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For odd $p$, let $y^2=f(x)$ be a nonsingular cubic Weierstrass equation of an elliptic curve over ${\mathbb{F}}_p$. Its Trace of Frobenius is $a_p=-{\sum }_x\chi (f(x))$, where $\chi$ is the Legendre symbol extended by $\chi (0)=0$. By the Euler criterion, reduce this sum to $f(x{)}^{(p-1)/2}$. The sum of a positive power $x^j$ over ${\mathbb{F}}_p$ is zero unless $p-1$ divides $j$, when it is $-1$. Since the polynomial has degree $3(p-1)/2<2(p-1)$, only its $x^{p-1}$ coefficient contributes.

For $p\nmid 2k$, the displayed integral equation has nonzero elliptic-curve discriminant modulo $p$, so it has good reduction. Let $\chi$ be the Legendre symbol, with $\chi (0)=0$. Counting the two, one, or zero possible ordinates over each $x$ givesIf $p\equiv 3(\mathrm{mod}4)$, then $\chi (-1)=-1$ and the terms for $x$ and $-x$ cancel. The term at zero vanishes. Thus the count is $p+1$.

For the converse, suppose $p\equiv 1(\mathrm{mod}4)$, and put $h=(p-1)/2$. The Euler criterion and the sum of powers over a finite field show that the Trace of Frobenius satisfiesIndeed, among the positive exponents in this polynomial, of degree $3(p-1)/2<2(p-1)$, only $p-1$ has a nonzero sum over ${\mathbb{F}}_p$, and that sum is $-1$. Since $(x^3+kx{)}^h=x^h(x^2+k{)}^h$, the coefficient formula for trace of Frobenius modulo p yieldsNeither $k$ nor the binomial coefficient vanishes modulo $p$. Hence $a_p\ne 0$, proving the point-count criterion for y squared equals x cubed plus k x:

We use the following formal logarithm fact, also useful for the next part. For a formal group law over ${\mathbb{Z}}_p$, its logarithm has coefficients $c_j/j$ with $c_j\in {\mathbb{Z}}_p$, obtained by integrating its integral invariant differential. If $p$ is odd and $t\in p{\mathbb{Z}}_p\setminus \{0\}$, every term of degree $j\ge 2$ has valuation strictly greater than $v_p(t)$, since $j{v}_p(t)-v_p(j)>v_p(t)$. Thus the logarithm converges and is nonzero at $t$. Its homomorphism into the additive characteristic-zero group proves torsion-freeness of the formal group over Qp for odd p. Consequently reduction injects the entire rational torsion subgroup at any odd prime of good reduction, including its $p$-primary torsion. In particular the torsion subgroup is finite, and its order divides $p+1$ at every good $p\equiv 3(\mathrm{mod}4)$.

By the Dirichlet theorem on primes in arithmetic progressions, choose a good prime $p\equiv 3(\mathrm{mod}8)$. Then $v_2(p+1)=2$, bounding the two-primary part by four. For any odd prime $\ell$ dividing the torsion order, the Chinese remainder theorem and the Dirichlet theorem on primes in arithmetic progressions give a good prime $p$ with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$, avoiding the finitely many primes dividing $2k$. But then $\ell \nmid p+1$, a contradiction. Therefore

The order four does occur. For $k=4$, the elliptic-curve addition formula gives $2(2,4)=(0,0)$, so $(2,4)$ has order four. More precisely, the only nonzero rational 2-torsion point when $k>0$ is $(0,0)$, so order four means a cyclic group with a rational half of this point. The duplication formula isSuch a half must have $x(P{)}^2=k$. Writing $k=s^2$, positivity of $y(P{)}^2=2kx(P)$ forces $x(P)=s>0$. Then $y(P{)}^2=2s^3$, so $2s$ is a rational square, equivalently $s=2t^2$ for an integer $t\ge 1$. Conversely $(2t^2,4t^3)$ is a half of $(0,0)$ when $k=4t^4$. The rational torsion on y squared equals x cubed plus positive k x is therefore cyclic of order four exactly for these $k$, and otherwise cyclic of order two.

For an odd prime $p$, the Legendre symbol isEuler criterion states that

The Gauss lemma says that, for $p\nmid a$, if $m$ of the least positive residues oflie in $(p/2,p)$, thenTo prove it, replace every residue above $p/2$ by its negative. The resulting absolute residues are distinct up to sign and therefore form a permutation ofMultiplying the congruences givesCancel the nonzero factorial and apply Euler's criterion.

For $a=2$, the residues are $2,4,\dots ,p-1$. The number above $p/2$ iswhose parity gives the second supplementary law for quadratic reciprocityConsequently, for odd primes,The congruence is also soluble for $p=2$.

For an odd prime $p$, the Legendre symbol isEuler criterion states thatIf $p\mid a$, both sides vanish. Otherwise choose a primitive root $g$ and write $a\equiv g^m$. Since $g^{(p-1)/2}\equiv -1$,The power $g^m$ is a square exactly when $m$ is even, proving the criterion. Taking $a=-1$ gives the first supplementary law for quadratic reciprocity

Because $I$ and $A$ commute, the matrix binomial theorem givesFor prime $p$, every intermediate coefficient $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ is divisible by $p$, proving the prime-power binomial congruence for a matrix

Forone has $(I+A{)}^2=2A$ and $(I+A{)}^4=-4I$. If $p=4k+1$, thenso $(-4{)}^k\equiv 1(\mathrm{mod}p)$. If $p=4k-1$, then $A^p=-A$ andso $(-4{)}^k\equiv 2(\mathrm{mod}p)$.

By Euler criterion, $(\frac{2}{p})\equiv 2^{(p-1)/2}(\mathrm{mod}p)$. In either case the preceding congruence gives $(\frac{2}{p})=(-1{)}^k$. Since $(p^2-1)/8$ has the same parity as $k$ for $p=4k\pm 1$, this is the second supplementary law for quadratic reciprocity

By Euler criterion,Choose a primitive root $g$. For any integer $m$ not divisible by $p-1$,For $p>3$, the exponent $m=(p+1)/2$ satisfies $0<m<p-1$. ThereforeTogether with part (i), this is the weighted complete Legendre-symbol sum.