= Solution
Use the standard complex orientations and the product orientation. Let $a,b\in H^2(S^2\times S^2;\mathbb Z)$ satisfy $a^2=b^2=0$ and $\langle ab,[S^2\times S^2]\rangle=1$, and let $x\in H^2(\mathbb{CP}^2;\mathbb Z)$ satisfy $\langle x^2,[\mathbb{CP}^2]\rangle=1$. These are the <cohomology rings> of the <product of two spheres> and the <Complex projective plane>.
For a map from the <Complex projective plane>, write $f^*a=rx$ and $f^*b=sx$. Naturality of the <cup product> gives $r^2x^2=f^*(a^2)=0$ and $s^2x^2=f^*(b^2)=0$. Since $x^2$ has infinite order, $r=s=0$. It follows that $f^*(ab)=0$, so \b[every such map has degree $0$].
In the reverse direction, write $g^*x=ra+sb$. Then
$$
g^*(x^2)=(ra+sb)^2=2rs\,ab,\qquad\boxed{\deg g=2rs\in2\mathbb Z.}
$$
Every even <mapping degree> really occurs. To construct it, identify $S^2$ with $\mathbb{CP}^1$ and use the <Segre embedding>
$$
\sigma:\mathbb{CP}^1\times\mathbb{CP}^1\longrightarrow\mathbb{CP}^3,\qquad
([z_0:z_1],[w_0:w_1])\longmapsto[z_0w_0:z_0w_1:z_1w_0:z_1w_1].
$$
The generator $x_3\in H^2(\mathbb{CP}^3;\mathbb Z)$ pulls back to $a+b$: restricting to either factor gives a projective line and hence coefficient $1$. By the <cellular approximation theorem>, $\sigma$ is homotopic to a <cellular map>. Its four-dimensional domain then maps into the four-skeleton $\mathbb{CP}^2$ of $\mathbb{CP}^3$. Denote that map by $g_0$; restriction of $x_3$ to $\mathbb{CP}^2$ is $x$, so $g_0^*x=a+b$ and $\deg g_0=2$.
For any integer $m$, choose a map $d_m:S^2\to S^2$ of <mapping degree> $m$. For $m>0$ one may use $z\mapsto z^m$ on the Riemann sphere, for $m<0$ use $z\mapsto\overline z^{\,|m|}$, and for $m=0$ use a constant map. Then $g_m=g_0\circ(d_m\times\operatorname{id})$ satisfies $g_m^*x=ma+b$, so \b[the possible degrees are exactly all even integers], with $g_m$ realizing $2m$. The <cellular approximation> here deforms this particular map into the skeleton; it does not require a retraction of $\mathbb{CP}^3$ onto $\mathbb{CP}^2$.
Finally, take the <connected sum of oriented manifolds> with both summands carrying their standard complex orientations. Its degree-two <cohomology> has generators $u,v$ with
$$
u^2=v^2=w,\qquad uv=0,\qquad\langle w,[\mathbb{CP}^2\mathbin\#\mathbb{CP}^2]\rangle=1.
$$
This follows by choosing the generators supported away from the two balls used to form the <connected sum>, so mixed <cup products> vanish while each square gives the common orientation class. Write
$$
h^*u=ra+sb,\qquad h^*v=ta+zb,\qquad P=\begin{pmatrix}r&t\\s&z\end{pmatrix}.
$$
If $\deg h=n$, the three <cup product> relations say
$$
P^{\mathsf T}\begin{pmatrix}0&1\\1&0\end{pmatrix}P=nI_2.
$$
Taking determinants gives $-(\det P)^2=n^2$, which forces $n=0$. Therefore \b[the only possible degree to the connected sum is $0$], realized by a constant map. This is a <degree constraint from intersection forms>: the indefinite <intersection form> of the sphere product cannot pull back a definite form with a nonzero degree multiplier.
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