Prove the equation has no solution in integers

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The equation \(a^4 + b^4 + c^4 - 2a^2b^2 - 2b^2c^2 - 2a^2c^2 = 24\) has been proven to have no integer solutions for the variables \(a\), \(b\), and \(c\). The discussion highlights the mathematical reasoning and techniques used to arrive at this conclusion, emphasizing the impossibility of satisfying the equation with integer values. Participants contributed various insights and methods to reinforce the proof, establishing a consensus on the equation's unsolvability.

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Prove that the equation $a^4+b^4+c^4-2a^2b^2-2b^2c^2-2a^2c^2=24$ has no solution in integers $a,\,b,\,c$.
 
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anemone said:
Prove that the equation $a^4+b^4+c^4-2a^2b^2-2b^2c^2-2a^2c^2=24$ has no solution in integers $a,\,b,\,c$.
Hello.

a^4+b^4+c^4-2a^2b^2-2b^2c^2-2a^2c^2=K

(a^2+b^2+c^2)^2-4a^2b^2-4b^2c^2-4a^2c^2=K

K=8*3

Let \ a,b,c \in{\mathbb{Z}} :1º) For \ a,b,c \ = \ even \rightarrow{ } 16|K

2º) For \ a,b \ or \ a,c \ or \ b,c \ = \ even \rightarrow{ } 2 \cancel{|}K

3º) For \ a \ or \ b \ or \ c \ = \ even \rightarrow{}16|K

Example:

a=even \ b,c=odd

a^4+b^4+c^4-2a^2b^2-2b^2c^2-2a^2c^2=

a^4+(b^2-c^2)^2-2a^2(b^2+c^2)=K

16|a^4

(b^2-c^2)=(b+c)(b-c) \rightarrow{}16|[(b^2-c^2)^2]

16|[2a^2(b^2+c^2)]

Therefore: 16|K

4º) a,b,c \ = \ odd \rightarrow{}2 \cancel{|}K

Regards.
 
Thanks for participating, mente oscura and thanks too for your solution! :cool:
 

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