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

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The discussion focuses on proving that the expression 2x⁴ + 2y⁴ + 2z⁴ is the square of an integer given that the sum of the integers x, y, and z equals zero. Participants engage in light-hearted banter while acknowledging each other's contributions, particularly praising a member named MarkFL for their solution. The conversation includes humorous remarks about coffee and lullabies, indicating a friendly atmosphere among the contributors. Despite the playful tone, the main mathematical inquiry remains central to the discussion. The thread highlights both the collaborative spirit and the focus on solving the mathematical problem.
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The sum of three integers $x,\,y,\,z$ is zero. Show that $2x^4+2y^4+2z^4$ is the square of an integer.
 
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My solution:

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.
 
MarkFL said:
My solution:

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

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
$2(x^4+y^4+z^4)= (x^2+y^2+z^2)^2$
 
anemone said:
(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
 
kaliprasad said:
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!
c0f0b88f59f9d9a94de7bb772231a994.jpg

And this is for my sweetest admin, MarkFL!:D
c777b22432d40a6093ab9308683bc82c.jpg
 
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