Consider $5x+3y=4$ and $3x+6y=1.$ List the set of primes for which this system of linear equations does not have a solution in the field $Z_p.$
122818
Let $p$ be a prime and consider the field $\Bbb Z_p$. List the primes for which the following system of linear equations does not have a solution in $\Bbb Z_p$:  $$ \begin{align} 5x + 3y &amp;= 4 \tag{1}\\ 3x + 6y &amp; = 1\tag{2} \end{align}$$  My try is as follows:  Determinant of the coefficient matrix is $21$. $21$ will be 0 if $p=3$ or $7$. So the answer will be $3$ and $7$. is it correct?
90960
Let $x$ and $y$ be integers not congruent to $0$ modulo $p$ where $p$ is a prime. Prove that if $p \equiv 3 \pmod{4}$ then $x^2+y^2 \not \equiv 0 \pmod{p}$.   I thought about proving this algebraically and by contradiction. So we say $x^2+y^2 = pz$ and have to show this is impossible.
106571
The question is from Zeitz's ''The Art and Craft of Problem Solving:"      Find all positive integer solutions $x,y,z,p$, with $p$ a prime, of the equation $x^p + y^p = p^z$.   One thing I noticed is that   $$ \frac{x^p + y^p}{x + y} = \sum_{i=0}^{p-1}x^{(p-1)-i}(-y)^i \implies (x+y) |p^z \implies x + y = p^n, \text{ }n &lt; z   $$  It is also not hard to determine all solutions for $p =2$. After this, however, I am at a loss. I don't know what restrictions I can impose to try to narrow down the solution set; for example, the class of solutions  $$ p=3, x = 3^n, y = 2\cdot3^n, z = 2 + 3n $$  for $n \ge 0$ show that $x+y = p^n$ cannot be sharpened. (Let me note that these are the only other solutions I have found besides the ones for $p = 2$.)  Could anybody give me a push in the right direction on this problem?
107737
Assume $0 &lt; a_0 \leq a_1 \leq \cdots \leq a_n$, show that equation   $$ p_n(z) = a_0 z^n + a_1 z^{n-1} + \cdots + a_{n-1} z + a_n = 0 $$   does not have root in $|z| &lt; 1$. Here $z$ is complex.   I started by considering the case in which $n = 2$, and tried to use the triangle inequality, while no progress was made. Any hints will be appreciated.
14947
Find all pairs of prime nummbers $p,q$ such that $p^3 - q^5 = (p+q)^2$.  It's obvious that $p&gt;q$ and $q=2$ doesn't work, then both $p,q$ are odd. Assuming $p = q + 2k$ we conclude, by the equation, that $k|q^3 - q - 4$ because $\gcd(k,q)=1$ (else $p$ is not prime) and $k=1$ has no solution.  I also tried to use some modules, but I couldn't.
45385
What I need to show is that  For $\gcd(ab,p)=1$ and p is a prime, the number of solutions of the equation $ax^2+by^2\equiv 1\pmod{p}$ is exactly $$p-\left(\frac{-ab}{p}\right)\,.$$  I got a hint that I have to use Legendre symbol from the answer. I think that I may count a solution one by one. What I did : $$(ax)^2 \equiv a-aby^2 \pmod{p}$$ It suffices to count $y$ such that $(\frac{a-aby^2}{p})=1$. I tried to use the complete residue system or a primitive root but it didn't work. The factorization also didn't work. I think that the pigeonhole principle may not work because it just says the existence. Thanks in advance.
51927
Prove that there doesn't exist integer $x \ge 3$ such that $x^2-1$ is prime.
88746
In the book, "Elementary Number Theory - 6th Edition" written by David M. Burton, I have a question. The problem is  If $p$ and $p^2+8$ are both prime numbers, prove that $p^3+4$ is also prime. (p. 58 problem 3.3 #20)  I don't know how to prove this. Thank you very much if you solve this question.
98853
Prove: Given a prime $p$, there are infinitely many $n\in \mathbb{Z}^+$, for   which $pn+1$ is a prime.   This is a simplified version of , so is there any elementary solution to this problem?  A proof by contradiction is recommended. Just like the proof of the existence of infinitely many primes by Euclid.
91210
Given a non-negative integer $p$. What is the number of solutions of $x_1+2x_2+3x_3 + \cdots + nx_n = p$, where the $x_i$'s are non-negative integers.   Can we answer this by using number of solutions of $x_1+x_2+x_3 + \cdots + x_m = q$ for any $m,q$?
