Why 26 is a Special Number

I found out a while back that the number 26 is the ONLY number out of all the natural numbers that is immediately after a perfect square and immediately before a perfect cube.  Indeed, 25 equals 5 squared, and 27 equals 3 cubed.  26 is sandwiched perfectly in-between these.  But how do we know there are no other instances of this anywhere?  If we list out ALL perfect squares and ALL perfect cubes, how do we know that no other number will be exactly 1 before and 1 after anything in the two lists?  That's a whole heck of a lot of numbers to test.  Especially since neither list ever ends.

Our first strategy is to write the equation x^2 + 1 = y^3 - 1 and simplify it to x^2 + 2 = y^3.  Okay, that part seems simple.  But where the heck do we go from here?  I was stuck.  So I went to the math stack exchange website, because it had the question along with several answers.  A link is right here:

https://math.stackexchange.com/questions/874226/y3-x2-2-rightarrow-y-3-27-is-the-only-cube-exceeding-a-square-by

But I really couldn't really understand the answers I saw for a long time.  Then after pondering over it for a long while and asking AI questions and stuff, I finally got the main idea.  I tried to post that main idea I had as an answer on the math stack exchange site to supplement the other answers, but it was not well received and was promptly deleted.  Probably because the audience for the site is not comfortable working in imprecise layman terms.  So I'm reposting what was deleted here:

"This is a very interesting question to me, and the path to its answer is even more interesting. Since my formal graduate level education is non-existent in this subject and I'm not comfortable at all with a lot of mathematical jargon, I had to think long and hard to understand the solution, even AFTER looking at the other answers here.

But the simplest way I can communicate the main idea behind the answer to the question is this: while it would be WONDERFUL to get this answer without leaving the domain of the integers, it's pretty darn tough to do. So what you do is you examine a domain that contains all the integers, but is actually LARGER than the integers and you get the solutions there. If you only have one solution in that larger domain, then you're only going to have one or less solutions in the integer domain. That's the main idea.

 

So what you do is you take this equation and look at its solutions in the domain that contains all integers AND all integers with scalar multiples of the square root of negative two added to them.

 

Numbers in this larger domain are of the form a+b(root -2) so rewrite the equation x squared plus 1 equals y cubed minus 1 where both x and y take on this form and then work with the algebra and see what happens. This is pretty much what Adam Hughes already did in his answer - better than I could ever do - but I know I had a hard time even comprehending his answer without understanding the main idea about the larger domains first.

 

For the equation you pose in your question in this a+b(root -2) domain, there's only one answer, and it is an integer, and so there's only one answer in the smaller domain of the integers.

 

This post may seem redundant to those comfortable with the material, but for novices I think it's a vital point to slow down on and emphasize in order to fully comprehend. And to put in plain English with little jargon if possible."

So with that main idea in place we factor the x^2 + 2 in the equation as (x+root(-2))*(x - root(-2)), and we can work in the larger domain.  So x^2 + 2 = y^3  becomes (x+root(-2))*(x - root(-2)) = y^3.

Then we can write the y in the a+b(root(-2)) form and take the cube of that form to see what happens. (a+b(root(-2)))^3 is straight-forward to expand but a lot of effort.  Thankfully the Adam Hughes answer in the site already does that part.  He has a good answer but for people who don't know what norms or integer rings are, they're going to be lost on how to even begin.  I certainly was.