Prove 2x⁴+2y⁴+2z⁴ is the square of an integer

[anemone]

The sum of three integers $x,\,y,\,z$ is zero. Show that $2x^4+2y^4+2z^4$ is the square of an integer.

 

[MarkFL]

My solution:

Spoiler

We have:

$$z=-(x+y)$$

Hence:

$$S=2x^4+2y^4+2z^4=2\left(x^4+y^4+(x+y)^4\right)$$

$$S=2\left(x^4+y^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\right)$$

$$S=2\left(2x^4+2y^4+4x^3y+6x^2y^2+4xy^3\right)$$

$$S=4\left(x^4+2x^3y+3x^2y^2+2xy^3+y^4\right)$$

$$S=4\left(x^2+xy+y^2\right)^2$$

$$S=\left(2\left(x^2+xy+y^2\right)\right)^2$$

If $x$ and $y$ are integers, then S must be the square of an integer.

 

[anemone]

My solution:

Spoiler

We have:

$$z=-(x+y)$$

Hence:

$$S=2x^4+2y^4+2z^4=2\left(x^4+y^4+(x+y)^4\right)$$

$$S=2\left(x^4+y^4+x^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\right)$$

$$S=2\left(2x^4+2y^4+4x^3y+6x^2y^2+4xy^3\right)$$

$$S=4\left(x^4+2x^3y+3x^2y^2+2xy^3+y^4\right)$$

$$S=4\left(x^2+xy+y^2\right)^2$$

$$S=\left(2\left(x^2+xy+y^2\right)\right)^2$$

If $x$ and $y$ are integers, then S must be the square of an integer.

(Yes) Well done, MarkFL! So, do you want a cup of coffee or me singing a lullaby for you? Hehehe...:p

 

[kaliprasad]

I feel I also deserve a cup of coffee as well because

Spoiler

We have
$(x^2 + y^2 + z^2)^2 = x^4 + y^4 + z^4 + 2x^2y^2 + 2 y^2 z^2 + 2 z^2 x^2 \cdots 1$

Now $(x^2 y^2 + x^2 z^2) = x^2((y+z)^2 – 2yz) = x^4 – 2x^2yz \dots2$ (as y+z = - x)

Similarly
$y^2 z^2 + y^2 x^2 = y^4 – 2y^2xz \cdots 3$
$z^2x^2 +z^2y^2 = z^4 – 2z^2xy \cdots4$

from (2) (3) and (4)
$2(x^2y^2 + y^2 z^2 + z^2 x^2) = (x^4 + y^4 + z^4 – 2xyz(x+y+z))$
= $x^4 + y^4 + z^4 ...5$
as x+y+z = 0
Putting value of $2(x^2y^2 + y^2 z^2 + z^2 x^2)$ from (5) in (1) we get the result
$2(x^4+y^4+z^4)= (x^2+y^2+z^2)^2$

 

[MarkFL]

(Yes) Well done, MarkFL! So, do you want a cup of coffee or me singing a lullaby for you? Hehehe...:p

Hmmm...one is a stimulant and the other a sedative...so perhaps I should have both so they will counteract one another. :D

 

[anemone]

I feel I also deserve a cup of coffee as well because ...

Of course you do! This is what I prepared for you, kali, my friend!

And this is for my sweetest admin, MarkFL!:D