It's 4am here and I need some sleep - choose 2 numbers from within the convergent range and do the same slight of hand with them. Clearly we can argue that x^^\inf = 4 has no convergence and so makes no sense but what if the equations are both in the converging region x € [e^−e, e^1/e] about [0.066, 1.445]:
{1} x^^\inf = 2
we've shown in the OP that x^2 = 2 is a solution.
Consider,
{2} x^^\inf = e^(2/e)
similarly to the OP we do
{3} x^ (x^^\inf) = x^ (e^(2/e))
but this is just
{4} x^^\inf = x^ (e^(2/e))
and substituting on the left from {2}
{5} e^(2/e) = x^ (e^(2/e))
gives us (I think)
{6} x = e^ (2e^ (-1-(2/e))) ; ~~1.42267
Hence we've shown that
{7} 2 ~~ 1.423
Which isn't as neat as 2==4 but equally troublesome.
There is no issue now with convergence is there (or there shouldn't be if I chose my numbers correctly)?
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Notations
^^ : ascii version of Knuths up-arrow notation
\inf : first countable infinity, aleph-0
€ : in the range, it's really a Euro symbol, but close enough
~~ : approximately equal to
===
Solution
y = x^2
= 9 give solution (3,9)
y = x^2
= 4 gives solution (2,4)
9=4?
They're just different points on the same curve ... doh! We're not showing that 9=4 we're showing that (3,9) and (2,4) are point solutions of y = x^2.
In the same way we have above 2 solution of y = x^^\inf.