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Prove that the equation $x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=24$ has no solutions in integers.

My work:
When $x,y,z$ are all even no solution exists as the LHS is divisible by 16 whereas RHS is not.
So now, remains two cases:
i) One of $x,y,z$ is odd and two are even.
ii) Two of $x,y,z$ are odd and one is even.

Case i) -say $x$ is odd. LHS stands as $x^4+16k_1+16k_2-8k_3x^2-32k_4-8k_4x^2$. We are now to consider, $x^4-2x^2y^2-2x^2z^2$ as other elements are divisible by $16$. If we can prove that this part is divisible by $16$ then we are done but I cannot do that.

As for Case ii) I could not think of how to show it divisible by $16$. So, I think I need to find out something else that works which I could not. Please help.

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Have you tried rewriting the equation as $(x^2-y^2-z^2)^2=4(6+x^2y^2)$? –  Andrea Mori Feb 26 at 12:55
    
Just throw it in and expand? –  Calvin Lin Feb 26 at 12:58
    
@AndreaMori I cannot see now, how that will help. –  Hawk Feb 26 at 12:59
    
@JoelReyesNoche Thank you for the identification. –  Hawk Feb 26 at 15:59

2 Answers 2

up vote 6 down vote accepted

The equation can be rewritten as $$ (x^2-y^2-z^2)^2=4(6+y^2z^2) $$ The LHS is a square, $4$ is a square, so also $6+x^2y^2$ needs to be a square. But also $y^2z^2$ is a square.

Are there two squares in $\Bbb Z$ whose difference is $6$?

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This is true...and really elegant solution. Thank you for the help. –  Hawk Feb 26 at 13:05
    
@CalvinLin : yes, of course! I made the correction, thank you. –  Andrea Mori Feb 26 at 16:27

Just throw it in and expand.

Case i) If only 1 of them is odd, show that the expression is odd, hence not 24.

Case ii) Set $ x = 2a+1, y = 2b+1, z = 2c $.

See Wolfram. It is a multiple of 16, and so we're done.

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I also need to prove the case i which I could not complete. –  Hawk Feb 26 at 13:01
    
@Hawk That case is pretty obvious, if you know what you're supposed to show. –  Calvin Lin Feb 26 at 13:03
    
Indeed...thanks for the help, though I should have been able to do this. –  Hawk Feb 26 at 13:04
    
for last case: LHS=$((x+y)(x-y))^2+z^4-2z^2(x^2+y^2),x+y=2p,x-y=2q,x^2+y^2=2r \implies $ LHS=$16t$ –  chenbai Feb 26 at 13:30

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