Calvin Lin:

Find the number of coprime pairs (a, b) of positive integers subject to 0 100, such that a|(b2-11) and b|(a2-11).

There would be a small correction to the problem. You say, a|b2-11. Then a = [ b^2 -11] / k1. Where k1 is a natural number. Then, [b- 11^(1/2)].[b+11^(1/2)] / k1 = a = prime number. Then either [b- 11^(1/2)] /k1 OR [b+11^(1/2)]/k1. Because a>1 according to the given conditions. But b is a natural number. And 11^(1/2) is not a natural number. Also k1 is a natural number. Then we have a contradition. If you input some square number instead of 11, then the problem may okay.

Note by Geerasee Wijesuriya
5 years, 1 month 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

Sort by:

Top Newest

Note that "coprime pairs of positive integers" means that gcd(a,b)=1 \gcd (a,b) = 1 .

It does not imply that aa and bb are prime numbers.

For example, (6,35) (6, 35 ) are coprime, but neither of the numbers are prime.

Calvin Lin Staff - 5 years, 1 month ago

Log in to reply

Calvin Lin: Actually, a and b are not needed to be prime numbers. According to my previous argument, then a =1. Then according to the problem's conditions b can take only 1, 2,5. Then answer should be 3.

Geerasee Wijesuriya - 5 years, 1 month ago

Log in to reply

Sorry but I do not understand your argument.

Note that (5,14) (5, 14) is also a solution, since 5211=14 5^2 - 11 = 14 and 14211=185 14^2 - 11 = 185 . This contradicts your claim that a=1 a = 1 .

Calvin Lin Staff - 5 years, 1 month ago

Log in to reply

@Calvin Lin No. Actually, according to my previous argument, [b- 11^(1/2)].[b+11^(1/2)] / k1 = a. But, if you put a square number(k^2) instead of 11, then gcd(a, b+k) or gcd(a, b-k) not equals to 1. Then according to the gcd operations, gcd(k,a) =1 and gcd(k,b) =1. (The operation is gcd(a,b) = gcd(a+m.b ,b) ). Therefore, if you choose k^2 instead of 11, such that gcd(k,ab) =1.

Geerasee Wijesuriya - 5 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...