# Some interesting questions

(1)Solve:- $\alpha^2+\alpha\beta+\beta^2=\gamma^2$ Where $\alpha,\beta,\gamma$ are positive distinct primes

(2) Find out all roots of:- $x^4+9x^3+1$

(3) Find out the discriminant of:-

$x^3+x^2+x+1$

(3) Solve the following differential equation:- $\frac{d}{dx}(y)=\frac{x^3+x^2y+xy^2+y^3}{x^3+x^2y+y^3}$

(4) Find out the sum of all values of $x$ such that:- $x^4+x^3+x^2+x+1$ is a perfect square Note by Aman Sharma
5 years, 10 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:

3) $ax^3 + bx^2 + cx +d$ has discriminant as $b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$

Thus given discriminant is -16

- 5 years, 10 months ago

Is there any general definition for "Determinant" for n-degree equation?

- 5 years, 10 months ago

- 5 years, 10 months ago

@U Z Is it possible to interpolate a function to calculate the number of terms in the discriminant? I tried to read wiki for this but couldn't understand it.

- 5 years, 10 months ago

Need some time , involving @Deepanshu Gupta too

- 5 years, 10 months ago

@U Z Oh! Ok thanks!

- 5 years, 10 months ago

$\dfrac{dy}{dx} = \dfrac{1 + \dfrac{y}{x} + (\dfrac{y}{x})^2 + (\dfrac{y}{x})^3}{1 + \dfrac{y}{x} + (\dfrac{y}{x})^3}$

$\dfrac{y}{x}= v$

$x\dfrac{dv}{dx} + v = \dfrac{1 + v + v^2 + v^3}{v^3 + v + 1}$

$x\dfrac{dv}{dx} = \dfrac{1 - v^4}{v^3 + v + 1}$

$\dfrac{v^3 + v + 1}{1 - v^4} dv = \dfrac{1}{x}dx$

$\dfrac{v(v^2+1)}{(1 + v^2)(1- v^2)} + \dfrac{1}{(1 + v^2)(1- v^2)} = \dfrac{1}{x}dx$

$\dfrac{v}{1 + v^2} + \dfrac{1}{2}( \dfrac{1}{1 + v^2} + \dfrac{1}{1 - v^2}) = \dfrac{1}{x}dx$

=$\dfrac{1}{2}ln{(1+v^2)} + \dfrac{1}{2}(tan^{-1}v) + \dfrac{1}{4}(ln(\dfrac{1 - v}{1 + v}) = lnx + lnC$

- 5 years, 10 months ago

4) $x^5 = 1$

$\dfrac{x^4}{x^5} + \dfrac{x^3}{x^5} + x^2 + x + 1 =0$

$x^2 + \dfrac{1}{x^2} + x + \dfrac{1}{x} + 1 =0$

$Keep~ y = x + \dfrac{1}{x}$

- 5 years, 10 months ago

How did you get $x^5 = 1$?

- 5 years, 10 months ago

You used the Fifth root of Unity here ? Right?

Where $w^5 = 1$

However, how will this give all possible values of x for which the given expression is a perfect square.

- 5 years, 10 months ago

No where in the question does it mention that $x$ is the fifth root of unity.

- 5 years, 10 months ago

That is what I asked @megh choksi. How does using $x^5 = 1$help in getting pefect squared values for the expresion. In a hurry Megh must have misread the question, and found its roots what is not wanted at all.

- 5 years, 10 months ago

Sorry I wrote a method to find its roots

- 5 years, 10 months ago

You know this - sum of GP

- 5 years, 10 months ago

But why you are finding root???? I mean how..will this give all values of x such that the given expression is perfect square.......i am not able to understand your solution..plz help

- 5 years, 10 months ago

Rhs can't be even as sum of 3 odd primes will give always a prime,

If $\alpha =2 , Then~ \beta$ will be an odd prime, and on RHS is also odd prime ,thus $\alpha=2~or\beta=2$

$4 + 2\alpha + \alpha^2 = \gamma^2$

$(\alpha + 1)^2 - \gamma^2 = -3$

$This ~can't~be ~possible~for~primes \to \boxed{(\alpha + \gamma + 1)(\alpha - \gamma + 1) = -3}$

- 5 years, 10 months ago

Hint:-complete the square by adding $\alpha\beta$ both sides

- 5 years, 10 months ago

For 2, plot $y=x^4+9x^3$ and $y=-1$ and check the points of intersection.

- 5 years, 10 months ago

Very nice thinking

- 5 years, 10 months ago

Thanks!

- 5 years, 10 months ago

For (1).. 3,5,7 satisfies the condition.... @megh choksi sum of odd primes is not always prime as evident.....

- 5 years, 10 months ago

You are right with 3, 5 and 7. I would like to answer to only question that I feel more interesting.

- 5 years, 10 months ago

2) For x^4 + 9 x^3 + 1 = 0 {Please write an equation.}

-0.489801491372372

-8.998627630183605

0.244214560777989 + j 0.408953683836186

0.244214560777989 - j 0.408953683836186

- 5 years, 10 months ago

Megh Choksi--- I think there is some mistake between the 3rd and 4th step!! as there is a term remaining in numerator that is v^3

- 5 years, 10 months ago

Thanks , so the expression becomes $x\dfrac{dv}{dx} = \dfrac{1 + v^3 - v^4}{1 + v^3 - v^4}$

$\dfrac{v^3 + v + 1}{1 + v^3 - v^4} dv = \dfrac{1}{x}dx$

$1 + \dfrac{v + v^4}{1 + v^3 - v^4} = \dfrac{1}{x}dx$

then its simple

- 5 years, 10 months ago

Am I right? let $\alpha$=a $\beta$=b $\gamma$=c Then, $a^2+ab+b^2-c^2$=o a=$\frac{-b \pm \sqrt{b^2-4b^2+4c^2}}{2}$...........(1) The discriminant should be a perfect square.$\Rightarrow 4c^2-3b^2$=0 or a perfect square. $\Rightarrow$ c=b= prime perfect square.(couldn't there be any other possibilities where b is not equal to c ?please check ) on substituting in (1) we can get values for $\alpha$ , $\beta$, $\gamma$.

- 5 years, 10 months ago

discriminant -16??

- 5 years, 10 months ago