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## Clever Fermat!

The last theorem of Fermat has been solved in
recent years so I hear. He must have been a
smart fellow to know of the future
mathematical techniques needed to prove it.
Or he was kidding in that margin entry.
Or he was mistaken in his insight.
Or he had a simpler proof than has just been
promugated and it is still to be discovered.
I am not going to give any proof myself but I
was thinking about it many years ago and tried
to visualise the subject. It might be
original though I doubt it.

Think of a family of all right angled triangles
on a base with the right side the opposite and
the hypotenuse coming from the left end of the
base. As the opposite gets higher the angle
goes from small to 45 degrees. This family
includes all possible integer solutions at
various scales. After 45 degrees the family is
being repeated so that it is superfluous to
consider them after 45 degrees. You may have
the simple 3,4,5 example or nearby ones with
sides of 20, 30, 1000 or more integers as
scaled. Indeed though you can specify the
irational 1,1,root2 triangle, you can get
ever close to it with sides of hundred of
integers long scaled. But lets not go there!
Now the tip of the opposite side is in effect
the "loci" of all possible right angled
triangles. Now think of all possible triangles
that obey the cubed rule. The base is still
one of the smaller sides, the longest side is
still as with the hypotenuse of the right
angled, with the opposite sloping inwards.
The loci of these cubed triangles now becomes
an "s" shaped line going ever upwards. And
extends beyond 45 degrees. Loci for higher
powers are slightly straighter lines until
the power of infinity which merges with the
right angled triangle loci straight up.

If you have managed to follow so far, you know
there are no integer solutions on the curved
loci of cubed and above loci. But why? Does
this give any insight?

2. Now think of all possible triangles
that obey the cubed rule.
Lost me here. What do you mean "obey the cubed rule"?

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Where the cube of the longest side is equal to
the sum of the cubes of the other two sides.
Likewise with higher powers.

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Lost me here. What do you mean "obey the cubed rule"?
In context, I presume he means that the lengths of the three sides are related by a3 + b3 = c3. I bow out at the next level: "Does this give you any insight?"

No.

Edit: Post #3 was added as I was typing this.

5. Originally Posted by Fortunate
In context, I presume he means that the lengths of the three sides are related by a3 + b3 = c3.
So one side has to be curved to fit into this equation. Presumably peteshimmon is using the opposite side to curve in. But that wouldn't be a triangle anymore.

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So one side has to be curved to fit into this equation. Presumably peteshimmon is using the opposite side to curve in. But that wouldn't be a triangle anymore.
I thought, in the case of cubes, for instance, that peteshimmon was proposing a set of triangles, all with the same horizontal base, with the vertices opposite the common base (the "top" vertex of each) tracing out a curve. So each triangle is a legitimate triangle, in which the "left-hand" side is the longest ("c" in my post #4 above). Thus, each point on the curve corresponds to a triangle in which the sum of the cubes of the two shorter sides equals the cube of the longest side. If any point of the curve yields a triangle in which the ratios of the lengths of each of the three pairs of sides are rational, the point would provide a counterexample to FLT.

7. OK, for some reason I thought OP is using a right triangle.

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If I were attempting to follow up on this line of thought, I might try to find equations for the curves, letting the base extend from (0,0) to (0,1) in Cartesian coordinates. All I can see is some nasty parametric equations and no obvious way to tell which points lead to rational ratios.

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OK, for some reason I thought OP is using a right triangle.
He did so for n=2 only (according to my interpretation of what he meant).

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I am glad you understand what I was trying to
describe, this was an exercise in seeing if I
could do that. But I do need people who know
the subject like you both. Anyway some other
old thoughts on the subject have occured
overnight and seem to have crystalised into
some sort of order. Lets try this:

I note that odd numbers remain odd when
squared, cubed etc. Likewise even numbers.
Now imagine an odd larger square of unit
squares. A smaller odd square sits in the
corner. Now for an integer solution the
remaining unit squares should sum to an
even number. And if it was a smaller even
square then the remains should sum to an
odd number. No problem;

even squared+ 2 times (odd times even) = even

odd squared+ 2 times (odd times even) = odd

Because the remaining square and two areas
add to odd or even in line with the required
other square, then integer solutions are
possible.

But with cubes;

even cubed + 3 times (even squared times odd)
= odd , when an even cube is needed...nope!

Odd cubed + 3 times (odd squared times even)
= even , when an odd cube is needed...nope!

