \equiv y \pmod{m}$ is by definition equivalent to  $m|(x−y)$. $x \equiv y \pmod{m}$ , $x,y\in \mathbb{Z}, m \in \mathbb{Z^+}$ iff $a \pmod {m} = b \pmod{m}$.  Question 1: Can we split it as follows based on module details above: $$8x + 51\pmod{10} = 8$$ $$51\pmod{10} = 1$$ Then, $$8x + 1\pmod{10} = 8$$ Question 2: Is it possible to subtract numbers from the module operation $1$ from both sides: $$8x \pmod{10} = 7$$ Question 3: Suppose we have the quantity $8x \pmod{10} = 7$, then what is the meaning that the inverse of $8 \pmod{10}$ is $7 \pmod{10}$? Given the inverse, how we can multiply both sides by it as follows as this was not covered in basic module operations in $$7\cdot 8x \pmod{10} = 7 \cdot 7 \pmod{10}$$ We can conclude that $$56x \pmod{10} = 7 $$
82626
Suppose,I have a number 1256. I want (1256 % 11)! That is 2 . But, here, if i try in this way,  ( (125 % 11 ) * 10 + 6 ) % 11 = 2. That is exactly the same answer 2 as above.  I am confused , how This two process gives the similar answer ??? Even if , I try this method for any number n,m  to find (n mod m) . How this works ?? Can anybody explain ?
99833
Currently working on this problem and I'm having trouble seeing how it goes from one line to the next.    $45x \equiv 63\mod 11$ goes to $x \equiv 8\mod 11$     Any help would be awesome thanks.  Please try to explain how it works to help me see thanks.
75685
The only solutions that I know of till now are $(x,y) = (1,1) \space , (19,7)$. We can note that: $$x^2-x+1 = y^3 \implies (2x-1)^2 = 4y^3-3$$ Thus, if odd prime $p \mid y$, then $(2x-1)^2 \equiv -3 \pmod{p}$ and thus, $-3$ is a quadratic residue. This implies that $p \equiv 1 \pmod{6}$ .  How can we further proceed into this problem?  Note: As mentioned in one of the links in the comments, if we instead replace $x$ by $x+1$, we get $x^2+x+1 = y^3$. Thus, the solutions for this is $(x,y) = (0,1) \space, (18,7)$.
113982
I can't figure this problem out.  Let $b_1, b_2, \cdots, b_{\phi(m)}$ be the integers between $1$ and $m$ that are relatively prime to $m$. $B$ is the product of these integers. Prove that either $B$ is congruent to $1$ or $B$ is congruent to $-1 \pmod{m}$. Could you give me some hint where to start?  EDIT: Please avoid use of group theory. I am an undergrad in my first semester of number theory.
42349
I would like to show that $a^n \mid b^n$ implies $a \mid b$  I thought I could convert it to congruences and work backwards, but as far as I remember, $a \equiv b \pmod{m}$ implies $a^n \equiv b^n \pmod{m}$, not the opposite, unless $m$ is prime. Is that right?  So I am not sure how to approach this one. Any ideas?  Thanks!
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Does $x^{y^z}$ equal $x^{(y^z)}$? If so, why?   Why not simply apply the order of the operation from left to right? Meaning $x^{y^z}$ equals $(x^y)^z$?  I always get confused with this and I don't understand the underlying rule. Any help would be appreciated!
16747
I'm having problems with an exercise from Apostol's Introduction to Analytic Number Theory.     Given $x$ and $y$, let $m=ax+by$, $n=cx+dy$, where $ad-bc= \pm 1$. Prove that $(m,n)=(x,y)$.   I've tried to give a proof, but I suspect it's wrong (or at least not very good). I would be very thankful for any hints/help/advice!  My proof:     We observe that since $ad-bc= \pm 1$, $ad=bc \pm 1$, and $(ad,bc)=1$.   Now, $(a,b) \mid m$ and $(c,d) \mid n$ but      $$(a,b) = ((d,1)a,(c,1)b)=(ad,a,bc,b)=(ad,bc,a,b)=((ad,bc),a,b)=(1,a,b)=1.$$      Similarly, we determine $(c,d)=1$. So, $1=(a,b) \mid m$ and $1=(c,d) &gt; \mid n$. But $(x,y)$ also divide $m$ and $n$. Since $(x,y) \geq (a,b)=(c,d)=1$, this implies that $(x,y)=m,n$. Hence   $(m,n)=((x,y),(x,y))=(x,y)$.
