= Solution
The short exact coefficient sequence
$$
0\longrightarrow\mathbb Z\xrightarrow{\times m}\mathbb Z
\longrightarrow\mathbb Z/m\longrightarrow0
$$
induces the integral <Bockstein homomorphism>
$$
\widetilde\beta:H^i(X;\mathbb Z/m)\longrightarrow H^{i+1}(X;\mathbb Z).
$$
If $\rho:H^{i+1}(X;\mathbb Z)\to H^{i+1}(X;\mathbb Z/m)$ is reduction modulo $m$, then
$$
\beta=\rho\circ\widetilde\beta.
$$
Equivalently, if an integral cochain $a$ lifts a modulo-$m$ cocycle and $\delta a=mb$, then $\widetilde\beta[a]=[b]$ and $\beta[a]=[b\bmod m]$.
For integral lifts $a,c$ of classes $x,y$, the <cup product> coboundary formula is
$$
\delta(a\smile c)=\delta a\smile c+(-1)^{|x|}a\smile\delta c.
$$
Dividing by $m$ and reducing modulo $m$ proves the <Bockstein derivation rule>
$$
\beta(x\smile y)=\beta(x)\smile y+(-1)^{|x|}x\smile\beta(y).
$$
Now assume that the stated closed five-manifold $M$ exists. Its top integral cohomology makes it connected and orientable. The <long exact sequence from a coefficient sequence> for multiplication by $p$ shows that
$$
\widetilde\beta:H^2(M;\mathbb Z/p)\xrightarrow{\sim}H^3(M;\mathbb Z)
$$
and that reduction $H^3(M;\mathbb Z)\to H^3(M;\mathbb Z/p)$ is an isomorphism. Hence
$$
\beta:H^2(M;\mathbb Z/p)\xrightarrow{\sim}H^3(M;\mathbb Z/p).
$$
The same coefficient sequence gives $H^4(M;\mathbb Z/p)=0$.
Choose $0\ne x\in H^2(M;\mathbb Z/p)$ and put $y=\beta x\ne0$. By <Poincare duality> over $\mathbb Z/p$, the pairing
$$
H^2(M;\mathbb Z/p)\otimes H^3(M;\mathbb Z/p)
\longrightarrow H^5(M;\mathbb Z/p)
$$
is nondegenerate. Both factors are one-dimensional, so $x\smile y\ne0$. But $x\smile x=0$ because $H^4(M;\mathbb Z/p)=0$, while the derivation rule and <graded commutativity of the cup product> give
$$
0=\beta(x\smile x)
=y\smile x+x\smile y
=2x\smile y.
$$
This contradicts $p>2$. Therefore no such manifold exists.
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