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\lhead{CS 355, FALL 2018}               
\rhead{Name: Hemanta K. Maji} %%% <-- REPLACE Hemanta K. Maji WITH YOUR NAME HERE

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\title{Homework 6}

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%%%%%%%%%%%% PROBLEM 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item {\bfseries Chinese Remainder Theorem.}
  Let $p$ and $q$ be distinct prime number. 
  Let $\alpha\in\{0,1,\dotsc,p-1\}$ and $\beta\in\{0,1,\dotsc,q-1\}$. 
  Then, we have seen earlier that there exists an integer $x$ such that it simultaneously satisfies $x=\alpha\mod p$ and $x=\beta\mod q$. 
  For brevity, we shall refer to this as $x= (\alpha,\beta)\mod(p,q)$. 
  
  In this problem, we shall prove a few interesting properties of the result and the fact that there exists a \ul{unique} $x\in\{0,1,\dotsc,N-1\}$, where $N=p\dot q$, that simultaneously satisfies the two equation. 
  \begin{enumerate}
  \item (5 points) Suppose that the integers $x$ and $y$ satisfy $x=(\alpha,\beta)\mod(p,q)$ and $y=(\alpha',\beta')\mod(p,q)$.
    Prove that the integer $x-y = (\alpha-\alpha',\beta-\beta')\mod(p,q)$. 
    
  {\bfseries Solution.}  \newline
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  \item (5 points) Suppose that the integers $x$ and $y$ satisfy $x=(\alpha,\beta)\mod(p,q)$ and $y=(\alpha',\beta')\mod(p,q)$.
    Prove that the integer $x\cdot y = (\alpha\cdot\alpha',\beta\cdot\beta')\mod(p,q)$. 
    
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  \item (5 points) Suppose $x$ and $x'$ are integers such that $x=(\alpha,\beta)\mod(p,q)$ and $x'=(\alpha,\beta)\mod(p,q)$.
    Prove that $N $ divides $(x-x')$, where $N=p\cdot q$. 
    
  {\bfseries Solution.}  \newline
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  \item (5 points) Prove that for every $\alpha\in\{0,1,\dotsc,p-1\}$ and $\beta\in\{0,1,\dotsc,q-1\}$ there exists a \ul{unique} $x\in\{0,1,\dotsc,N-1\}$ such that $x=(\alpha,\beta)\mod(p,q)$.
  
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  \item (5 points) Prove that for every element $x\in\{0,1\dotsc,N-1\}$ there exists \ul{unique} $(\alpha,\beta)$ where $\alpha\in\{0,1,\dotsc,p-1\}$ and $\beta\in\{0,1,\dotsc,q-1\}$ such that $x=(\alpha,\beta)\mod(p,q)$. 
   
  {\bfseries Solution.}  \newline
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  \end{enumerate}

 
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\item {\bfseries Proving $\bbZ_N^*$ is a group.} 
  Let $p$ and $q$ be two prime numbers, and $N=p\cdot q$. 
  Define 
    $$\bbZ_N^* = \left\{x\colon 0\leq x<N, \gcd(x,N)=1\right\}$$ 
  Let $\times$ be integer multiplication $\mod N$. 
  We shall prove that $(\bbZ_N^*,\times)$ is a group. 
  
  Our starting point is the result of Problem 1.e. that shows that every integer $x\in\{0,1,\dotsc,N-1\}$ has a unique $(\alpha,\beta)$ associated with it, such that $\alpha\in\{0,\dotsc,p-1\}$, $\beta\in\{0,\dotsc,q-1\}$, and $x=(\alpha,\beta)\mod (p,q)$. 
  \begin{enumerate}
  \item (5 points) Prove that $x\in\bbZ_N^*$ if and only if $x=(\alpha,\beta)\mod (p,q)$, such that $\alpha\in\{1,\dotsc,p-1\}$ and $\beta\in\{1,\dotsc,q-1\}$. 
  Remark: This result proves that $\abs{\bbZ_N^*}=(p-1)(q-1)$.  
  {\bfseries Solution.}  \newline
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  \item (5 points) (Closure) Suppose $x=(\alpha,\beta)\mod(p,q)$ and $y=(\alpha',\beta')\mod(p,q)$.
    Prove that $x\times y\in\bbZ_N^*$.

