= Solution
An <isogeny of elliptic curves> over $k$ is a nonconstant morphism $\phi:E_1\to E_2$ defined over $k$ and preserving the identity. Such a morphism is a <group homomorphism>: apply the <Mumford rigidity lemma> to $\phi(P+Q)-\phi(P)-\phi(Q)$, whose restriction to either coordinate axis is zero. A nonconstant morphism of <smooth projective curves> is finite and surjective, so its <degree of an isogeny> is the positive integer
$$
\deg\phi=[k(E_1):\phi^*k(E_2)].
$$
Degrees multiply under composition. A <separable isogeny> has degree equal to the number of geometric points in its <kernel of an isogeny>. In positive <characteristic of a field>, inseparable isogenies also occur, for example the <Frobenius isogeny>; their total degree includes inseparable multiplicity, so counting geometric kernel points alone is insufficient. Every isogeny of degree $d$ has a <dual isogeny> $\widehat\phi$ satisfying $\widehat\phi\phi=[d]$ and $\phi\widehat\phi=[d]$.
Pointwise addition and negation make the <homomorphism group of elliptic curves> $H=\operatorname{Hom}_k(E_1,E_2)$ an <abelian group>. Its zero is the constant identity map; every other element is an isogeny, because its nonconstant image is the whole target curve. Define $q(0)=0$ and $q(\phi)=\deg\phi$ otherwise. We prove the <positive definiteness of degree on elliptic-curve homomorphisms>, including arbitrary characteristic.
Let $L=\mathcal O_{E_2}(O)$, a <line bundle associated to a divisor> of degree one. On $E_2\times E_2$ write $m(P,Q)=P+Q$, $d(P,Q)=P-Q$ and let $\pi_i$ be the projections. The <divisor proof of the degree parallelogram law> gives
$$
m^*L\otimes d^*L\cong\pi_1^*L^{\otimes2}\otimes\pi_2^*L^{\otimes2}.
$$
Here is a direct justification. The degree-two Weierstrass coordinate $x$ has a double pole at $O$, and $x(P)=x(Q)$ precisely when $Q=P$ or $Q=-P$. Consequently
$$
\operatorname{div}(x(P)-x(Q))=D_++D_--2(\{O\}\times E_2)-2(E_2\times\{O\}),
$$
where $D_+$ is the diagonal and $D_-$ the graph of negation. Since $D_+=d^{-1}(O)$ and $D_-=m^{-1}(O)$, this <principal divisor> gives the displayed <line bundle> identity. This remains valid in characteristic two: the general <Weierstrass equation of an elliptic curve> of a <smooth algebraic curve> has a nontrivial generic involution $(x,y)\mapsto(x,-y-a_1x-a_3)$, so the two graphs still occur. This identity is also the symmetric-line-bundle form of the <Theorem of the square>.
Pull back the bundle identity along $(\phi,\psi):E_1\to E_2\times E_2$ and take degrees of <line bundles>. For a nonzero map, $\deg(\phi^*L)=q(\phi)$; for a zero map its pullback is trivial and has degree zero. Thus the <degree parallelogram law> is
$$
q(\phi+\psi)+q(\phi-\psi)=2q(\phi)+2q(\psi).
$$
Using bundles rather than pulling back the rational function itself is essential when $\phi=\pm\psi$, where that function can vanish identically. Negation preserves degree, so $q(-\phi)=q(\phi)$. Applying the parallelogram identity to $n\phi$ and $\phi$ gives the <quadratic degree recursion for elliptic multiplication>, and induction from $q(0)=0$ yields $q(n\phi)=n^2q(\phi)$ for every integer $n$.
Define the <bilinear degree pairing for elliptic-curve homomorphisms> by
$$
B(\phi,\psi)=\frac{q(\phi+\psi)-q(\phi-\psi)}4
=\frac{q(\phi+\psi)-q(\phi)-q(\psi)}2.
$$
It is symmetric and odd in each argument. Two applications of the <parallelogram law> give
$$
\begin{aligned}
B(x+z,y)+B(x-z,y)&=2B(x,y),\\
B(x+z,y)-B(x-z,y)&=2B(z,y).
\end{aligned}
$$
Adding proves $B(x+z,y)=B(x,y)+B(z,y)$, and symmetry proves additivity in the other argument. Also $B(\phi,\phi)=q(\phi)$. Therefore $q$ is an integer-valued <quadratic form>. The integral polarization is $2B(\phi,\psi)=q(\phi+\psi)-q(\phi)-q(\psi)$; $B$ itself may be half-integral.
Finally $q(\phi)>0$ for every nonzero $\phi$, since an isogeny has positive finite degree. If $n\phi=0$ with $n\ne0$, the identity $q(n\phi)=n^2q(\phi)$ forces $\phi=0$, so $H$ is a <torsion-free abelian group>. On any finite-dimensional rational span in $H\otimes\mathbb Q$, the pairing has a rational matrix and is positive on every nonzero rational vector. By density it is nonnegative on real vectors; if that rational matrix were singular, its <kernel> would contain a nonzero rational vector, contradicting strict positivity. Thus its real extension is a <positive-definite quadratic form>. This establishes the required positivity as well as the quadratic identity. For example, on the subgroup of multiplication maps $[n]$, $q([n])=n^2$; on the <endomorphism ring of an elliptic curve> with <complex multiplication> it is the imaginary-quadratic field norm, explaining the lattice geometry behind isogeny degrees.
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