[proofplan] Choose a bundle metric on $E$ and use the orientation to define a fibrewise rotation by $\pi/2$, thereby turning $E$ into the underlying real bundle of a complex line bundle $L$. Under the standard normalization of the Euler class for oriented real two-plane bundles, $e(E)=c_1(L)$. The complexification of $E$ then splits as $L\oplus \overline L$, so the Whitney product formula and the conjugation rule for Chern classes give $c_2(E_{\mathbb C})=-e(E)^2$. Finally, the definition of Pontryagin classes through the complexification gives $p_1(E)=-c_2(E_{\mathbb C})=e(E)^2$. [/proofplan]

[step:Use the orientation and a metric to make $E$ into a complex line bundle] Since $M$ is paracompact, choose a smooth bundle metric $h$ on $E$. For each $p\in M$, let \begin{align*} J_p:E_p\to E_p \end{align*} be the unique $h_p$-orthogonal [linear map](/page/Linear%20Map) satisfying $J_p^2=-\operatorname{id}_{E_p}$ and such that, for every nonzero $v\in E_p$, the ordered basis $(v,J_pv)$ is positively oriented. In any positively oriented local $h$-orthonormal frame $(e_1,e_2)$, this endomorphism is given by $J e_1=e_2$ and $J e_2=-e_1$, so the assignment $p\mapsto J_p$ is smooth. Define $L$ to be the complex line bundle whose underlying real bundle is $E$ and whose multiplication by $i$ is the bundle endomorphism $J:E\to E$. Thus, for $a,b\in\mathbb R$ and $v\in E_p$, the scalar multiplication is \begin{align*} (a+ib)v:=av+bJ_pv. \end{align*} With the standard Euler-Chern normalization for oriented real two-plane bundles, the Euler class of $E$ is the first Chern class of this associated complex line bundle: \begin{align*} e(E)=c_1(L) \end{align*} in $H^2_{\mathrm{dR}}(M)$. [/step]

[step:Identify the complexification with the line bundle and its conjugate] Let \begin{align*} E_{\mathbb C}:=E\otimes_{\mathbb R}\mathbb C \end{align*} be the complexification of the real vector bundle $E$. Extend $J$ complex-linearly to a bundle endomorphism \begin{align*} J_{\mathbb C}:E_{\mathbb C}\to E_{\mathbb C}. \end{align*} For each $p\in M$, the complex [vector space](/page/Vector%20Space) $(E_{\mathbb C})_p$ decomposes into the $i$- and $-i$-eigenspaces of $(J_{\mathbb C})_p$: \begin{align*} (E_{\mathbb C})_p=(E_{\mathbb C})_{p,i}\oplus (E_{\mathbb C})_{p,-i}. \end{align*} The map \begin{align*} \Phi_p:L_p\to (E_{\mathbb C})_{p,i} \end{align*} defined by \begin{align*} \Phi_p(v)=v\otimes 1-J_pv\otimes i \end{align*} is complex-linear and bijective. Similarly, the $-i$-eigenspace is naturally identified with the conjugate complex line $\overline L_p$. These fibrewise maps vary smoothly in local frames, so they assemble to a complex vector bundle isomorphism \begin{align*} E_{\mathbb C}\cong L\oplus \overline L. \end{align*} [/step]

[step:Compute the second Chern class of the complexification]By the [Whitney product formula for Chern classes](/theorems/7052) applied to the complex vector bundle isomorphism $E_{\mathbb C}\cong L\oplus \overline L$, one has \begin{align*} c(E_{\mathbb C})=c(L)c(\overline L). \end{align*} Since $L$ and $\overline L$ are complex line bundles, \begin{align*} c(L)=1+c_1(L) \end{align*} and \begin{align*} c(\overline L)=1+c_1(\overline L). \end{align*} By [citetheorem:9778] applied to the complex line bundle $L$, the first Chern class of the conjugate line bundle satisfies \begin{align*} c_1(\overline L)=-c_1(L). \end{align*} Therefore \begin{align*} c(E_{\mathbb C})=(1+c_1(L))(1-c_1(L)). \end{align*} Multiplying in the graded-commutative [cohomology ring](/theorems/2271) $H^*_{\mathrm{dR}}(M)$ gives \begin{align*} c(E_{\mathbb C})=1-c_1(L)^2. \end{align*} Since $E_{\mathbb C}$ has complex rank $2$, the degree-four component is \begin{align*} c_2(E_{\mathbb C})=-c_1(L)^2. \end{align*} Using $c_1(L)=e(E)$, this becomes \begin{align*} c_2(E_{\mathbb C})=-e(E)^2. \end{align*}[/step]

[guided]The point of passing to $L\oplus \overline L$ is that Chern classes are easy for line bundles and behave multiplicatively under direct sums. The isomorphism from the previous step lets us apply the Whitney product formula: \begin{align*} c(E_{\mathbb C})=c(L\oplus \overline L)=c(L)c(\overline L). \end{align*} For a complex line bundle, all Chern classes above degree one vanish, so its total Chern class is completely determined by its first Chern class: \begin{align*} c(L)=1+c_1(L). \end{align*} The same statement applies to $\overline L$: \begin{align*} c(\overline L)=1+c_1(\overline L). \end{align*} Now we use the effect of conjugation on Chern classes. By [citetheorem:9778] with $k=1$, applied to the complex line bundle $L\to M$, one has \begin{align*} c_1(\overline L)=-c_1(L). \end{align*} Substituting this into the Whitney product computation gives \begin{align*} c(E_{\mathbb C})=(1+c_1(L))(1-c_1(L)). \end{align*} The degree-two terms cancel, and the degree-four term is the negative square: \begin{align*} c(E_{\mathbb C})=1-c_1(L)^2. \end{align*} Because $E_{\mathbb C}$ has complex rank $2$, its second Chern class is exactly the degree-four component of its total Chern class. Hence \begin{align*} c_2(E_{\mathbb C})=-c_1(L)^2. \end{align*} Finally, the complex line bundle $L$ was built from the oriented two-plane bundle $E$, and the Euler-Chern normalization gives \begin{align*} c_1(L)=e(E). \end{align*} Therefore \begin{align*} c_2(E_{\mathbb C})=-e(E)^2. \end{align*}[/guided]

[step:Convert the Chern class computation into the Pontryagin class identity] By the definition of Pontryagin classes through the complexification, equivalently by [citetheorem:9780] with $k=1$, the first Pontryagin class of $E$ satisfies \begin{align*} p_1(E)=(-1)^1c_2(E_{\mathbb C}). \end{align*} Thus \begin{align*} p_1(E)=-c_2(E_{\mathbb C}). \end{align*} Using the computation from the previous step, \begin{align*} p_1(E)=-(-e(E)^2). \end{align*} Hence \begin{align*} p_1(E)=e(E)^2. \end{align*} This is the desired identity in $H^4_{\mathrm{dR}}(M)$. [/step]