Do there exist 4 rationals $(a,b,c,d)$ which satisfy the following two relations?,
$a^2+b^2+c^2+d^2=1$ and $2a+b+c+d=0$?
I spent a lot of time with it and tried the criteria of quadratic equations having rational root but it doesn't seem to be working! Any help or comments would be appreciated , Thanks!
From the second equation, $a=-(1/2)(b+c+d)$ which may be put into the first, and after multiplying by $4$ we have $(b+c+d)^2+4b^2+4c^2+4d^2=4.$ Now since the $4$ on the right is $2^2$ we may divide the equation by $4$, at the same time resetting $b,c,d$ to respectively $b/2,c/2,d/2.$ [so we do not divide the coefficients $4$ by $4$, rather we divide the variables].Thus we have arrived at $$(b+c+d)^2+4b^2+4c^2+4d^2=1.\tag{1}$$ Now consider the equation $w^2+4x^2+4y^2+4z^2=1,$ which has the rational solution $(w,x,y,z)=(1,0,0,0).$ We carry out the usual steps to get a rational parametrization of the solutions: Put $w=1-k,\ x=rk,\ y=sk,\ z=tk.$ This leads to $k=2/(1+4r^2+4s^2+4t^2),$ and then if we let $D=1+4r^2+4s^2+4t^2$ we have $w=(D-2)/D,x=2r/D,y=2s/D,z=2t/D.$ Now comparing to $(1)$ we see we want a solution for which $w=x+y+z,$ which leads to $$4r^2+4s^2+4t^2-2r-2s-2t=1.\tag{2}$$ Now let $a=4r-1,b=4s-1,c=4t-1,$ and note that $(2)$ is equivalent to $a^2+b^2+c^2=7.$ But this is not possible for rational $a,b,c$ as it leads, on clearing denominators, to an integer equation of the form $p^2+q^2+r^2=7s^2,$ and as is known an integer of the form $7s^2$ is not the sum of three squares.
Added: The usual theorem for a number $t$ to not be the sum of three squares is that $t$ be of the form $4^n(8m+7).$ Now if the above $7s^2$ were the sum of three squares, one could write $s=2^u\cdot w$ with $w$ odd. Then $s^2=4^u\cdot w^2,$ and since $w$ is odd, $w^2$ is $1$ mod 8, which makes $7w^2$ of the form $8m+7$, and thus $7s^2=4^u(8m+7)$ so is not the sum of three squares.