Deriving proof that $x^2 = 2^x$ has three real solutions I'm new to proofs and although I find this proof fascinating, I'm hoping someone could help me understand how the seasoned mathematician arrived at this result... I'm assuming it involves some abstraction of the Lambert W function.
This is a solution from the Book of Proof by Professor Richard Hammack.
Prove the equation $x^2 = 2^x$ has three real solutions
Proof:
By inspection, the numbers $x=2$ and $x=4$ are two solutions of this equation. But there is a third solution. Let $m$ be the real number for which $m2^m = \frac{1}{2}$. Then the negative number $x=-2m$ is a solution, as follows.
$x^2=(-2m)^2 = 4m^2 = 4(\frac{m2^m}{2^m})^2 = 4(\frac{\frac {1}{2}}{2^m})^2 = 2^2\cdot[ 2^{-(m+1)}]^2 = 2^x$.
Therefore we have three solutions $2$, $4$ and $m$.          $\Box$
Again, I understand the calculations within the proof. But what mental process might unfold to make this realization?
 A: This is a terrible argument. The desired solution pops out of nowhere in a totally unnecessary way. A much more conceptual approach, which you could take without knowing in advance what answer it will spit out, is to think about what the graphs of $y = x^2$ and $y = 2^x$ look like (because you're trying to figure out when they cross). The first graph is a parabola, and the second graph is an exponential. At $x = 0$, the parabola is below the exponential, and at $x = 2$, they meet. Since exponentials eventually grow much faster than polynomials, there must be at least one more positive solution when the exponential overtakes the parabola again, which turns out to be $x = 4$. 
But there must also be at least one more negative solution, where the parabola crosses the exponential while it's exponentially decaying. It doesn't particularly matter what the exact form of this solution is; you don't really learn anything from expressing it in terms of the Lambert W function. It's straightforward to narrow down an interval it must belong to using the intermediate value theorem; for example, it must be between $-1$ and $0$, and even must be between $-1$ and $- \frac{1}{2}$. 
You can actually see the graph at WolframAlpha, which instantly tells you with no work that there are three solutions. With a little more work you can even prove that the only three solutions you can see on this graph are in fact the only solutions. 
