Find the roots

Number Theory Level 2

The number of real roots of ${ \left( \eta +3 \right) }^{ 4 }+{ \left( \eta +5 \right) }^{ 4 }=16$ is?

The answer is 2.

3 solutions

Pi Han Goh
Feb 13, 2015

Let $x= \eta + 4$ , then we have $(x-1)^4 + (x+1)^4 = 16$

Use Binomial Expansion, $(A \pm 1)^4 = A^4 \pm 4 A^3 + 6 A^2 \pm 4 A + 1$

Expand and simplify gives $x^4 + 6x^2 - 7 = 0 \Rightarrow (x^2+7)(x^2-1) = 0 \Rightarrow \eta + 4 = \pm \sqrt{7} i, \pm 1$

So there's $\boxed{2}$ real roots.

nice solution , thanks

Utkarsh Bansal - 6 years, 4 months ago

I have a silly doubt, doesn't that expression violate Fermat's last Theorem?

Swapnil Das - 5 years, 10 months ago

No. We are not dealing with integers.

Pi Han Goh - 5 years, 10 months ago

So is the theorem limited to integers?

Swapnil Das - 5 years, 10 months ago

@Swapnil Das – Please review the theorem again.

Pi Han Goh - 5 years, 10 months ago

The two solutions both amount to a^4 + 0^4 =a^4, which is pretty consistent with Fermat's last theorem. The Theorem only applies to natural numbers of course.

Jason Short - 5 years, 9 months ago

Isabella Martin
Aug 20, 2015

n: -5 and n: -3

Jason Short
Aug 20, 2015

a^4+b^4=16 has only four real solutions, (0,+-2) and (+-2,0). Two of those solutions give a solution for n.

Find the roots

The answer is 2.

3 solutions

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