# Tagged Questions

45 views

### Computing question: A quadratic which gives primes [closed]

This is about Project Euler Problem 27. The question is: Considering quadratics of the form $n^2 + an + b$, where $\lvert a \rvert < 1000$ and $\lvert b \rvert < 1000$ Find the product ...
90 views

### Distribution of integer solution pairs (x,y) for $2x^2 = y^2 + y$

I am looking for integer pairs $(x,y)$ that respect $$2x^2 = y^2 + y$$ For example $(6,8)$ is a solution for that. Simple solution is to enumerate on $x$ or $y$ and test if the corresponding ...
108 views

### number of integral values for which $x^2+19x+92$ is a perfect square.

number of integral values of x for which $x^2+19x+92$ is a perfect square=? I have no idea how to do this. Please help.
186 views

### Quadratic expression that generate primes

I recently learned that there exist quadratic expression that generate some primes and some of these equations generate more primes than others. In the following video, the person shows the following ...
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### Finding a prime $p$ to solve a quadratic congruence $\pmod{p}$

I have a congruence of the form $$ax^2+bx \equiv -1 \pmod{p},$$ where $p$ is an odd prime and $a,b \in \mathbb{Z}$. Given $a$ and $b$, is there a general method to finding $p$ such that the above ...
290 views

### Quadratic Equation Modulo an even composite

I am familiar with using the quadratic formula and Tonelli-Shanks with Hensel's Lifting Lemma to solve a quadratic equation, but how do I solve a quadratic equation in an even modulus? I can't use the ...
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### is there an analytic solution to $n^2+kn-d=m^2$ m,n integers

For $k=24,d=-17;m=8,n=3$, completing the square gives $(12+n)^2=m^2+161$ Where $161$ just happens to be the product of two primes $(q=7,p=23)$, so for large $k,m,n$ factoring may be very slow. ...
318 views

### Does $x^2 \equiv 211\pmod{ 159}$ have a solution?

Note that 159=3*53. The answer to this question is yes. I managed to find two of the solutions. They are $x=23,136$ but there are two more. The main question that I have is whether there is an easier ...