A Diophantine Problem By Mohammed Imran

The answer is 0. Since $x^2+y^2+z^2+3(x+y+z)+5=0$ , it implies that $(2x+3)^2+(2y+3)^2+(2z+3)^2 =7$ .If there exists rational solutions x,y,z satisfying this condition,then there exists a,b,c,m which are integral such that they satisfy $a^2+b^2+c^2=7m^2$ .We will use Fermat's method of infinite decent.If $(a0 , b0 , c0 , m0)$ is the least integral solution,then we will prove that there exists $(a1 , b1 , c1 , m1)$ such that m1<m0 and (a1 , b1 , c1 , m1) is a solution to $a^2+b^2+c^2=7m^2$ . If $m0$ is odd then: $a0^2+b0^2+c0^2=7(mod 8)$ which is impossible.So $m0$ is even,which implies that $a0 , b0 , c0$ are all even(The proof is easy).And hence $a0 , b0 ,c0 and m0$ are even.Hence if we have $a1=a0/2 ; b1=b0/2 ; c1=c0/2 ; m1=m1/2$ then (a1 , b1 , c1 , m1) becomes a least integral solution.A contradiction.

Q.E.D