= Solution
Here is an elementary <polynomial pencil with four square members> argument. Suppose first that $u,v$ are linearly dependent. Their <coprimality of polynomials> then forces both to be constant. Otherwise write the four distinct members as $L_j=\alpha_j u+\beta_jv=s_j^2$. They are nonzero and pairwise <coprime polynomials>: a common nonconstant factor of two members would divide both $u$ and $v$.
Put $d=\max(\deg u,\deg v)>0$. At most one member of the pencil has <degree of a polynomial> smaller than $d$, since cancellation of its leading coefficient determines a unique projective pair. Choose a member $L_k$ of minimal degree $e$ and any independent member $L_l$ of degree $d$. The <polynomial>
$$
W=L_k^{\prime}L_l-L_kL_l^{\prime}
$$
is nonzero: otherwise the <rational function> $L_k/L_l$ would have zero <derivative>, hence would be constant in characteristic zero. Its degree is at most $d+e-1$; if both members have degree zero the original $d>0$ assumption has already failed.
Replacing this pair by any other independent pair changes $W$ only by a nonzero scalar. Since $L_j=s_j^2$, each $s_j$ divides $W$. The pairwise <coprimality of polynomials> therefore gives $\prod_j s_j\mid W$. But three members have degree $d$, so
$$
\deg W\ \geq\sum_{j=1}^4\deg s_j
\ \geq\frac{3d+e}{2}
\ >d+e-1,
$$
a contradiction. \b[Consequently $u$ and $v$ are constant.] The possibility that a member is zero was already covered by linear dependence.
To apply this to an <elliptic curve>, complete the square in its <Weierstrass equation of an elliptic curve> and work over $\mathbb C$. Nonsingularity gives three distinct roots $e_1,e_2,e_3$, so the equation becomes $y^2=\prod_{j=1}^3(x-e_j)$. A nonconstant <rational parametrization of an algebraic curve> would have $x=u/v$ with <coprime polynomials>. Clearing denominators gives
$$
(yv^2)^2=v(u-e_1v)(u-e_2v)(u-e_3v).
$$
The four factors are pairwise <coprime polynomials>. A <rational function> whose square is a <polynomial> is itself a <polynomial>, by comparing numerator and denominator in lowest terms. <Unique factorization>, and the fact that every nonzero complex constant has a square root, make each of these four factors a square in $\mathbb C[t]$. They correspond to four distinct projective pairs. The result just proved forces $u,v$ to be constant, and the equation then forces $y$ to be constant as well. \b[This proves the <nonparametrizability of an elliptic curve>.]
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