Waste less time on Facebook — follow Brilliant.
×

Help!!!

Pls help me solving this problem.

Note by Shivamani Patil
3 years, 1 month 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

\(N^{2}-1=(N+1)(N-1)\) So no such primes!! (Except when N-1=1)

Similarly for \(N^{2}-4=(N+2)(N-2)\) No primes!! (Except when N-2=1, Oh but wait! It implies N=3 and \(3^{2}-1=8\) which is not a prime!)

What I think is for all perfect squares a, at most a single prime may exist!

For non perfect squares, I am getting an intuition that infinite primes may exist!

Pranjal Jain - 3 years, 1 month ago

Log in to reply

what about \({ n }^{ 2 }-3\)

Shivamani Patil - 3 years, 1 month ago

Log in to reply

\(4^{2}-3=13\) so it may be a prime number! And I guess maybe ∞ such prime numbers

Pranjal Jain - 3 years, 1 month ago

Log in to reply

Well if we talk about your question if we consider \(N^{2}\)-2, we will get the primes when N=odd number except 1 and if we consider \(N^{2}\)-5 , we will get the primes when N= even number except 2...

In my view this works the best!!!

Jaiveer Shekhawat - 3 years, 1 month ago

Log in to reply

You made a good observation that if \(N\) is even, then \( N^2 - 2 \) is even and hence not a prime if \( N > 2 \).

However, this does not imply that if \(N\) is odd, then \( N^2 - 2 \) must be a prime. For example, \( 11 ^ 2 -2 = 119 = 7 \times 17 \). We can show that if \( N \equiv 11 \pmod{14} \) , then \( N^2 - 2 \) is a multiple of 7 and hence not prime.

Calvin Lin Staff - 3 years, 1 month ago

Log in to reply

7*19=133.

Shivamani Patil - 3 years, 1 month ago

Log in to reply

@Shivamani Patil Yeah! @Calvin Lin Typoed! 119=7×17

Pranjal Jain - 3 years, 1 month ago

Log in to reply

@Pranjal Jain @shivamani patil @Pranjal Jain Thanks! Fixed the typo.

Calvin Lin Staff - 3 years, 1 month ago

Log in to reply

well infinitely many prime numbers are of the form 6k±1, it will cover all the primes....

Jaiveer Shekhawat - 3 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...