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:
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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