27820
In an Introduction to Abstract Algebra by Thomas Whitelaw, he gives examples of the congruence mod operation, such as $13 \equiv5 \pmod4$,  and $9 \equiv -1 \pmod 5$. But when I first learned about the modulo operation my junior year, I would have told you that $13 \equiv 1 \pmod 4$, and that $9 \equiv 4 \pmod 5$.     Is this just a difference in the definition of modulo? Or is this pretty typical (to not take it to the largest factor, or one over)?
63022
Given that $a \equiv b \pmod n$ and $c\equiv d \pmod n$, I need to prove that $ac \equiv bd \pmod n$  So far, I've only managed to deduce that $a+b \equiv c+d \pmod n$. I don't know if this is usable, but it's there, at least.  Any hint would be greatly appreciated!
26708
I'm currently learning how to find the inverse of a modulo with the Extended Euclid Algorithm and I stumbled upon a problem when finding an inverse when the $m&gt;p$ as for $m \equiv 1 \pmod{p}$  For example, in $$ 240x \equiv 1 \pmod{17} $$ what is the inverse? What are the steps to find it?
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To find last two digits  of $2^{100}$  I have just learned about modular arithmetic and I wanted to solve this problem. I only know about equivalence classes and about  $a=b \pmod n$. I have also learned about multiplication and addition of classes.  Can someone explain to me step by step on how to apply modular arithmetic to this problem?  Thanks
123173
The set $Z_n = \{ 0, 1, . . . , n-1 \}$ for $n ≥ 1$ is a group under addition modulo n. For any $j &gt; 0$ in $Z_n$, the inverse of $j$ is $n - j$. This group is usually referred to as the group of integers modulo n.  My understanding of this is as following, please let me know if I have gotten the main idea.  Let's pick $Z_3$ as an example.  Then the members of this group is $0,1,2$ let $j=2$ then $n-j=3-2=1$  it says the inverse of $j$ is $n - j$, by definition, an inverse in a group is such ab=identity. And here the identity is 0 under addition.   therefore, 2+1=3 and 3 mod 3 =0 which is the identity.   This applies to any n in $Z_n$  So, it satisfies as a group?
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I am looking for ways to prove that $\mathbb{Z}_m=\{0,1,2,\dots,m-1\}$ is a field under addition and multiplication $\bmod\ m$ iff $m$ is a prime.   I tried it this way: If $m$ is a prime $p$,it is obvious that most importantly, we need to prove that every non-zero element in $\mathbb{Z_p}$ has an inverse in $\mathbb{Z_p}$.We note that if $a,b\in \mathbb{Z_p}$, then $a\cdot b\in \mathbb{Z_p} \dots \boxed{*}$. Next, $a^p\equiv a \bmod p$ for prime $p$(the proof follows by induction on $a$).As  $\displaystyle 0\leq a\leq p-1$,$a\not =p$. Thus $$a^{p-1}\equiv 1 \bmod p$$ for prime $p$ which,together with $\boxed{*}$ guarantees the existence of an inverse in $\mathbb{Z_p}$.Proving the other properties is simple.  Conversely,if $\mathbb{Z_m}$ is a field,then every non-zero element of $\mathbb{Z_m}$ has an inverse.Note that inverse of an element is unique.For if $a\cdot b\equiv a\cdot c\equiv 1 \bmod m$,then $(a\cdot b)\cdot c\equiv c\equiv (a\cdot c)\cdot b\equiv b \bmod m$ i.e $m|(b-c)\implies b-c=0(0\leq a\leq m-1,a\in \mathbb{Z_m})$.So, inverses must be unique and every element is either an inverse of itself or has in inverse in the field.That allows us to conclude that $[(m-1)!]