So the problem with cubes is the need to
multiply by 3 which does not even out odd
numbers.

Maybe this is it

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even cubed + 3 times (even squared times odd) = even

12. Seeing this thread is going more off-topic I would like to post my own
observations:

Euler proved the closed form for this infinite summation

Infinite sum [n = 1] ( 1 / n^2 ) = pi^2/6 = appx. 1.645
in a proof which uses the trigonometric identities and some other identities.

And this function of Infinite sum [n = 1] ( 1 / n^q ) is named the Riemann Zeta function for the number q. It diverges for q=1, has closed forms involving the Bernoulli numbers for even integers and for its complex numbers goes into the Riemann Hypothesis, or so I have understood.

But how can we know the closed form for a higher iteration such as:

Infinite Sum [n=1] 1/ n^n

Using Mathematica this number is appx 1.291285997...

I checked some maths website and there seemed to be no discovered closed form for it, yet.

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Seeing this thread is going more off-topic I would like to post my own
observations:
I'm not sure what you mean by "off-topic." The OP described some curves related to FLT. Post #10 adresses another issue related to the difficulty of finding integer solutions to FLT in the case n = 3.

(snip)

Infinite Sum [n=1] 1/ n^n

Using Mathematica this number is appx 1.291285997...

I checked some maths website and there seemed to be no discovered closed form for it, yet.
Infinite Sum [n=1] 1/ n^n is probably a transcendental number (not the root of a polynomial with integer coefficients). Transcendental numbers cannot be expressed in terms of radicals of integers (indeed, neither can most roots of polynomials with integer coefficients). We have given "names" to a few transcendental numbers, such as pi and e, but no matter how many we name, most transcendental numbers will not have "closed" expressions in terms of numbers we have named employing all the functions and operations we have named because a countable number of functions and operations applied to a countable number of "named" numbers can only yield a countable number of results, and the number of transcendental numbers is uncountable.

Edit: My last sentence is not worded well. I am only considering expressions of finite length.

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Yeah well...I was happy for a few hours! But
had my doubts out shopping then found
Fortunates correction. Not even (ouch) half
right. Would have been neat though! The
mathematical path is very badly paved in my
opinion But Fermat might have thought of
something like this. It must be a well trodden
path looking for something. It does seem that
the curved loci will not contain integer
solutions because of the curvature. This is
a guess though.

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Originally Posted by peteshimmon
It does seem that
the curved loci will not contain integer
solutions because of the curvature. This is
a guess though.
Hi,
We don't need the sides to be integers. We just need the ratios of the lengths of the sides to be rational. If the ratios are rational, we can find integers by multiplying by a multiple of the denominators of the ratios.

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Two small points I note from my abortive
odd/even proof of FLT. One, two smaller cubes
will never sum to less than 2 away from the
nearest larger cube. Two, all integer right
angled triangles have sides two odd, one even.
No doubt trivial points long since realised in
the history of FLT. But I admit to being just
a dabbler here!

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Originally Posted by peteshimmon
Two, all integer right
angled triangles have sides two odd, one even.
This is true after they have been "reduced" to lowest terms as follows: divide all three sides by the greatest common factor of the three sides. For instance, the right triangle with sides 24, 32, and 40 would be reduced to one with sides 3, 4, and 5. A slightly more subtle fact is that the hypotenuse cannot be the even side. This is because the square of an even number will always be a multiple of 4, whereas the square of an odd number will always leave a remainder of 1 when divided by four. Thus, the sum of the squares of two odd numbers will always leave a remainder of two when divided by four and can never equal the square of an even number, which would always leave a remainder of zero when divided by four.

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Yes I was almost going to add in my previous
post that the largest side will be odd but I
did not have the reasoning you have and mine
was wrong again. A small table I have just
produced shows all sums of cubes of numbers
from 1 to 10. (like the tables showing
all distances between many cities). I find
that the cubes of 5 and 6 sum to two less than
7 cubed but the cubes of 6 and 8 sum to one
less than 9 cubed. So I was wrong again.
A very tricky subject this.

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I will just add some more thoughts here. If
you understood my loci diagram you might have
noticed the loci for cubes would have to curve
into its corner avoiding the loci for
isosceles triangles which is a semi circle
centred at the left side. Now one wonders
how the loci for powers between 2 and 3 might
look. It seems they would cover the same
space as all powere above 3. Perhaps there is
infinite space available as loci have no width.
Or perhaps loci cross.

illuminating as it seems the proof of no
integer solutions for cubes was found long ago.
I wonder what the problem was as you have to
go through cubes to get to higher powers! But
Mathematicians want their general proofs it
seems.

