Find all solutions that aren’t congruent to each other in modulo $105$ of   $53x \equiv 62 \pmod{105}$  I’ve know that this equation has $1$ solution, because $(53 , 105) = 1$. But, how can I find that solution without brute-forcing?
15178
When I am faced with a simple linear congruence such as  $$9x \equiv 7 \pmod{13}$$ and I am working without any calculating aid handy, I tend to do something like the following:  "Notice" that adding $13$ on the right and subtracting $13x$ on the left gives: $$-4x \equiv 20 \pmod{13}$$  so that $$x \equiv -5 \equiv 8 \pmod{13}.$$  Clearly this process works and is easy to justify (apart from not having an algorithm for "noticing"), but my question is this: I have a vague recollection of reading somewhere this sort of process was the preferred method of C. F. Gauss, but I cannot find any evidence for this now, so does anyone know anything about this, or could provide a reference? (Or have I just imagined it all?)  I would also be interested to hear if anyone else does anything similar.
122878
I wish to find the number of solutions to the congruence $x^n \equiv 1 \mod p$ where $n$ is any natural number, $p$ prime. I tried using primitive roots and Fermat but I don't know where to go from here.
115222
For example $151x - 294\equiv 44\pmod 7 $. How would I go about solving that?  The answer says to simplify it into the $ax\equiv b\pmod c $ form first, but I have no clue on how to get rid of the $294$... help?
70704
For a fixed $n$ and $M$, I am interested in the number of unordered non-negative integer solutions to $$\sum_{i = 1}^n a_i = M$$  Or, in other words, I am interested in the number of solutions with distinct numbers. For $n = 2$ and $M = 5$, I would consider solutions $(1,4)$ and $(4,1)$ equivalent, and choose the solution with $a_1 \ge a_2 \ge ... \ge a_n \ge 0$ as the representative of the class of equivalent solutions.  I know how to obtain the number of total, ordered, solutions with the "stars and bars" method. But unfortunately, I cannot just divide the result by $n!$ since that would only work if all the $a_i$ are distinct.
32668
$$\begin{align*} x&amp;\equiv 1\pmod2\\ x&amp;\equiv 2\pmod3\\ x&amp;\equiv 3\pmod4\\ x&amp; \equiv 4\pmod5\\ x&amp;\equiv 5\pmod6\\ x&amp;\equiv 0\pmod7\\ \end{align*}$$  So the solution says we can eliminate $x\equiv 5(\bmod6)$ because the first two cases cover it, but I don't really know how it does. How do we solve it in cases like this where the moduli are not mutually relatively prime.
87594
How to list all integer solutions of   $m^2-n^5 = m - n$  Here $m$ and $n$ are some positive integers.  Also, I want to know the name of this type equations (if name exist).   Regards Rosy
41953
I can't solve the following:  Find the number of positive-integer solutions that satisfy $x_1\cdot x_2\cdot x_3\cdot x_4=1,000,000$.  Thanks.
90886
I'm still making my way along in Niven's Intro to Number Theory, and the title problem is giving me a little trouble near the end, and I was hoping someone could help get me through it.  Now $x^8\equiv 16\pmod{2}$ is solvable with $x\equiv 0\pmod{2}$, so I assume $p$ is an odd prime. From a theorem earlier in the text,      If $p$ is a prime and $(a,p)=1$, then the congruence $x^n\equiv a\pmod{p}$ has $(n,p-1)$ solutions or no solution according as $a^{(p-1)/(n,p-1)}\equiv 1\pmod{p}$ or not.   So since $(16,p)=1$, the problem reduces to showing that $16^{(p-1)/(8,p-1)}\equiv 1\pmod{p}$ holds for all $p$. I note that $(8,p-1)$ can only take values $2,4,8$. For $2$, the above equivalence is then $4^{p-1}\equiv 1\pmod{p}$, which is true by Fermat's little Theorem. For $4$, it is then $2^{p-1}\equiv 1\pmod{p}$, which again holds by FlT. However, the case where $(8,p-1)=8$ is throwing me off. At best I see that $16^{(p-1)/8}\equiv 2^{(p-1)/2}\pmod{p}$, but I'm not sure how to show this is congruent to $1$ modulo $p$. Maybe there's a more elegant way to do it without looking at cases. Thanks for any insight.
