Reaction between alumina and aqueous sodium hydroxide, which one's right?
Quoting from chemguide (emphasis mine):


  With hot concentrated sodium hydroxide solution, aluminium oxide reacts to give a solution of sodium tetrahydroxoaluminate.
  
  $$\ce{Al2O3(s) + 2NaOH(aq) + 3H2O(l) -&gt; 2NaAl(OH)4}$$
  
  Note: You may find all sorts of other formulae given for the product from this reaction. These range from $\ce{NaAlO2}$ (which is a dehydrated form of the one in the equation) to $\ce{Na3Al(OH)6}$ (which is a different product altogether). 
  
  What you actually get will depend on things like the temperature and the concentration of the sodium hydroxide solution. In any case, the truth is almost certainly a lot more complicated than any of these.
  
  The version I am using is perfectly acceptable and is consistent with the aluminium chemistry you will find elsewhere on the site.


You can see that products of the reaction depends on the concentration of sodium hydroxide (Basic pH) but  other factors like temperature and pressure also come into play.

If you do not add water and only change the concentration of sodium hydroxide, product will also be different:

$$
\begin{align}
\ce{Al2O3 + 6 NaOH &amp;-&gt; 2 Al(OH)3 + 3 Na2O}\\
\ce{Al2O3 + 2NaOH &amp;-&gt;[\pu{900 .. 1100 ^\circ C}] 2 NaAlO2 + H2O}
\end{align}
$$

Sodium aluminate is generally $\ce{NaAlO2}$ but its aqueous solution contains many other species like $\ce{[Al(OH)4]−}$, $\ce{[Al(OH)6]^3-}$ or perhaps the $\ce{[Al(H2O)2(OH)4]−}$. They are all considered to be aluminate species. So, if you are getting sodium tetrahydroxoaluminate or sodium hexahydroxoaluminate, they are no different compounds but are all hydrated form of sodium aluminate (addition of water + change in the concentration of sodium hydroxide).