On Friday, 25 August 2023 at 07:30:23 UTC+10, Arindam Banerjee wrote:

Some fallouts from the "proposed" proof or disproof of Fermat's Last Theorem, presented below, and recently published in sci.math
Let us note that FLT (which says that a^n + b^n = c^n cannot happend for a b c n as nonzero positive integers and n>2) has not been proved yet. So who can get the first a b c for n-3...? Or n=4? Could be a few million digits long!
arcsin(sqrt(x^n)) + arcsin(sqrt(1-x^n)) = pi/2 where 0<x<1 and n can have any value had been worked out numerically to many decimal places as part of the disproof of the FLT. It was presented earlier in my facebook page, and also in sci.math.
The main point of that attempt, was to show that the terms a^n b^n c^n can be expressed as (a^(n/2))^2, (b^(n/2))^2, (c^(n/2))^2 to any place of decimals.
The values (a^(n/2)), (b^(n/2)), (c^(n/2)) could be constructed to form a triangle.
***
Effectively, that reduces a multi-dimensional construction to a two-dimenstional construction.
***
If the included angles sum to pi/2, then it is a right angled triangle.
If they do not, it is not a right angled triangle.
Assume that there is a value set of such a b c and n where a right angled triangle with such sides could be formed. a b c have to be nonzero postive integers.
In which case the included angles would be arcsin(a/c)^n/2 and arcsin(b/c)^n/2.
If x=a/c and going by our assumption that a right angled triangle is formed, thus disproving the FLT - which cannot let that ever happen - then for this assumption that it does form a right angled triangle we will have
(a^(n/2))^2 + (b^(n/2))^2 = (c^(n/2))^2
or a^n + b^n = c^n and dividing both sides by c^n
(b/c)^n = 1 - (a/c)^n = 1 - x^n
And thus we have the included angles - again, based on the assumption that a right angled triangle can be formed - to be
arcsin(a/c)^n/2
or arcsin(sqrt(x^n)
and
arcsin(b/c)^n/2
which is arcsin(sqrt(1 - x^n))
arcsin(sqrt(x^n)) + arcsin(sqrt(1-x^n)) sum up to pi/2 numerically, for all values of n, and 0<x<1, to many decimal places. This has been found by Excel spreadsheet using their ASIN() function. It is also easy to show analytically how this has to

happen, given any right angled triangle.

To me, this is an indication of the possible disproof of the FLT. When you flatten the n dimensions to just two, and apply Pythagoras' theorem to the two dimensional space, you *may* get some value of a b c and n that match exactly, while countless

others will not match exactly.

The number set is infinite. Beyond the scope of computation, that way. Out of that infinity of numbers, for any value of a b c and n, there could be solutions exactly satisfying the relation a^n + b^n = c^n. On the other hand, that may not happen.
I am not sure if this proves or disproves anything relating to FLT. If x is not indefinite, then perhaps integer values of a b c can be found by brute force computer methods leading to disproof of FLT. For even n, such as n=4, there are chances of

getting exactness in terms of a^(n/2) b^(n/2) c^(n/2) values. Maybe there will be better luck with n=4 as opposed to n=3 - and that is a hint!

Now for the fallouts, in terms of possible new mathematical relations: arcsin(sqrt(x^n)) + arcsin(sqrt(1-x^n)) = pi/2
boils down to
arcsin(sqrt(y)) + arcsin(sqrt(1-y)) = pi/2 where 0<y<1
(for any value of n as y=x^n is also 0<y<1 for 0<x<1)
***crucial concept step for the "disproof" of FLT, just above, as it indicates the irrelevance of n for the FLT relation, in the context of infinite numbers. ***
Here note that n need not be an integer.
Let y = sin^2(t), then substituting we get
arcsin(sqrt(sin^2(t)) + arcsin(sqrt(1-sin^2(t)) = pi/2
that is
arcsin(sin(t)) + arcsin(sqrt(cos^2(t)) = pi/2
or
t + arcsin(cos(t)) = pi/2 where 0<t<pi/2
Now taking sin on both sides and expanding
sin(t + arcsin(cos(t))) = sin(pi/2) = 1
or sin(t)cos(arcsin(cos(t)) + cos(t)sin(arcsin(cos(t)) = 1
or sin(t)cos(arcsin(cos(t)) + cos^2(t) = 1
implying
sin(t) = cos(arcsin(cos(t))
as sin^2(t) + cos^2(t) = 1
Anyway, I **hope** some new approaches and mathematical relations like the above have been made, and some new math functions found.
Arindam Banerjee
25 Aug 2023, Melbourne

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