1) Determine all pairs of real numbers $(x,y)$ such that

$x^6 = y^4 + 18$

$y^6 = x^4 + 18$

2) Let $a,b,c$ be integers such that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=3$. Prove that $abc$ is a perfect cube. Note by Dev Sharma
4 years, 2 months ago

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.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. 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 1paragraph 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} $\boxed{123}$

Sort by:

First of all replace $({x}^2, {y}^2)$ by a and b. Then it is easy to see that if $a \geq b$ then ${a}^3 \geq {b}^3$ but that would also mean that ${a}^2 \leq {b}^2$ comparing the rhs of the two equations . But we have $a \geq b$ so we also have ${a}^2 \geq {b}^2$ . Hence a = b now we write the equation as ${a}^3 - {a}^2 - 18 = 0$ it is easy to see that a = 3 satisfies the equation and so we get $x = y = \pm \sqrt {3}$ we then get another quadratic factor whose roots are obtained and then we get two more values of x as ;- $\sqrt {-1 \pm \sqrt{-5}}$ so we have only two real solutions x = y = $\pm \sqrt{3}$

- 4 years, 2 months ago

Nice way @akash deep

- 4 years, 2 months ago

Hey can u plz post the solution for the second question for the cases in which a,b,c may be negative. I have done it upto natural numbers. Please post the solution.

- 4 years, 2 months ago

@Pi Han Goh @Nihar Mahajan @Surya Prakash Adarsh Kumar

- 4 years, 2 months ago

- 4 years, 2 months ago

I have posted a question related to the first one with the name; "inspired by dev sharma's note" u can check that out.

- 4 years, 2 months ago

For question no.2

By AM-GM we see that:-

$\Large{\frac { a }{ b } +\frac { b }{ c } +\frac { c }{ a } \ge 3\sqrt [ 3 ]{ \frac { a }{ b } \times \frac { b }{ c } \times \frac { c }{ a } } =3}$

So we see the minimum value of expression and its value are same therefore $a=1,b=1,c=1$ for the equality.

We see $abc=1$ which is a perfect cube!

- 4 years, 2 months ago

@Harsh Shrivastava , please check it. @Dev Sharma

- 4 years, 2 months ago

I too did it the same way but the question is not restricted to natural numbers it is for integers which can be negative also. So we can't use AM -GM .

- 4 years, 2 months ago

Yes that's a problem

- 4 years, 2 months ago