2022/021) find rational x,y z, such that $x^2 + y^2 + y^2 + x + y + z =1

 we have 

$x^2 + y^2 + z^2 + x + y + z = 1$

or $x^2 + x + y^2 + y + z^2 + z = 1$

we shall complete the square so let us multiply 4 to avoid fraction

$(4x^2 + 4x ) + (4y^2 + 4y) + (4z^2 + 4z) = 4$

adding 1 to each term on the left

 $(4x^2 + 4x + 1) + (4y^2 + 4y + 1) + (4z^2 + 4z+ 1) = 7$

or $(2x+1)^2 + (2y +1)^2 + (2z+1)^2 = 7$

multipying by LCM of denomiantor of 2x+1,2y+1,2z+1 we get

$p^2 + q^2 + r^2 = 7m^2$ for integers p,q,r,m.

this has a least m which has a solution

let m be even say 2n

then  all a,b,c have to be even because RHS is divisible by 4 and if any one to 3 numbers is odd then working in mod 4 and summing we get a remainder of 1to 3 when divided by 4.

so all are even

so m cannot be even

so m is odd.

because m is say m = 2k+1

so $7m^2 = 7(2k+1)^2 = 7 * 4k^2 + 7 * 4k + 1$

ot $7m^2 = 7 * 4k(k+1) + 7$

which leaves a remainder 7 when divided by 8

a number if odd on squaring leaves a remainder 1 when divided by 8 and if even then remainder is 0

sum of 3 such numbers (0 or 1) cannot be 7.

so the equation has has no solution and hence original equation has no solution.