no. of triplet.

[man111]

no. of tripilet [tex](x,y,z)[/tex] satisfying The equation [tex]x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=24[/tex]

[martin123456]

I think that there are no such triplets in [tex]\mathbb{Z}[/tex].
The equation is equivalent to
[tex]x^4-2x^2(y^2+z^2)+(y^2-z^2)^2-24=0[/tex]
[tex]D_1=D/4=(y^2+z^2)^2-(y^2-z^2)^2+24=4(y^2z^2+6)[/tex]
[tex]\Rightarrow \exist a \in \mathbb{Z}[/tex] [tex]y^2z^2+6=a^2 \Rightarrow (a+yz)(a-yz)=6[/tex]
But this is impossible because [tex]a \pm yz[/tex] have the same parity.

[man111]

Thanks martine for nice solution..

can u explain me the last line.....

[martin123456]

well,
a+b and a-b have the same parity because
1. if a+b is even then a+b=2x and so a-b=2x-2b so that is even too
2. if a+b is odd then a+b=2x+1 and so a-b=2x+1-2b so that is odd too

[allier]

Another way to solve this problem is to realize that the LHS factors into [tex](x+y+z)(x-y+z)(x-y-z)(x+y-z)[/tex]. All of these have the same parity since the difference between any two of them is even. However, 24=8.3; thus, at least one of the factors is even and at least one is odd (8=2.2.2).

Pretty much the same idea as Martin's but with direct factorization.