I need to verify and give some more details to the following proof-sketch.
Theorem: A positive-definite binary quadratic form of discriminant $1$ represents an odd prime $p$ if and only if $p \equiv 1 \pmod 4$.
Sketch proof: If $p ≡ 1 \bmod 4$, then $m^2 ≡ −1 \bmod p$ has a solution. Let $m^2 = −1 + np$.(Can we choose m to be even?) Define the following quadratic form: $f(x, y) = px^2 + 2\frac{m^2}{2}xy + ny^2$. This form has discriminant $1$, is positive definite and respresents $p$ (choose $(x,y) = (1,0)$. Every form which is equivalent to $f$ represents the same integers. Now, for discriminant $d = 1$, there is only one equivalence class. Hence $f$ is equivalent to $f_2(x, y) = x^2 + y^2$. Hence $f_2$ respresents $p$. Conversely, for $p \equiv 3 \pmod 4$: these are not represented by $f_2$, hence not represented by any other positive-definite binary quadratic form of discriminant $1$.
This is what I did so far:
If $p ≡ 1 \bmod 4$, then $m^2 ≡ −1 \bmod p$ has a solution. Let $m^2 = −1 + np$.(Can we choose m to be even?)
This follows from the First Supplement to Quadratic Reciprocity. I think that we can not choose $m$ to be even since a quadratic residue has to be coprime to its modulus.
This form has discriminant $1$, is positive definite and respresents $p$ (choose $(x,y) = (1,0)$.
This is just a straight-forward calculation.
Every form which is equivalent to $f$ represents the same integers. Now, for discriminant $d = 1$, there is only one equivalence class.
This should be standard properties of binary quadratic forms.
Hence $f$ is equivalent to $f_2(x, y) = x^2 + y^2$.
Here I have to find an invertible matrix $A$ such that $f(A(x,y) = f_2(x,y))$ for all $(x,y)$. However, I can not find such an $A$. Could you help me?
Conversely, for $p \equiv 3 \pmod 4$: these are not represented by $f_2$..
I do not get why they should not be represented by $f_2$. Could you help me?