MHB Prove the equation has no solution in integers

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Integers
AI Thread Summary
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.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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 $a,\,b,\,c$.
 
Mathematics news on Phys.org
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:
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top