MHB 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 discussed with a focus on proving it has no integer solutions for a, b, and c. Participants have engaged in exploring various mathematical approaches and reasoning to demonstrate the impossibility of finding such integers. The conversation highlights the complexity of the equation and the need for rigorous proof techniques. Overall, the consensus is that the equation does not yield integer solutions. The discussion emphasizes the importance of mathematical proof in resolving such equations.
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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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