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
5 years, 5 months ago

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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 = $\pm \sqrt{3}$

- 5 years, 5 months ago

Nice way @akash deep

- 5 years, 5 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.

- 5 years, 5 months ago

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

- 5 years, 5 months ago

- 5 years, 5 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.

- 5 years, 5 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!

- 5 years, 5 months ago

@Harsh Shrivastava , please check it. @Dev Sharma

- 5 years, 5 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 .

- 5 years, 5 months ago

Yes that's a problem

- 5 years, 5 months ago