Waste less time on Facebook — follow Brilliant.
×

Is this proof of 4th Landau's problem correct?

Dirichlet's theorem

For any two coprime positive integers \(a\) and \(d\), there are infinitely many primes of the form \(a + nd\), where \(n\) is a non-negative integer.

Argument: There are infinitely many prime of the form \(n^2 + 1\).

Proof:

Let \(n > 0\). For \(n = 1, n^2 + 1 = 2\) which is prime. Otherwise, if \(n^2 + 1 \neq 2\) then \(n^2 + 1 = (n)(n) + 1\) and \(\gcd(n,1) = 1\). Thus, by Dirichlet's theorem there are infinitely many prime of the form \((n)(n) + 1\).

Note by Paul Ryan Longhas
1 year, 5 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Wait this is still an unsolved problem right? The proof may seem correct because \(n\) is used in two different contexts in the expression \((n)(n)+1\), one \(n\) is given/fixed while the other \(n\) not. Dirichlet can only tell us that there's infinitely many \(m\) such that \(nm+1\) is a prime, but it doesn't guarentee that \(m=n\).

Xuming Liang - 1 year, 5 months ago

Log in to reply

Put n=8 Then from n^2+1 We get 8^2 +1 = 64+1 =65 Which is not prime

Aaryan Saha - 1 year, 5 months ago

Log in to reply

There are infinitely many primes of the form \(n^2+1\) doesn't mean that all the numbers of the form \(n^2+1\) are primes. For example, there are infinitely many positive integers which can be termed as prime numbers and so there are infinitely many primes. Does this mean that all positive integers are prime?

Tapas Mazumdar - 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...