I once figured out an interesting way to approach Fermat’s Last Theorem, where he postulates that ax + bx could never equal cx, if x was a whole number equal to or larger than 3.

The problem is, of course, solving for thre variables of an exponent simultaneously.

So why not re-express it as (a-b)x = an expanding equation in terms of a and b = (a+b)x<? Then we only have to solve for two variables, not three.

If x = 2, we get (a-b)2 + 4ab = (a+b)2

If x = 3, we get (a-b)3 + 3a2b + 3 ab2 = (a+b)3

If x = 4, we get (a-b)4 + 4a3b + 6a2b2 = 4ab3 = (a+b)4

and so on.

Now look at the constants.

1
11
121
1331
14641
and so on…

Yes, it’s Pascal’s Triangle!

Now all someone would have to prove was that the central expanding equation between (a-b)n and (a+b)n could never precisely equal some perfect excponential zn, and viola! Fermat’s Last Theorem would be solved by means of mathematics commonly known in Fermat’s day, which would lead credence to his claim, on the page that he proposed it, that he had an elegant proof for it, but it wouldn’t fit in the margin of the page, so he wrote it down elsewhere - an elsewhere that was never found!