Prove that \(512^3 + 675^3 + 720^3 \) is a composite number.

I want to solve this problem be without computer assisted solution.

Note by Chung Kevin
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

Note first that \(a = 512 = 2^{9}, b = 675 = 3^{3}5^{2}\) and \(c = 720 = 2^{4}3^{2}5,\) and thus \(2c^{2} = 3ab.\)

Next, note that

\(a^{3} + b^{3} + c^{3} = a^{3} + b^{3} - c^{3} + 2c^{2}c = a^{3} + b^{3} - c^{3} + 3abc = (a + b - c)(a^{2} + b^{2} + c^{2} - ab + ac + bc).\)

Thus \(a + b - c = 512 + 675 - 720 = 467\) divides the given expression, i.e., the given expression is composite.

Brian Charlesworth - 3 years, 1 month ago

Log in to reply

Nice observation with \( 2c^3 = 3ab \)!

Calvin Lin Staff - 3 years, 1 month ago

Log in to reply

512^3-512+675^3-675+720^3-720+(512+675+720) by fermats little theoram it is divisible by 3

Shiwang Gupta - 3 years, 1 month ago

Log in to reply

Hm, can you explain in detail? I'm pretty sure that the number is not a multiple of 3.

Calvin Lin Staff - 3 years, 1 month ago

Log in to reply

The problem here is the fact that hcf is one. So, I feel that fermats little theorem must be used

Achuthan Raja Venkatesh - 3 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...