35054
Which of the following primes satisfy the congruence   $$a^{24}\equiv6a+2\pmod{13}$$  1) 41  2) 47  3) 67  4) 83  I am interested in Theorem statement, corollary, or Trick or Logic which solves this problem within one minute. Thank you in Advance
46053
I'm not sure if it's correct, but what I have so far is; $$21n^5 + 10n^3 + 14n ≡ (1 + 0 - 1) ≡ 0 \mod 5$$ but I'm having trouble solving it in $\bmod 3$. I have: $$21n^5 + 10n^3 + 14n ≡ (0 + (?) + 2),$$ I'm not sure how to solve the $10n^3$ part of the congruence in $\mod 3$. Also, if I have any mistakes please let me know.
95986
i just started solving mordell's equations and get a little bit stuck. For example:  $ y^2 = x^3 - 3$.  I know that $x$ must be odd, for if $x$ is even $y^2 \equiv 5 \pmod{8}$. So $x \equiv \{1,3\} \pmod{4}$. Now note the following if we add 4 at both sides:  $y^2 + 4 = (x+1)(x^2 -x +1)$. The right hand side is the same as $(x-1)^2 + x$  I now know $(x^2 -x +1)$ is divisible by a prime divisor $3 \pmod{4}$ if $x \equiv 3 \pmod {4}$ and divisible by $1 \pmod{4}$ if $x \equiv 1 \pmod{4}$.  So we get $y^2 + 4 \equiv 0\pmod{p}$, which is solvable if $\left(\frac{-2}{p}\right)$ = 1. (I don't think this step is valid). this is only true if the following occurs:  $p \equiv 1 \pmod{4}$ and $p \equiv +- 1\pmod{8}$.  or  $p \equiv 3 \pmod{4}$ and $p \equiv +- 3\pmod{8}$.  so $p \equiv \{1,3\}\pmod{8}$  but $y^2 + 4$ is either $4 \pmod{8}$ or $5 \pmod {8}$, so there are no solutions?  Am i solving this the correct way or is there a shorter path?  Kees
77243
I'm trying to learn how to use the Chinese Remainder Theorem (CRT), and in order to give some context:  We search for all $x ∈ Z$, where $Z$ is the set of integers.  $x≡a_1\pmod{m_1}$  $x≡a_2\pmod{m_2}$  ...  $x≡a_k\pmod{m_k}$  The easy case (which I can solve) is if all $m_i$, where $i=1,2,...,k$ are pairwise coprime.  Example:  $x≡4\pmod 5$  $x≡5\pmod 6 $  $x≡3\pmod 7$  Then the first equation is satisfied iff $x=4+5s$, for some $s ∈ Z$.  These $x$ also satisfy the second equation iff $4+5s≡_6 5 ↔ -s≡_6 1 ↔ s=-1+6t$, for some $t ∈ Z$. Thus $x=4+5(-1+6t)=-1+30t$.  Lastly, these $x$ also satisfy the third equation iff $-1+30t ≡_7 3 ↔ 2t ≡_7 4 ↔ t ≡_7 2 ↔ t = 2+7n$, for some $n ∈ Z$. Thus $x=59+210n$.  Now to my issue, I have the problem:  $x≡2\pmod 4$  $x≡3\pmod 5$  $x≡5\pmod 6$  Here $\gcd(4,6)=2$, so they are not coprime and I don't know how to solve this. Can someone please solve it and explain why the problem becomes more difficult to solve when $m_i$ are not pairwise coprime.
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.
104049
Prove that there does not exist an integer solution for the diophantine equation $x^5 - y^2 = 4$.   It's obvious that $x$ and $y$ are of the same parity. We can also claim that if $x$ is odd, then it is $1 \pmod 4$. Also, if $x$ and $y$ are even, then $y \equiv 2 \pmod 4\text{ since } x^5 \text{ is a multiple of } 32$ and if $y$ were a multiple of $4$ then it would be at a distance of at least $16$ from $x^5$ or be equal to it. These are my observations.  How should I proceed with the proof? Please give hints.
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?
115622
How many nonnegative integer solutions are there to the pair of equations  \begin{align}x_1+x_2+\dots +x_6&amp;=20  \\ x_1+x_2+x_3&amp;=7\end{align}  How do you find  non-negative integer solutions?
113858
So I'm trying to find a solution for the equation $$(65!)^2\equiv x\:(\text{mod}\: 131)$$ I'm given the hint to use Wilson's Theorem which states, that for a prime $p$ we have $(p-1)!\equiv -1\:(\text{mod}\:p)$.   The only thing I've managed to so far is insert my number in Wilson's Theorem: So I get $(130!)\equiv -1\:(\text{mod}\: 131)$. From here on I have no idea what to do.   Any help would be appreciated. Apologies for any english errors.