^2\equiv 1 \bmod m$ which can only mean that either  $(m-1)!\equiv 1 \bmod m$ or $(m-1)!\equiv -1 \bmod m$. If $m$ is not a prime, it has a prime factor, say $p$. $ 0\leq p\leq m-1\implies p|(m-1)!\implies p|1$or $-1$ which is absurd, a contradiction.Thus $m$ is prime.  I feel that this method is too involved and complicated. Can anyone please suggest some other method?(Some context: This is a problem from Finite Dimensional Vector Spaces by Paul Halmos. I am learning linear algebra for the first time.It seems some of the comments have been posted on the assumption that I know rings and fields; I don't. )
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How does $a\equiv b\pmod n$ implies $a$ and $b$ have the same remainder when divided by $n$?     I don't understand the huge jump from modular equivalence to having equal remainders.   I see by definition  $a\equiv b\pmod n\,$ implies $n\mid (a-b)\,$ so $\,a-b=nk\,$ so $\,a=b+nk$.     But I do not see how this implies that $a$ and $b$ have the same remainder when divided by $n$.
54300
Yeah so I'm kind of stuck on this problem, and I have two questions.   Is there a way to define a number mathematically so that it cannot be a multiple of $7$? I know $7k+1,\ 7k+2,\ 7k+3,\ \cdots$, but that will take ages to prove for each case.  Is this a proof? $$(7k + a)^3 \equiv (0\cdot k + a)^3 \equiv a^3 \bmod 7$$    Thank you.
19057
I know how to compute modular multiplicative inverses for co-prime variables $a$ and $b$, but is there an efficient method for computing variable $x$ where $x &lt; b$ and $a$ and $b$ are not co-prime, given variables $a$, $b$ and $c$, as described by the equation below?   $ a x \equiv c \mod b $   For example, given  $ 154x \equiv 14 \mod 182 $, is there an efficient method for computing all the possibilities of $x$, without pure bruteforce?  Please note that I'm not necessarily asking for a direct solution, just a more optimized one.    I do not believe that the Extended Euclidean Algorithm will work here, because $a$ and $b$ are not co-prime.   Edit:  Follow up question, since the first one had a shortcut:  Could the be computed efficiently as well?  $12260x \equiv 24560 \mod 24755$.  $107$ needs to be one of the computed answers.
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Let $D$ be an integer such that it is not a perfect square in   $\mathbb{Z}$.       If $D \equiv 1$ mod $4$, the set $$T:= \left\{\left( \begin{matrix} a &amp; b  \\ \frac{(D-1)b}{4} &amp; a+b \\ \end{matrix} \right) : a, b \in  \mathbb{Z} \right\}$$ is a subring of $M_2(\mathbb{Z})$ which is isomorphic   to the ring $$O := \left\{a+b\frac{1+\sqrt{D}}{2}: a, b \in \mathbb{Z} \right\}.$$   My problem is to find an isomorphism. However, I find this relation : $$\det \left( \begin{matrix} a &amp; b  \\ \frac{(D-1)b}{4} &amp; a+b \\ \end{matrix} \right)$$ $$= a(a+b)-\frac{b^2(D-1)}{4}$$ $$= \left(a+b\frac{1+\sqrt{D}}{2}\right)\left(a+b\frac{1-\sqrt{D}}{2}\right)$$  and $D \equiv 1$ mod $4$ $\iff$ $\left(\frac{1+\sqrt{D}}{2}\right)\left(\frac{1-\sqrt{D}}{2}\right) = -k$ for some $k \in \mathbb{Z}$  Does someone could give me a little hint to help me complete this problem?