Prove the equation has no solution in integers

[anemone]

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 $a,\,b,\,c$.

 

[mente oscura]

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.

Spoiler

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

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

$$K=8*3$$

$$Let \ a,b,c \in{\mathbb{Z}}$$ :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.

 

[anemone]

Thanks for participating, mente oscura and thanks too for your solution!