As to an example from number theory, Mumford style drawings of arithmetic schemes really helped me to get the difference between split, ramified, and inert ideals in (rings of integers over) algebraic number fields.
I work a lot on this myself, and study it some. I know visualization is not just something you are either born with or not. You can learn by effort. But I believe there is no substitute for trying, practicing, the very things you want to visualize.
I doubt that visualization per se, is an "expertise," which can be transferred from say differential equations to category theory (though these are both places i have worked on it). Whether I am right or not, what I mean to say is that visualizing one part of math only helps with others in the sense that success at one case encourages you to believe you can succeed at others.
I have a new result that might be a step towards proving there are infinitely many solutions. We show that there is an infinite sequence of values for which $F(s)$ is at most 1.
Consider $s_k = 2^{2^k} - 5$ for $k \ge 3$.
We have stated that a number is in $S$ if it is not $4 + p$ for prime $p$, and is not $3q$ for prime $q$. We also stated that a product is in $P$ iff exactly one divisor pair has sum in $S$. We will show that $s_k \in S$ and at most one of the products $l (s_k - l)$ is in $P$.
$s_k$ is not a multiple of 3, so the latter condition for inclusion in $S$ is always satisfied.
Subtract 4 to get $2^{2^k} - 9$. This number is not prime. It is divisible by 7 is $k$ is even, and 13 if $k$ is odd. This can be proven by induction. Therefore $s_k \in S$ for all $k \ge 3$.
Consider ways to sum to $s_k$. Since $s_k$ is odd, all pairs of summands must consist of one even and one odd number. Let the even number be $a'$, the odd number $b$. Let $v_p(x)$ be the p-adic valuation of $x$. Define $m = v_2(a')$ and $a = 2^{-m} a'$. We know $m \in [1, 2^k)$ and $a$ is odd. We can write $s_k = 2^m a + b$.
If $a \ge 3$, write $s'_k = 2^m b + a$.
Add $s_k$ and subtract it:
$s'_k = 2^m b + a + (2^m a + b) - (2^{2^k} - 5)$
Factor and subtract 4:
$s'_k - 4 = (2^m + 1)(a + b) - (2^{2^k} - 1)$
Since $m < 2^k$, we know $2^m + 1$ shares a nontrivial factor with $2^{2^k} - 1$. This factor divides $s'_k - 4$, and thus $s'_k - 4$ is composite. Furthermore, $s'_k$ is congruent mod 3 to $(-1)^m s_k$, so it is not a multiple of 3. $s'_k \in S$, so the product $2^m a b$ with $a \ge 3$ has at least two factorizations with sum in $S$, and cannot be in $P$.
If $a = 1$, we have $b = 2^{2^k} - 5 - 2^m$. For $m \ge 3$, we can examine $b$ modulo $y = 2^{2^{v_2(m-2)}} + 1$.
Write $2^m$ as $4 * 2^{m-2}$, and then write $2^{m-2}$ as $2^{2^{v_2(m-2)} m'}$ with $m'$ odd. Modulo $y$, $2^m = 4(-1)^{m'} = -4$.
Since $2^k > m$, we know $2^{2^k}$ is $2^{2^{v_2(m-2)}}$ raised to some even power. Thus, modulo $y$, $b = 1 - 5 + 4 = 0$. We found a divisor of $b$.
When $m$ is even, the alternate factor sum $y + 2^m (b/y)$ is 1 mod 6, and thus is in $S$, so $2^m b \not \in P$.
When $m$ is odd, $y = 3$, so we know $b$ is a multiple of 3. Define $r = v_3(b)$, $z = 3^{-r} b$. We know $r$ is at least 1 and $z$ is congruent to either 1 or 5 mod 6. If $z \ge 5$, we can write the factor sums $s'' = z + 3^r * 2^m$ and $s'''= 3^r + 2^m z$.
If $z$ is 1 mod 6, $s''$ is 1 mod 6. If $z$ is 5 mod 6, $s'''$ is 1 mod 6. Either way, one of the two values is in $S$, so $2^m 3^r z \not \in P$.
$z = 1$ occurs when there is a solution to the diophantine equation $2^{2^k} - 5 = 2^m + 3^n$.
I am to believe that there is exactly one solution for $k \ge 3$. When $k = 3$, we have $2^8 - 5 = 251 = 8 + 243$. In this case, we can consider the alternate factor sum $27 + 72 = 99$, which is in $S$. A result told to me here claims there can be no other cases where $z = 1$ can occur.
We have eliminated $a \ge 3$ unconditionally and $a = 1$ when $m \ge 3$. We are left with only one possible pair of summands: 4 and $2^{2^k} - 9$. I don't have a proof that $4(2^{2^k} - 9)$ is or is not in $P$. By my counts it certainly can be, if 3 derived numbers are all prime. That is a rare event, and it would be reasonable for it never to occur given how quickly $2^{2^k}$ grows. However, our initial claim was that at most one derived product is in $P$. We have shown that.
This completes the proof.
Best Answer
This is known as Klyachko's Car Crash Theorem. It was proved in order to prove a theorem about finitely presented groups. In fact, the result allows the ants to move at arbitrary nonzero speeds so long as they make infinitely many loops around their 2-cell. The conclusion is that there's either a collision between ants in the interior of an edge, or else there is a `complete collision', which means that there's a collision at a vertex of all ants from adjacent edges.
EDIT: Oh, it's actually important that there are two complete collisions, which is somewhat harder to prove than one (a collision in the middle of an edge is also a complete collision.)
You can read an expository article by Colin Rourke here.