Question

Prove that 1 cannot be represented by binary quadratic form

Answer 1

If $F(x,y) = ax^2+bxy+cy^2$ is the positive definite reduced form, then first, it is clear that $F(x,y) \ge (a-|b|+c)\min(x^2,y^2)$ (just enumerate the cases). Now, if $xy\ne 0$, then $F(x,y)\ge a$ since $c-|b|\ge 0$. Further, if $a<c$, then $F(x,y) > a$.

Next, note that if $y=0$, then $F(x,y) = ax^2 \ge a$ unless $x=0$, and if $x=0$, then $F(x,y) = cy^2\ge c$ unless $c=0$. It follows that $F(x,y) \ge a$ unless $(x,y) = (0,0)$.

If $c\ne a$, then $f(x,y) > a$ when $xy\ne 0$, so $F$ can achieve its minimum value only when $x=0$ or $y=0$; since $a<c$, this can occur only when $x=\pm 1$.

So the minimum value is $a$, and if $a\ne c$, it is achieved only for $(x,y) = (\pm 1, 0)$.