Let a,b be Natural numbers where a > 1. Also p is a prime number. If $ax^2 + bx + c = p$ for two distinct integer values of x, then prove that $ax^2 + bx + c \neq 2p$ for any integral values of x.  So this seems to be a pretty straightforward question and I got the final expression that is $p= a (x-y)(x-z)$ where y and z are roots of equations $ax^2 + bx + c - p$, so I got p has three factors while a prime number can only have 2 factors. But I couldn’t understand what would happen if $x-y = x-z = 1$. So essentially I need to try and prove that $y \neq z$. Thank you.
123904
We had received some questions on Quadratic equations, But I wasnt able to do one. Here it goes:  Let $a,b$ be natural numbers $a&gt;1$. Also, $p$ is a prime number. If $ax^2+bx+c=p$ for 2 distinct integral values of $x$. Then the number of integral roots of the equation $ax^2+bx+c=2p$ is ? Well I know the answer is $0$, But I am not able to get it properly.
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.
122716
$P(x) = 0$ is a polynomial equation having at least one integer root, where $P(x)$ is a polynomial of degree five and having integer coefficients.  If $P(2) = 3$ and $P(10)= 11$, then prove that the equation $P(x) = 0$ has exactly one integer root.  I tried by assuming a fifth degree polynomial but got stuck after that.  The question was asked by my friend.
47264
Let $a$ and $b$ be positive integers such that $a | b^2, b^2 | a^3, a^3 | b^4, b^4 | a^5, \cdots $ Prove that $a = b$.  My way is as follows: Let $A=v_p(a), B=v_p(b)$ be the exact power of a prime $p$ dividing $a$ and $b$ respectively. Then by assumption, $(2n-1)A \leq 2nB, n=1,2,3,\cdots.$ Hence $A \leq (2n/(2n-1))B$. Taking limit as $n$ tends to infinite yields $A \leq B.$ Likewise, $2nB \leq (2n+1)A, n=1,2,3,\cdots,$ by a similar argument, we have $B \leq A.$ So that $A=B$ and we are done. But I am rather unsatisfied with the proof because it involves limit. I wanna ask if there is any other alternative proof, thank in advances.
116479
Prove that if $p$ and $p+2$ are both prime integers greater than $3$, then $6$ is a factor of $p+1$.  Computing a few primes greater than $3$ modulo $6$ shows the following pattern $5,1,5,\dots$ Thus $$\begin{cases}p\equiv1,5 \pmod{6} \\ p+2 \equiv3,7 \equiv3,1 \pmod{6}\end{cases}$$ Also $p+1 \equiv 2,6 \equiv2,0 \pmod{6}$. I'm not sure how to continue from here... I would like to use modular arithmetic to show this. Is it necessary to remember that every prime can be written as $6n+1$ or $6n+5$ here?
49660
I wanted to prove the following question in an elementary way not using Bertrand postulate or analytic estimates like $x/\log x$. The question is  $$ p_{n+1}^2&lt;p_1p_2\cdots p_n,\qquad(n\geq4) $$  I made the following argument. Does anyone have some opinion or simpler ideas to complete.  We consider two cases: Case 1: $N=p_1p_2\cdots p_n-1$ is composite: then there will be a prime factor  $q\leq\sqrt{N}$, and of course $q$ is not any of the $p_i$'s. therefore $p_{n+1}\leq q\leq\sqrt{N}$. So $p_{n+1}^2&lt;N+1$.  Case 2: $N=p_1p_2\cdots p_n-1$ is prime. Then???
57988
I was working on number theory homework, and this problem was the second part to  problem.  Let $a,b,c,d,m$ be constants and $n$ and $k$ be coprime such that there are infinitely many values of $x$ that satisfy  $$ax+b=mn$$ $$cx+d=mk$$  Find the values of $ad-bc$ such that the above is true.   Now, according to the answer suggested above, these equations are true if $n \equiv b/m\pmod a,$ and $k \equiv d/m\pmod c,$ but I can't really find any insight on how to find the values of $x$ that would help with this part of the problem.
22046
I got a question to show that :      If $p$ is prime number, then   $$x^p - x \equiv x(x-1)(x-2)(x-3)\cdots (x -(p-1))\,\,\text{(mod }\,p\text{)}$$    Now I got 2 steps to show that the two polynomials are congruent:   Show they have same roots. Show the leading term's coefficients on both sides are the same.    For step 2, I know it is obvious $x^p$ has coefficient 1 on both sides.  For step 1, but I am not sure how to find all the solutions for $x^p -x$ modulo $p$.
82511
For a prime $p \equiv 1$ or $3$  (mod $8$), show that the equation $x^{2} + 2y^{2} = p$ has a solution.   I'd like to see an elementary proof.
