I'm trying to understand how to know the numbers of solutions for x^2=a(mod m). I know that if a has roots they are solutions but how could I know the rest of them? Is there a need to use the Chinese reminder here? if so how?
120715
Let $a\in \mathbb{Z}/p\mathbb{Z}$ and $m$ is an odd number. Then show that the number of solutions to $x^2\equiv a\pmod m$ is $$\prod\limits_{p|m}\left(1+\left({\tfrac {a}{p}}\right)\right)$$  N.B I know I have to use CRT but how to get the expression?
100189
Find the number of solutions to the equation $$2011^x+2012^x+2013^x-2014^x=0$$  The answer seems to be zero, but I have no idea why. Please avoid considering complex solutions and other scary things.
92378
As suggested on Mathoverflow () I am transfering this question to math-stackexchange:  I am looking for integer solutions of the following algebraic equations:  $ 2a^{2} + 2b^{2} - c^{2} - d^{2} = 0 $ Preferentially the solutions should obey $a+b+c+d=0$.   By inspection I found the following solutions: $(a,b,c,d)=(1,0,1,1)$ , $(a,b,c,d)=(0,1,1,1)$, $(a,b,c,d)=(1,1,-2,0)$  and $(a,b,c,d)=(1,1,0,-2)$.   Additional solutions can be generated by swapping the sign of $a,b,c,d$ or  by scaling the solutions that I have given by an integer.   As a theoretical physicist I am rarely working with these diophantine equations and hence I am wondering what other solutions of the equation exist that I have not taken into account.   I am looking forward to your responses.
67429
if $m$ is an integer greater than 2, and a primitive root modulo $m$ exists, prove that the only incongruent solutions of $x^2 \equiv 1 \mod m$ are $x \equiv \pm 1 \mod m$.  I know that if a primitive root mod $m$ exists, then $m = 1, 2, 4, p^m,$ or $2p^m$, where p is an odd prime and $m$ is a positive integer.  Obviously I can check the cases where $m = 1, 2, 4$ by hand but I am having trouble proving the other two cases.
58191
I am reviewing the proof of the Chinese Remainder Theorem and I am confused at one of the concluding sentences in my textbook.  I understand why there is a solution for relatively prime $m,n$ such that for arbitrary integers $s,t$ that $ x \equiv s$ (mod m) and $x \equiv t$ (mod n) but the concluding sentence says that now the solution is  $ x$  mod(mn) . I do not understand this last statement.
24810
Consider the function   $$ f(x) = 2\cos((x^2+x)/6) - 2^x - 2^{-x}. $$  Clearly $x= 0$ is a root of equation so number of real roots can not be 0 but how to figure out question how many ? From where does this observation start?
80232
Find all the primes $p$ so that $$x^3+x \equiv 1 \pmod p \tag{1}$$ has integer solutions.   We consider $x$ and $y$ are the same solutions iff $x\equiv y \pmod p.$  We can prove that $(1)$ cannot have exactly $2$ solutions unless $p=31$. When $p=31$, the solutions of $(1)$ are $x \equiv {17\text{ (double), } 28} \pmod {31}.$  My question is: when does $(1)$ have no solution, and when does $(1)$ have $1$ solution and when does $(1)$ have $3$ solutions? (As Ma Ming has pointed out, an equation of degree $3$ has no more than $3$ solutions.)  I have proved that if $p \not = 31$, and $a$ is a solution of $(1)$, then $(1)$ has $3$ solutions iff  $$\left(\frac{a-1}{p} \right)=\left(\frac{a+3}{p} \right),$$ here $\left( \frac{a}{p} \right)$ is the Jacobi symbol. Thanks in advance!
28656
For any real numbers $a$ and $b$ and even natural number $n$, Show $x^n+ax+b=0$ has most two solutions for all $x$ in $\mathbb{R}$.   let $f(x)=x^n+ax+b$. Set $f(x)$ and $a$ be $0$, then we have $x^n+b=0$. So $x=\pm b^{1/n}$. $f$ has two solutions. To yeld a contradiction. Now assume that $f$ has more than two solutions. As $f$ is a polynomial on $\mathbb{R}$, it is differentiable on $\mathbb{R}$; thus $f'(x)=nx^{n-1}+a$. Set $f'(x)=0$, then $0=nx^{n-1}+a\Rightarrow x=\left(\frac{-a}{n}\right)^{1/(n-1)}$. At $x=\left(\frac{-a}{n}\right)^{1/(n-1)}$, $f(\left(\frac{-a}{n}\right)^{1/(n-1)})$ is the only extreme value of $f$ which contradicts with $f$ has more than two solutions.    Can someone check this solution? I am not sure right or not. If not right, please give me a hint or suggestion. Thanks
7289
I understand there is a method for solving simultaneous modular equations. For example; $$x = 2 \mod{3}$$ $$x = 3 \mod{5}$$ $$x = 2 \mod{7}$$ We find numbers equal to the product of every given modulo except one of them - giving $5 \cdot 7$, $3 \cdot 7$ and $3 \cdot 5$. We then find the multiplicative inverses of these numbers with modulo equal to the number missing from the product. The numbers found are then 2, 1 and 1 in this case. The value of x is then given by: $$x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1 = 233 = 23 \mod{3\cdot5\cdot7}$$  But I do not understand how this method correctly gives the value of $x$. I understand that the Chinese remainder theorem proves that there is a unique value of $0\le x \lt 3\cdot5\cdot7 \mod{3\cdot5\cdot7}$ but can someone please explain why this method finds this value of x?
