m+n+m+n+m+n+...=6\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+...}}}}}}=6

Let's solve a fun problem relating infinite square roots,

Find the number of pairs of positive integers (m, n) such that m+n+m+n+m+n+...=6\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+...}}}}}}=6 .

First, we observe that

m+n+m+n+m+n+...=6\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+\sqrt{m+\sqrt{n+...}}}}}}=6 is equal to

m+n+6=6\sqrt{m+\sqrt{n+6}}=6 since the expression is infinite, and we can ONLY DO THIS IFF IT'S INFINITE, take note. So, the problem reduces to find the number of pairs of positive integers (m,n)(m,n) that satisfies the equation,

m+n+6=6\sqrt{m+\sqrt{n+6}}=6

We can first say that m<36m<36 since n+6\sqrt{n+6} and nn are positive.

Let's square both sides and simplify the expression a little bit, and at last it will become

m272mn+1290=0m^2-72m-n+1290=0 and it's a quadratic equation in mm which has real solutions if the discriminant, Δ0\Delta \ge 0.

Δ=7224(1)(n+1290)0\Delta = 72^2-4(1)(-n+1290) \ge 0

24+4n0\Rightarrow 24+4n \ge 0

n6\Rightarrow n \ge -6 which is trivial since n>0n>0.

Also, using the quadratic formula, we get,

m=72±24+4n2=36±6+nm=\frac{72\pm\sqrt{24+4n}}{2}=36\pm\sqrt{6+n} since m<36m<36 we shall take the negative sign, this leaves

m=366+nm=36-\sqrt{6+n} . Thus, let 6+n=k<36\sqrt{6+n}=k <36, where kk is a positive integer, then n=k26n=k^2-6.

For n,m>0n,m>0, k=3,4,5,...,35k=3,4,5,...,35, at last, there are 3333 pairs of positive integers (m,n)(m,n).

Note by ChengYiin Ong
1 week, 2 days ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

There are no comments in this discussion.

×

Problem Loading...

Note Loading...

Set Loading...