  {\bfseries Solution.}  \newline
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  \item (8 points) (Existence of identity) Find an element $e\in\bbZ_N^*$ such that $e=(\alpha,\beta)\mod (p,q)$ and for all $x\in\bbZ_N^*$ we have $e\times x=x$. 
  (That is, $e$ is the identity element)  
  
  {\bfseries Solution.}  \newline
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  \item (8 points) (Multiplicative Inverse) Suppose $x=(\alpha,\beta)\mod(p,q)$ and $x\in\bbZ_N^*$. 
    What is the element $y\in\bbZ_N^*$ such that $x\times y =e$? 
    {\bfseries Solution.}  \newline
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  \end{enumerate}

  
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%%%%%%%%%%%% PROBLEM 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item {\bfseries An Observation about Solving Equations.}
  Let $p$ and $q$ be distinct primes, and $N=p\cdot q$. 
  Suppose there exists one solution $x\in \{0,1,\dotsc,N-1\}$ such that $x^2 = a\mod N$. 
  Define 
    $$S(a)=\left\{X \colon X\in\{0,1,\dotsc,N-1\},X^2=a\mod N\right\}$$
  That is, $S(a)$ is the set of all solutions of $X^2=a\mod N$, where $X\in\{0,1,\dotsc,N-1\}$. 
  \begin{enumerate}
  \item (8 points) If $a\in\bbZ_N^*$ then prove that $\abs{S(a)}=4$. 
   
  {\bfseries Solution.}  \newline
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  \item (8 points) If $a$ is divisible by $p$ or $q$, then prove that we have $\abs{S(a)}=2$.
   
  {\bfseries Solution.}  \newline
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  \item (8 points) If $a=0$, then prove that we have $\abs{S(a)}=1$.
   
  {\bfseries Solution.}  \newline
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  \end{enumerate} 
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%%%%%%%%%%%% PROBLEM 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item {\bfseries Proving Bijection of $X^i$.} (25 points)
  Suppose $p$ and $q$ are primes, and $N=p\cdot q$. 
  We define $\times$ as integer multiplication $\mod N$. 
  The objective of this problem is to prove that the function $X^i\colon\bbZ_N^*\to\bbZ_N^*$ is a bijection, if $i$ is relatively prime to $(p-1)$ and $(q-1)$. 
  
  Suppose $X\in \bbZ_N^*$ such that $X=(\alpha,\beta)\mod(p,q)$, $\alpha\in\bbZ_p^*$, and $\beta\in\bbZ_q^*$. 
  Suppose $Y$ is a different element $\in \bbZ_N^*$ such that $Y=(\gamma,\delta)\mod(p,q)$.  
  
  If possible let $i$ be relatively prime to $(p-1)$ and $(q-1)$, and $X^i=Y^i$. 
  If this condition is true, then we have $(\alpha^i,\beta^i) = (\gamma^i,\delta^i)\mod(p,q)$. 
  This statement is equivalent to $0=(\alpha^i-\gamma^i,\beta^i-\delta^i)\mod(p,q)$. 
  By problem 3.c. we know that this equation has a unique solution $\alpha^i=\gamma^i\mod p$ and $\beta^i=\delta^i\mod q$. 
  
  Now, all that remains is to prove the following result. 
  Suppose $\alpha,\gamma$ are distinct elements in $\bbZ_p^*$. 
  If $\gcd(i,p-1)=1$, then it is impossible to have $\alpha^i=\gamma^i\mod p$. 
  In your proof, you can assume that $\bbZ_p^* = \{g^0,g^1,\dotsc,g^{p-2}\}$, for some $g\in \bbZ_p^*$. 


  {\bfseries Solution.}  \newline
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\end{enumerate}
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%%%%%%%%%%%% PLEASE LIST COLLABORATORS BELOW  %%%%%
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{\bfseries Collaborators :} \newline 
% ENTER THEIR NAMES HERE  

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