25251
We want to find the solution $x$ to the congruence system  $$\begin{align} x &amp;\equiv r_1 \!\!\!\pmod{\!m_1}\\ x &amp;\equiv r_2 \!\!\!\pmod{\!m_2}\end{align},\ \ {\rm where}\ \  \gcd(m_1, m_2) = 1$$  These can be rewritten as  $$\begin{align} x &amp;= r_1 + m_1 j\\ x &amp;= r_2 - m_2k\end{align}$$  for unknown integers $j, k$. So we set those two equations as equal and rearrange them to  $$m_1j + m_2k = r_2 - r_1$$  Now assume we perform $\text{egcd}(m_1, m_2)$, which gives us the solution to  $$m_1u + m_2v = \gcd(m_1, m_2)$$  Let the integer $\,h\,$ make $\,h \gcd(m_1, m_2) = r_2 - r_1$.  Then $\qquad hm_1u + hm_2v = h\gcd(m_1, m_2) =  r_2 - r_1 = m_1j + m_2k$  At this point I am lost because I don't think we can just assume  $hm_1u = m_1j$ and extract $j = hu$.   How do I get the values of $j$ or $k$ so I can get the value of $x$?
60238
For any integer $a,b$ let $N_{a,b}$ denote the number of positive integer $x&lt;1000$ satisfying $x= a \mod 27$ and $x=b \mod 37$. Then which of them is correct         there exist $a,b$ such that $N_{a,b} =0$    for all $a,b$, $N_{a,b}=1$   for all $a,b$, $N_{a,b}&gt;1$   there exist $a,b$ such that $N_{a,b}$ and  there exist $a,b$ such that $N_{a,b}=2$.      Since we know that by Chinese Remainder Theorem if $m_1, m_2, .., m_k$ are pairwise relatively prime positive integers, and if $a_1, a_2, .., a_k$ are any integers, then the  simultaneous congruences $x ≡ a_1 (mod\;m_1), x ≡ a_2 (mod\;m_2), ..., x ≡ a_k (mod\;m_k)$ have a solution, and the solution is unique modulo $m$, where  $m = m_1m_2⋅⋅⋅m_k$. Then how we will approach by acoording to this theorem. So please someone help me and explain it.
46840
1.find a for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ such that for every m real, f has real roots 2.find m such that for every a real, f has real roots  My ideea is to demonstrate that $\delta=4(m-a)^2 - 4(3am-2) = 4(m^2-5am+a^2+2) &gt;0$ But I have no ideea how to do this.
57117
Find out all solutions of the congruence $x^2 \equiv29 \mod 5$.  [Hint:Find the solutions of this congruence $\mod 5$ , $\mod 7$ , and $\mod 7$ , and then use the Chinese Remainder Theorem.]
10072
Possible Duplicate:          Coudl someone explain me the following: Why should $x^2+1=0$ have uncountable infinite many solutions $x\in\mathbb H$?  In my opinion it has only 4 solutions, namely $i^2=j^2=k^2=ijk=-1$ ?
38537
Solve for $z$: $$z^2+|z|=0$$ I found this question on my textbook but I unable to derive an answer to it as a modulus of $z$ is present. I am unable to take an approach please guide me ..thanks
90684
Question  What is the minimum possible value of $a^{2}+b^{2}$ so that the polynomial $x^{4}+ax^{3}+bx^{2}+ax+1=0$ has at least 1 root?  Attempt  I divided by $x^{2}$ and got $$x^{2}+\frac{1}{x^{2}}+2+a\left(x+\frac{1}{x}\right)+b-2=0$$ by letting $$x+\frac{1}{x}=X$$ the equation becomes: $$X^{2}+aX+(b-2)=0$$ $$\therefore X=\frac{-a\pm \sqrt{a^{2}-4b+8}}{2}$$ but I am not sure how to continue. If the polynomial has 1 root doesn't that is should be a double root? Are we counting multiplicity or not?
96127
I was working through some number theory problems , when I came across the following question :     Find all solutions of $a^2+b^2=2c^2$   My Solution (Partial) :        We can rewrite the above equation as : $c^2 = (a^2 + b^2)/2 $   So $\Rightarrow$ $a^2 , c^2 , b^2$ are in an Arithmetic Progression $\Rightarrow$ there exist infinite solutions   WLOG , let $a^2$ be $t$ , $b^2$ be $s$ and $c^2$ be $m$   The equation can be re-written as $ t + s = 2m$   Now , what I was thinking was that should I solve this Diophantine equation for $t$ &amp; $s$ in terms of $m$ ; substitute their values back into the equation and find out $m$ ; and then finally put this value of $m$ back into the values of $t$ and $s$ to derive a general solution      Can someone help me out ? Maybe a hint ...