I know that if $({t\over p})=1$, then $t$ is a quadratic residue, which means that $x^2 \equiv t$ mod p has solutions. So is there any skills or methods that I can use to solve the equation given a prime $p$ and $({t\over p})=1$? Thanks a lot!
124268
Given a prime $p$ and number $a$, how do I find the smallest $n$ such that $n^a\equiv 1 \pmod p$? Is there a non trial-and-error method?  Edit: obviously removing $n=1$!
27657
According to the theorem, if  $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{p-1} =\frac{r}{q}$$  then we have to prove that $r\equiv0 \pmod{p^2}$.  (Given $p&gt;3$, otherwise $1+\dfrac{1}{2}=\dfrac{3}{2}$, $3 \not\equiv  0 \pmod 9$.)  I guess there's a $(\bmod p)$ solution for this, but I don't really get how to start it.
86327
Is there an efficient way (algorihtm) to compute the solutions of the congruence $x^n\equiv a \pmod p$?   $n\in \mathbb{N}$ $a\in\mathbb{Z}_p$ $p$ is a large prime number   Note that:   By efficient I mean computationally efficient even when the given prime $p$ is appropriate for use in cryptographic protocols (hard to find discrete logarithm in $\mathbb{Z}_p$) We do not know the prime factors of $p-1$.   It’s easy to find one solution when $\gcd(n,p-1)=1$.  I think this should be as hard as finding a Discrete Logarithm or as Factoring. Any ideas ? Thank you for your time.
7948
I split this into $x^2\equiv 1\pmod {7}$ and $x^2\equiv 1\pmod {13}$.  For $x^2\equiv 1\pmod {7}$, i did: $$ (\pm1 )^2\equiv 1\pmod{7}$$ $$(\pm2 )^2\equiv 4\pmod{7}$$ $$(\pm3 )^2\equiv 2\pmod{7}$$ Which shows that the solutions to $x^2\equiv 1\pmod {7}$ are $\pm1$.  For $x^2\equiv 1\pmod {13}$, i did: $$ (\pm1 )^2\equiv 1\pmod{13}$$ $$(\pm2 )^2\equiv 4\pmod{13}$$ $$(\pm3 )^2\equiv 9\pmod{13}$$ $$ (\pm4 )^2\equiv 3\pmod{13}$$ $$(\pm5 )^2\equiv {-1}\pmod{13}$$ $$(\pm6 )^2\equiv 10\pmod{13}$$Which shows that the solutions to $x^2\equiv 1\pmod {13}$ are $\pm1$.  Thus, I concluded that the solutions to $x^2\equiv 1\pmod {91}$ must be $\pm1$. I thought that $\pm1$ were the only solutions, but apparently I am incorrect! How do I go about finding the other solutions to this congruence?
3696
I have a problem in solving my number theory homework. My question is as follows:  Let $p$ be an odd prime. Prove that $a$ is a quadratic residue mod $p$ if and only if the $I_{g}(x)$ (index with respect to any primitive root of $p$) is even.  Please edit my writing. Thanks. Does anyone know where to start? Thank you very much for everything!
94191
Suppose that p ≡ 3 (mod 4)   and $r = \frac {p-1}2$  Show that $r! \equiv (−1)^k \pmod p$  where k is the number of non-quadratic residues modulo p which are smaller than $\frac p2$  I know from Wilson's Theorem for a prime p:  $(p−1)! \equiv −1 \pmod p$  I don't know where to start from ..   I managed to show that   $r! \equiv \pm 1 \pmod p$  any help or hint will be appreciated
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?
111049
I need to determine all primes $p$ for which $5$ is a quadratic residue modulo p.    I think I'll need to use quadratic recprocity laws to do this, i.e., I need to need to find numbers $p$ where $x^2$ is congruent to $5 \bmod p$. I'm ok doing this for single values of $p$. But how do I find all primes for which this holds?   Thanks.
