How to solve this

Determine all integers n which satisfy

\(((25/2+((625/4)-n)^\frac{1}{2})^\frac{1}{2})+((25/2-((625/4)-n)^\frac{1}{2})^\frac{1}{2})\)

is an integer

Note by Suci Mayeza
4 years, 11 months ago

No vote yet
5 votes

  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

If \[ b \; = \; \sqrt{a + \sqrt{a^2-n}} + \sqrt{a - \sqrt{a^2-n}} \] then \[ b^2 \; = \; a + \sqrt{a^2-n} + 2\sqrt{n} + a - \sqrt{a^2-n} \; = \; 2a + 2\sqrt{n} \] so that \(b^2 \ge 2a = 25\) and \[ n \; = \; \big(\tfrac12b^2 - a\big)^2 \; = \; \big(\tfrac{b^2-25}{2}\big)^2 \] Thus we need \(b\) to be an odd integer greater than or equal to \(5\). and so \[ n \; = \; \big(\tfrac{(2m+3)^2-25}{2}\big)^2 \; = \; (2m^2 + 6m - 8)^2 \; = \; 4(m-1)^2(m+4)^2 \] for any \(m \ge 1\).

Mark Hennings - 4 years, 11 months ago

Log in to reply

even n=0 wrks gud.. !!

Ramesh Goenka - 4 years, 11 months ago

Log in to reply

n is 144 !!! nt for any n > 4 ; say as 5 will nt yield an integer .. liam reconsider your solution.. !!

Ramesh Goenka - 4 years, 11 months ago

Log in to reply

square the expression to get \(25+\sqrt{n}\), so we try to find all n such that that expression is a perfect square. this gives \(11^2, 24^2, 35^2, ..., (m^2-25)^2\) for all int m > 4

Liam Lawson - 4 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...