75582
Why does $x^2+41=y^3$ have no integer solutions?  I know how to find solutions for some of the Mordell's equations $(x^2=y^3+k)$ (using $\mathbb{Z}[\sqrt{-k}]$, and arguments of the sort). Still, I can't find a way to prove this particular equation has no integer solutions.   I know it doesn't because there are lists of solutions for $k \in [-100,100]$, like this one  and for what I've seen, the argument one uses to say a given Mordell equation has not solution is based on congruences, like in .  Still, I can't seem to figure out why the one with $-41$ (meaning $x^2=y^3-41$) has no solutions. There's a theorem in the Apostol book (p191) that says that such an equation has no solution if $k$ has the form $k=(4n-1)^3-4m^2$, with $m$ and $n$ integers such that no prime $p\equiv -1 \pmod 4$ divides $m$. I've tried with this but can't find $m$ and $n$.  Any ideas?
119709
How can one prove that numbers of the form $4x^2+1$ can only be divided by primes of the form $4y+1$ (e.g. there is no $x$ for which $7$ divides $4x^2+1$)?   On a quick lookup, the statement is given as the first comment in  and of course, manual testing concurs with this
43693
I have to prove that the equation $x^2+y^2+z^2 = n$ only has a solution in the 2-adic integers $\mathbb{Z}_2$ if and only if $n \neq 4^m(8k+7)$, with $k,m \in \mathbb{Z}_{&gt;0}$. In the previous exercises, I already showed that this equation has a solution in $\mathbb{Z}_p$ for all odd primes by using the multivariate Hensel-Rychlik lemma but I have no clue how I could solve this problem.
28115
I saw this question     Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $b ∈ \mathbb{Q}^m$. Suppose   that the system of linear equations $Ax = b$ has a solution in   $\mathbb{R}^n$. Does it necessarily have a solution in $\mathbb{Q}^n$?   and I thought I'd give an interesting, possibly wrong, approach to solving it. I'm not sure if such things can be done, if not maybe you can help me refine.    I considered the form of the equality as  $$ A^{(1)}x_1+\cdots+A^{(n)}x_n=b, $$  where $A^{(i)}$ is a column vector of $A$. I then noticed that for $x_i\in\mathbb{R\setminus Q}=\mathbb{T}$ then, and this is where I think I'm doing something forbidden, each $x$ has the represenation  $$ x_1  =  k_{11}\tau_1+\cdots+k_{1p}\tau_p \\  x_2= k_{21}\tau_1+\cdots+k_{2p}\tau_p \\ \vdots \hspace{4cm} \vdots \\ x_n=k_{n1}\tau_1+\cdots+k_{np}\tau_p,$$  where $\tau_i$ is a distinct irrational number, $k_{ij}\in\mathbb{R}$, and $p$ is the number of such distinct irrational numbers. I wound this out, but there may be a discrepancy with $p$ and $m$. I feel this method can lead me to the answer, but I'm not sure where to go from here.  EDIT$^1$:  I end up getting something like this, I believe, after substitution:  $$ A(k^{(1)}\tau_1+\cdots+k^{(p)}\tau_p)=b$$  Here, $k^{(i)}$ is the vector  $$\begin{pmatrix} k_{1i} \\ \vdots \\ k_{ni} \end{pmatrix}.$$  EDIT$^2$:  I think there is no discrepancy with $p$ and $m$ because $A\in M_{m\times n}(\mathbb{Q})$, $K\in M_{n\times p}(\mathbb{R})$, and $\tau\in M_{p\times 1}(\mathbb{T})$, so  $$ (m\times n)\cdot (n\times p)\cdot (p\times 1) = m\times 1. $$
20547
Prove if $p$ is a prime then $p \mid \binom pk$ for $k\in\{1,\ldots,p-1\}$  I don't really know where to begin with this one.   I can see that I have to use the fact that $p$ is prime somewhere - the same is not true for composite numbers, for example $4\nmid 6=\binom42$.  I have checked that this is true for the first few primes:    $3$ divides $\binom 31=\binom32=3$ $5$ divides both $\binom 51=\binom 54=5$ and $\binom 52=\binom 53=10$ $7$ divides $\binom71=\binom76=7$, $\binom72=\binom75=21$ and $\binom73=\binom74=35$.
8423
If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$.  I.e. if $4^n+2^n+1$ is prime, prove that $n=3^m$, where $m\in\mathbb N\cup\{0\}$.  I don't know how I could solve this, I don't have any ideas. I know that $4^n+2^n+1$ is obviously odd. Assume $m=0$, i.e. $n=1$. Then $3$ is prime. So that doesn't disprove what we're trying to show.  So assume $m&gt;0$. Then $4^n+2^n+1&gt;3$, so $4^n+2^n+1$ is either $3k+1$ or $3k+2$, where $k\in\mathbb N$. So either $4^n+2^n$ or $4^n+2^n-1$ is divisible by 3. I think it's not something that could help out at solving this. Just a little observation. I don't have any ideas.
12833
Let $a,b,c$ be $3$ distinct integers, and let $P$ be a polynomial with integer coefficients.Show that in this case the conditions $$P(a)=b,P(b)=c,P(c)=a$$ cannot be satisfied simultaneously.  Any hint would be appreciated.