17561
For which prime numbers p does the congruence $x^2 + x + 1 \equiv 0 \pmod{p}$ have solutions?  We've recently learnt about quadratic reciprocity in class, however I am not sure how to tackle this problem. I have tried starting with the $b^2-4ac$ (discriminant) but that hasn't really helped.
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
62230
Help needed in understanding proof: Every odd prime $p$ has exactly $(p-1)/2$ quadratic residues and $(p-1)/2$ quadratic nonresidues.   We assume there exist $k$ incongruent quadratic residues and each yield two solutions of the equation $x^2 \equiv a (\mod p)$. Also each of these solutions are incongruent, right ? So we have $2k$ solutions to quadratic congruences.  Then there are $p-1$ squares of the least residues, $1$ through $p-1$. But these need not be incongruent (consider $p = 7$ and $1^2, 6^2$). Then $2k = p-1$, why ?  Could someone explain to me in details what the reasoning is ?
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.]
106183
Find the number of solutions of $x^2=x\mod m$ for any integer $m$.  I was thinking about bringing the $x$ over to the other side and somehow apply chinese remainder thm.
122809
So I'm trying to find all solutions of $x^2 \equiv$ $a$ mod $p$ and for some reason the formulas that are suggested everywhere online (for example, ) say that, if p is an odd prime and you have a congruence like this:    Then you can find the solutions with this (if there are any):    That's what I was taught as well, but when entering examples into Wolfram Alpha, I've noticed that there are actually more solutions - these formulas just give you one of them! For example:   If I need to find the solutions of: x^2 ≡ 8 (mod 23) and I use the above formulas, I get the solution: x = +- 13, but  (and it does). So why is that? Why do these formulas only give you 1/2 of the answer and why can't I find a way to find the whole answer (excluding the way where you draw the table with x^2 mod p for x from 0 to p and)?  I realize that I'm probably doing something wrong and I'm not very good at this in the first place, but I've spent hours searching for an answer and couldn't really find anything (that I could understand, at least).
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.
44874
$$\text{ If } p \neq 2 \text{ is a prime, we know that: }$$ $$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$$  According to this,prove that: $$p \equiv 1 \pmod 4 \Rightarrow x^2+1\equiv 0 \pmod p \text{ has a solution}$$ $$\text{Which is the solution?}$$ Solve: $\displaystyle{ x^2+1 \equiv 0 \pmod{29} \text{ , } x^2+1 \equiv 0 \pmod{37}}$  Inversely,prove that $x^2+1 \equiv 0 \pmod p \text{ has a solution } \Rightarrow p \equiv 1 \pmod 4$  I have thought the following:   $$\left(\left(\frac{p-1}{2}\right)!\right)^2-(-1)^{\frac{p+1}{2}} \equiv 0 \pmod p \Rightarrow \left(\left(\frac{p-1}{2}\right)!\right)^2+(-1)^{\frac{p+3}{2}} \equiv 0 \pmod p $$   As $p=4k+1$, $(-1)^{\frac{p+3}{2}}=(-1)^{\frac{4k+4}{2}}=1$  So,we have: $$\left(\left(\frac{p-1}{2}\right)!\right)^2+1 \equiv 0 \pmod p$$  So,we conclude that $x^2+1 \equiv 0 \pmod p \text{ has a solution , this one: } (\frac{p-1}{2})!$  Is it right or have I done something wrong?   So,are the solutions of $x^2+1 \equiv 0 \pmod{29} \text{ and } x^2+1 \equiv 0 \pmod{37}$,these one: $\displaystyle{(14)! \text{ and } (18)! \text{ respectively}}$ ? Are these the only solutions?? How can I show the inverse?
30278
Let p be an odd prime and let a,b ∈ Z such that p ∤ a. Prove that the congruence $(x^2-a)(x^2-b)(x^2-ab)$ ≡ 0 (mod p) is always solvable.   Not sure where to begin here.