19912
This is a question from Putnam and Beyond, and I have a question about the proof. The question is:  Show $x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ is irreducible over $\mathbb{Z}[X]$.  The proof first shows that the polynomial $P(x) = x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ only has zeros of absolute value $&gt; 1$. Then, it assumes it can be factored into two polynomials, namely, $P(X) = Q(X)R(X)$, and since $p = P(0) = Q(0)R(0)$, one of $Q(0), R(0) = \pm 1$, so WLOG $Q(0)$ has a zero of absolute value $&gt; 1$ which contradicts the fact that $P(X)$ only has zeros of absolute value $&gt; 1$. I don't see why $Q(0) = \pm 1$ means that it has a zero with absolute value greater than one. (The following is the proof in question.) &quot;We will show that all the zeros of $P (x)$ have absolute value greater than $1$. Let $y$ be a complex zero of $P(x)$. Then $0 = (y − 1)P(y) = y^p + y^{p−1} + y^{p−2} +···+ y − p.$ Assuming $|y| ≤ 1$, we obtain $p = |y^p + y^{p−1}+ y^{p−2}+···+ y| ≤ \sum_{i=1}^p |y|^ i ≤  \sum_{i=1}^p 1 = p.$ This can happen only if the two inequalities are, in fact, equalities, in which case $y = 1$. But $P (1) &gt; 0$, a contradiction that proves our claim. Next, let us assume that $P (x) = Q(x)R(x)$ with $Q(x)$ and $R(x)$ polynomials with integer coefficients of degree at least $1$. Then $p = P (0) = Q(0)R(0)$. Since both $Q(0)$ and $R(0)$ are integers, either $Q(0) = ±1$ or $R(0) = ±1$. Without loss of generality, we may assume $Q(0) = ±1$. This, however, is impossible, since all zeros of $Q(x)$, which are also zeros of $P (x)$, have absolute value greater than $1$. We conclude that $P(x)$ is irreducible.&quot;
38510
Prove: If a prime number $p\in \mathbb N$ is from the form $p=4k+3,k\in \mathbb N$, then its also a prime number in $\mathbb Z[i]$,i.e. if $p|(z_1\cdot z_2)$ then $p|z_1$ or $p|z_2$.  I dont have any idea how to solve it, so I am looking for some hints.  Thanks in advance :)
35496
Let N denote the number, and suppose $N=a^p \times b^q \times c^r \cdots$, where $a,b,c,\cdots$, are different prime numbers and $p,q,r,\cdots$ are positive integers.Then it is clear that each term of the product   $$(1+a+a^2+\cdots+a^p)(1+b+b^2+\cdots+b^q)(1+c+c^2+\cdots+c^r)\cdots$$  is a divisor of the given number,and that no other number is a divisor; then how could we use this to show/find that the number of ways in which this $N$ can be resolved into two factors which are prime to each other is $2^{n-1}$,where $n$ is the number of different prime factors in $N$?
38627
Let $a$ and $b$ be integers. Show that if $(a,b) = 1$ then $(a,a + b) = 1$.   I'm stuck on how to do this proof. I tried using the fact that $ax+by=1$ but that didn't seem to help.
4827
I came across this problem and I believe Lagrange's theorem is the key to its solution.  The question is:     Let $p$ be an odd prime. Prove that there is some integer $x$ such that $x^2 \equiv −1 \pmod p$ if and only if $p \equiv 1 \pmod 4$.   I appreciate any help.  Thanks.
98720
I proved that if $n$ is a prime, then $p(x)=x^{n-1}+\cdots+x+1$ is irreducible over $\mathbb Z$. But, I don't know how to prove that if $p(x)$ is irreducible over $\mathbb Z$, then $n$ is prime. Can you give me a hints?
102129
Can someone provide a proof that every quadratic residue, when working in $\mathbb Z_p$, where $p$ is a prime, has exactly two roots? Indeed, there cannot be only one root as for any $a^2$, we know $a$ and $-a$ are both roots. So then we need only show that there cannot be more than two- does this follow from some argument for the number of roots a quadratic can have modulo a prime? I'm not well-acquainted with the theorems pertaining to prime moduli.
72096
Suppose the quadratic polynomial  $P(x) = ax^2 + bx +c $ has positive coefficients $ a, b, c $ in an arithmetic progression in that order. If $P(x) = 0$ has integer roots $\alpha $ and $\beta $, then $\alpha + \beta + \alpha\beta $ equals?  1) $3$  2) $5$  3) $7$  4) $14$  It's a single choice correct problem. All genuine answers are welcome :)  The question is from a famous Indian Scholarship test 'KVPY' for high school students. The official website doesn't provide any solutions to it that is why I am asking for help.  My go on the question   we know that in a quadratic equation $ax^2 + bx +c$ with roots $\alpha, \beta$ the sum of roots is $\alpha +\beta = \frac{(-b)}{a} $ and product is $\alpha\beta = \frac{c}{a} $  So $\alpha + \beta + \alpha\beta  = \frac{(-b+c)}{a}$ and since $a,b,c$ are in arithmetic progression $\frac{(a+c)}{2} = b$  Something that I tried was also this -  $  a, b, c     $   are in AP (arithmetic progression)  $a-b , 0, c-b$ are in  AP (subtracting by $b$)  $1-\frac{b}{a}, 0, \frac{(c-b)}{a}$ are in AP  (dividing by $a$)  Now I can't proceed further. After a bit of plugging and chugging I still cant get an integer as an answer...