Can an elliptic curve have discrimant one? Can the discriminant, $4a^3 +27b^2$ of an elliptic curve $$E: y^2=x^3+ax+b$$ be equal to 1.
I believe that this should not be possible otherwise the curve would have good reduction at all primes $p$, but I only believe this to not be the case, I do not know this as a fact. 
 A: Note that if $4a^3+27b^2=1$ then $b$ must be odd. Set $b=2c+1$ then
$$4a^3+108c^2+108c+27=1$$ or
$$4a^3+108c^2+108c=-26$$but the left hand side is divisible by $4$ and the right hand side is not.
More slickly, consider the original equation modulo $4$ and you get $b^2\equiv-1$, which is easily shown to be impossible.
A: The fact that elliptic curves over $\mathbb{Q}$ cannot have everywhere good reduction is an exercise in Silverman's "The Arithmetic of Elliptic Curves" (8.15 in Chapter VIII). He offers this hint:

(Hint. Suppose that there is a Weierstrass equation with integer coefficients and discriminant $\Delta= \pm 1$. Use congruences modulo 8 to show that $a_1$ is odd, and hence $c_4 \equiv 1 \bmod 8$. Substitute $c_4 = u \pm 12$ into the formula $c_3^4 − c_2^6 = \pm 1728$. Show that $u$ is either a square or
  three times a square. Rule out both cases by reducing modulo 8.)

If you are working over a number field, then you can have elliptic curves with everywhere good reduction. For instance, $y^2 + y = x^3 - 30x + 63$ over $\mathbb{Q}(\sqrt[4]{3},\beta)$ where $\beta^3-120\beta+506=0$ has a model with everywhere good reduction and discriminant a unit. 
