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Algebra Level 2

$\large \alpha + \beta + \gamma + \delta = \alpha^7 + \beta^7 + \gamma^7 + \delta^7 = 0$

Suppose $\alpha, \beta, \gamma$ and $\delta$ are real numbers satisfying the equation above, find $\alpha(\alpha + \beta)(\alpha + \gamma)(\alpha + \delta)$ .

The answer is 0.

2 solutions

Khang Nguyen Thanh
Apr 10, 2016

This is a tricky solution.

Choose $\alpha=\beta=\gamma=\beta=0$ , we have $\alpha(\alpha+\beta)(\alpha+\gamma)(\alpha+\delta)=0$

How do you know that $\alpha=\beta=\gamma=\delta=0$ is the only solution?

Pi Han Goh - 5 years, 2 months ago

This is not the only solution. Any ordered quadruples of $(\alpha,\beta,\gamma,\delta) = (a,b,-a,-b)$ satisfies the equation, and all of them will yield the same result.

I am unable to prove whether an ordered quadruple of $(\alpha,\beta,\gamma,\delta) = (a,b,-c,-d)$ where $a\neq c,\;a \neq d,\;b \neq c,\;b \neq d$ exists that satisfies both equations. If such an ordered quadruple exists, then I believe that this expression will have a non-zero value

Hung Woei Neoh - 5 years ago

My solution isn't a legal solution and $\alpha=\beta=\gamma=\delta=0$ isn't the only solution.

But because it is a solution, we can choose $\alpha=\beta=\gamma=\delta=0$ to get value of $\alpha(\alpha+\beta)(\alpha+\gamma)(\alpha+\delta)$ .

I think that this trick cannot be used if the question change to: "Find the sum of the possible values of $\alpha(\alpha+\beta)(\alpha+\gamma)(\alpha+\delta)$ ".

Khang Nguyen Thanh - 5 years, 2 months ago

If it was given that $\alpha^6+\beta^6+\gamma^6+\delta^6=0$ , only then could we assume that $\alpha=\beta=\gamma=\delta=0$ as then the LHS would be nonnegative

Abdur Rehman Zahid - 5 years, 2 months ago

Done it by applying your technique @Khang Nguyen Thanh

Atanu Ghosh - 5 years, 2 months ago

If I may ask, what precisely is that technique.

Ian Limarta - 5 years, 2 months ago

If you think about the question the author posed, you just have to choose $\alpha, \beta, \gamma, \delta \in \mathbb{R}$ that satisfy the equation. Letting them all be zero is a solution to the equation. Therefore, you have that $\alpha(\alpha+\beta)(\alpha+\gamma)(\alpha+\delta)=0$ .

Nowras Otmen - 5 years, 2 months ago

Can this be done using Newton's identities?

Nihar Mahajan - 5 years, 2 months ago

Hung Woei Neoh
May 29, 2016

Any ordered quads of the form $(\alpha,\beta,\gamma,\delta) = (a,b,-a,-b)$ will satisfy the equation.

Eg. $(\alpha,\beta,\gamma,\delta) = (1,2,-1,-2)$

$\alpha+\beta+\gamma+\delta = 1+2+(-1)+(-2) = 0\\ \alpha^7+\beta^7+\gamma^7+\delta^7 = 1^7+2^7+(-1)^7+(-2)^7 = 0$

In this case, the value of

$\alpha(\alpha+\beta)(\alpha+\gamma)(\alpha+\delta)\\ = a(a+b)(a-a)(a-b)\\ =a(a+b)\color{#3D99F6}{(0)}(a-b)\\ =\boxed{0}$

The thing is, I am unable to show that no ordered quads of

$(\alpha,\beta,\gamma,\delta) = (a,b,-c,-d)$ where $a+b = c+d,\;a\neq c,\;a \neq d,\; b \neq c,\; b \neq d$

exists that satisfies the equation $\alpha^7+\beta^7+\gamma^7+\delta^7=0$

Eg. $(\alpha,\beta,\gamma,\delta) = (1,4,-2,-3)$ . It satisfies $\alpha+\beta+\gamma+\delta=0$ but does not satisfy $\alpha^7+\beta^7+\gamma^7+\delta^7=0$

Can someone show this, or prove that such an ordered quad exists?

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