Proving the Congruence of Primes of the Form x² + 5y² (mod 20)

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The discussion centers on proving that an odd prime \( p \) expressible as \( p = x^2 + 5y^2 \) must satisfy \( p \equiv 1 \text{ or } 9 \text{ (mod 20)} \). Participants emphasize the use of basic genus theory for the proof, while clarifying potential misunderstandings regarding the interpretation of the problem statement. The conversation highlights the importance of precise mathematical expression and encourages further exploration of the topic.

PREREQUISITES
  • Understanding of quadratic forms, specifically \( x^2 + 5y^2 \)
  • Familiarity with modular arithmetic, particularly modulo 20
  • Basic knowledge of genus theory in number theory
  • Experience with prime number properties and their representations
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  • Study the properties of quadratic forms and their classifications
  • Learn about modular arithmetic and its applications in number theory
  • Explore genus theory and its implications for prime representations
  • Investigate other forms of primes and their congruences
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Mathematicians, number theorists, and students interested in advanced topics related to prime numbers and quadratic forms.

Shobhit
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If $p$ is an odd prime expressible as
$$p=x^2+5y^2$$
where $x,y$ are integers, then prove that $p \equiv 1 \text{ or }9\text{ (mod 20)}$.
 
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Re: Primes of the form $x^2+5y^2$

Very nice, I like this one. The proof follows immediately from basic genus theory, and I am not going to post it down here (spoiler purpose).

For beginners, I would give a hint :

A prime of the particular form is of the form 1, 3, 7 or 9 modulo 20

Balarka
.
 
Re: Primes of the form $x^2+5y^2$

Shobhit said:
If $p$ is an odd prime expressible as
$$p=x^2+5y^2$$
where $x,y$ are integers, then prove that $p \equiv 1 \text{ or }9\text{ (mod 20)}$.

It must be necessarly x even and y odd or x odd and y even. In the first case You set x = 2 n and y = 2 m + 1 so that is...

$\displaystyle p = 4\ n^{2} + 20\ m^{2} + 20\ m + 5\ (1)$

... and then...

$\displaystyle p \equiv 4\ n^{2} + 5\ \text{mod}\ 20\ (2)$

Observing (2) it is evident that $\displaystyle p \equiv 1\ \text {mod}\ 20$ or $\displaystyle p \equiv 9\ \text {mod}\ 20$. The same result You obtaining supposing x odd and y even...

Kind regards

$\chi$ $\sigma$
 
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Re: Primes of the form $x^2+5y^2$

How does (2) follows from (1)?
 
Re: Primes of the form $x^2+5y^2$

mathbalarka said:
How does (2) follows from (1)?

I made a mistake writing $\displaystyle p \equiv 4\ n^{2}\ \text{mod}\ 20$ instead of $\displaystyle p \equiv 4\ n^{2} + 5\ \text{mod}\ 20$, so that I corrected it... very sorry! (Angry) ...

Kind regards

$\chi$ $\sigma$
 
Re: Primes of the form $x^2+5y^2$

Ah, I see, I have interpreted the question wrong. I interpreted is as : p is an odd prime expressible as x^2 + 5y^2 if and only if p is 1 or 9 modulo 20.

But perhaps OP must have intended exactly this? Can Shobhit clarify whether he did a typo or not?
 
Re: Primes of the form $x^2+5y^2$

mathbalarka said:
Ah, I see, I have interpreted the question wrong. I interpreted is as : p is an odd prime expressible as x^2 + 5y^2 if and only if p is 1 or 9 modulo 20.

But perhaps OP must have intended exactly this? Can Shobhit clarify whether he did a typo or not?

Balarka, the question is correct. You had interpreted it correctly.
 
Re: Primes of the form $x^2+5y^2$

Shobhit said:
Balarka, the question is correct. You had interpreted it correctly.

I see, thank you for clarifying.

One can carry on with my hints on the #2 then.
 
Last edited:

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