Solving polynomial equations in spectra?

Here is a simple argument that would show many finite complexes can not be `integral' in your sense.

If $Sq^{2^k}$ acts nontrivially on $H^*(X;\mathbb Z/2)$ then $Sq^{2^{k+1}}$ will act nontrivially on $H^*(X \wedge X;\mathbb Z/2)$. But if $X$ were integral then there would be an upper bound on $k$ such that $Sq^{2^k}$ acts nontrivially on $H^*(X^{\wedge d};\mathbb Z/2)$ for some $d$. Thus if $X$ were integral, all nontrivial Steenrod operations would have to vanish on its mod 2 cohomology.

A similar argument would work at odd primes, using the operations $\mathcal P^{p^k}$. And this answers your question 2 in the negative, as, for all primes $p$, and all $n$, there are certainly type $n$ complexes at $p$ with nontrivial Steenrod operations acting on their cohomology.

Here's a slightly fleshed-out version of Nicholas Kuhn's argument. I wasn't able to verify the exact statement he uses, but a variant thereof.

For a spectrum $X$, let $f(X)$ be the maximal $k \in \mathbb N$ such that $Sq^k$ acts nontrivially on $H^\ast(X)$ (note that we're not using $Sq^{2^k}$). If $X$ is finite, then $f(X) < \infty$. Then clearly

• $f(X \oplus Y) = \max(f(X),f(Y))$

• $f(\Sigma X) = f(X)$

Let us show that:

• $f(X_1\wedge X_2) = f(X_1) + f(X_2)$

By the Cartan formula, $Sq^k(x_1 \otimes x_2) = \sum_{k_1 + k_2 = k} S^{k_1}(x_1) \otimes Sq^{k_2}(x_2)$. If $k > f(X_1) + f(X_2)$, then in each summand, at least one tensor factor has $Sq^{k_i}$ acting for $k_i > f(X_i)$, so $Sq^k$ vanishes. Thus $f(X_1 \wedge X_2) \leq f(X_1)+f(X_2)$. If $k = f(X_1) + f(X_2)$, then every summand has a tensor factor for which $Sq^{k_i}$ acts with $k_i > f(X_i)$ except for $k_i = f(X_i)$, so $Sq^k = Sq^{f(X_1)} \otimes Sq^{f(X_2)}$. So if $x_i \in H^\ast(X_i)$ are such that $Sq^{f(X_i)}(x_i) \neq 0$, then $Sq^k(x_1 \otimes x_2) \neq 0$. Thus $f(X_1 \wedge X_2) \geq f(X_1) + f(X_2)$.

Now suppose that $X$ is integral, i.e. $X^n \cong \oplus_{i=0}^{n-1}\oplus_j a_{ij} \Sigma^j X^i$. Then by the above three formulas, we have $n f(X) = f(X^n) = f(\oplus_{i=0}^{n-1} \oplus_j a_{ij} \Sigma^j X^i) = \max_{i=0}^{n-1} i f(X) = (n-1) f(X)$. Therefore $f(X) = 0$, i.e. the Steenrod algebra acts trivially on $H^\ast(X)$.