Anyway, playing in the sandbox it might be fun
to use computers to generate larger tables as
I describe in the previous post and look for
patterns in the differences from perfect cubes
and squares. I suppose mathematicians did this
years ago as soon as they got their hands on
machines for general use (and fun!).

And to finish, cubes made up of two smaller
cubes using unit cubes will have 6 other blocks
to fill the space. So any thoughts of odd/even
leading to logical impossibilities are out.
pity.

20. Originally Posted by peteshimmon
illuminating as it seems the proof of no
integer solutions for cubes was found long ago.
I wonder what the problem was as you have to
go through cubes to get to higher powers!
Please explain. Fermat's theorem is general. The cubes problem is relatively toy-like, isn't nearly as powerful a statement, and has far less value solved.

In what way do "you have to go through cubes to get to higher powers"?

21. Originally Posted by peteshimmon
If you have managed to follow so far, you know
there are no integer solutions on the curved
loci of cubed and above loci. But why? Does
this give any insight?
As others have mentioned or alluded to, you mean "rational solutions" rather than "integer solutions".

But I think they would make interesting curves. As you start with a base of one with exponent 2, your family of points is defined by the triangle with sides (1, a, sqrt(12+a2)) where a goes from 0 to 1. Every possible triangle that satisfies the pythagorean theorem (has a right angle), will be similar to one of that family of triangles. Similarly, the family (1, a, (13+a3)1/3) where a goes from 0 to 1 will contain all possible combinations where the three values are related by cubes. We now know that the locus of all those points (for integer exponents three and up) completely avoid points with three rational components.

22. If you care to actually plot these curves, using (1, a, (1n+an)1/n) where a goes from 0 to 1, then this should do it:

Divide the interval from 0 to 1 into m intervals, then a = M/m where M goes from 1 to m are the values that we are going to plot.

Let c = (1n+an)1/n

Then plot the point (x, y) where
y = 2(1+a4+c4-2a2-2c2-2b2c2)
x = sqrt(a2-y2)

I think I got that right

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Well I know if you re-arrange my loci plots
and make the base the longest side then the
loci of all right angled triangles is a
semi circle with the base the diameter.
(Ah..gimme that old time geometry!).
And I suppose the loci of higher powers
becomes flatter curves within. And the ones
between 2 and 3 flll the space between the
semi circle and the cube loci.

I should have said "seems to me" you have to
go through cubes to get to higher powers. It
seems that if "rational" answers are
precluded at the power 3 then implicitly
they are also lost at higher powers.

24. Originally Posted by peteshimmon
Well I know if you re-arrange my loci plots
and make the base the longest side then the
loci of all right angled triangles is a
semi circle with the base the diameter.
But you duplicate some, which isn't true the way you did it in the OP
Originally Posted by peteshimmon
I should have said "seems to me" you have to
go through cubes to get to higher powers. It
seems that if "rational" answers are
precluded at the power 3 then implicitly
they are also lost at higher powers.
"Seems"? "implicitly"?

Sure, you can say that now, now that you know the answer...

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Your guess is as good as mine!

26. Originally Posted by hhEb09'1
Then plot the point (x, y) where
y = 2(1+a4+c4-2a2-2c2-2a2c2)
x = sqrt(a2-y2)

I think I got that right
Of course I didn't

y = .5 * sqrt(1+a4+c4-2a2-2c2-2b2c2)
Originally Posted by peteshimmon
Your guess is as good as mine!
Certainly not going to argue about that now

27. Originally Posted by peteshimmon
I should have said "seems to me" you have to go through cubes to get to higher powers.
OK: to you.

I thought you might have been restricting yourself to certain methods that required proving for 3 before, or at least in parallel with, proving for n.

I don't see anything inherent in Fermat's Theorem that requires "going through cubes to get to higher powers".

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Originally Posted by peteshimmon
I should have said "seems to me" you have to
go through cubes to get to higher powers. It
seems that if "rational" answers are
precluded at the power 3 then implicitly
they are also lost at higher powers.
Well, multiples of 3 would be precluded if 3 was precluded. For instance,
if x, y, and z satisfy x6 + y6 = z6, then
x2, y2, and z2 would satisfy the relationship for n = 3.

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