fermat's last theorem redux

[Thelema]

so wile's has proved fermat's last theorem, that for any a,b,c,n, integers, ab,b,c,>1, n>2 then

a^n + b^n = c^n is false.

Which got me thinking... 'cos we can prove that for any number a, for n=2 there exists a b and c such that a^2 + b^2 = c^2

Which means of course, for any 2n, there exists a b, c such that a^2n + b^2 = c^2.

I started to wonder, what power values could the right hand values of the equation take?

I found the following, and I leave this to you as a proof:

a^2 + b^2 = c^4 for an infinite number of solutions.

a^2 + b^2 = c^8 for exactly one solution.

Can you find the one solution?

Free 4 tabs of etifoxine and 2 tabs of pantogam for the person that gets it.

edited: clarity error rectified

Edited June 28, 2013 by Thelema

[DELETED ACCOUNT]

I gave this a go, but got lost... do you have an analytical solution? I'd be very impressed if so!

Edited June 28, 2013 by antoncan

[Thelema]

sorry, let's change that

I found the following, and leave this to you as a proof:

a^2 + b^2 = c^4 for an infinite number of solutions

a^2 + b^2 = c^8 for exactly one solution

I have edited the previous.

Strangely, I have also found a single solution for

a^2 + b^2 = c^12

and I postulate that there is a single solution for all c^4n, n>1.

In case you're wondering, the c^12 is 450117362^12.

But I won't make it so easy for the c^8 solution ( it's farrrr simpler, I guarantee that c is less than 50)

Sadly, I have now reached the limit of excel spread-sheets and most online computational tools.

Some other interesting things came up....

1) if not c^2 then must be a^2.

2) Also, if you map the progress of the infinite solutions of a with c where a^2 + b^2 = c^4 then strangely the ratio of successive c's and a's approach quite quickly (3 + √8). For Instance, the first few (a,c) sets are (7,5) (41,29) (239,169):

41/7= 5.857 and 29/5=5.8; 239/41=5.829 and 169/29=5.828 etc.......and (3 + √8) = 5.82843

Edited June 28, 2013 by Thelema

[DELETED ACCOUNT]

Hmmm, me thinks you might be getting some integer overflow errors...

Also, if a^2 + b^2 = c^4 has an infinite number of solutions (for different values of c), why does a^2 + b^2 = c^8 not also have an infinite number of solutions, if c in the first is simply c^2 in the second?

[Thelema]

because not all c^4 are c^8! and in fact, only one of them is, that satisfies a^2 + b^2. I think you got your thinking muddled, but you can certainly say that all c^8 are c^4!.

By the way, I was wrong about the c^12, that was an integeroverflow error. I had to go on an online super calculator to check it, but I found a neat little theorem in the process. If you need help, look up the "pell sequence". That might help you on your way.

[Thelema]

just to show I'm not making this shit up, the first non trivial one in the sequence is 7^2 + 24^2 = 5^4

Way cool!

[DELETED ACCOUNT]

Haha, I know you're not making this up, and neither am I!

Here are some solutions to a^2 + b^2 = c^8, for values of (a,b,c):

175, 600, 5

220, 585, 5

336, 527, 5

375, 500, 5

2800, 9600, 10

3520, 9360, 10

5376, 8432, 10

6000, 8000, 10

... so I'm guessing there are infinitely many more triples like this which satisfy the equation which are simply larger variants based on such ratios, which is kind of what I was thinking earlier.

Kind of like how there are an infinite number of integers (0, 1, 2, ... ∞), and reals (0, ..., 0.01, ..., 0.1, ..., 1, ..., ∞), but there are 'more' reals than integers (no integers between [0,1], but an infinite number of reals between [0,1]), just like there are 'more' solutions to c^4 than there are to c^8s -- still infinitely many.

Likewise, with a^2 + b^2 = c^12, we have:

1185, 15580, 5

4375, 15000, 5

5500, 14625, 5

...these are far fewer in between than, say, with c^8, which in turn are far fewer than for c^4, and again for c^2.

Cool problem this, a bit of a head scratcher!

By the way, have you seen the proof that Wiles produced for Fermat's Last Theorem? It's crazzzyyy complex... I could never really understand it. There's a cool documentary about it though!

[Thelema]

OK ! WOW!

You definitely win a prize for this. PM me with your addy.

I was under the impression that the only solution was based on 169 being the only non trivial pell number that is a perfect power was

239^2 + 285606^2 = 13^8

But by reducing yours by common factors, and accounting for the fact we've already established an infinite ^4 -

you got

1) 44^2 + 117^2 = 5^6

2) 336^2 + 527^2 = 5^8

3) 237^2 + 3116^ = 5^10

Which is pretty darn interesting and got me thinking.. .so you also got (3,4,5) and (7,24,5)

There's clearly a pattern: 4, 24, 117, 527, 3116

Which if you type into oeis is AO67312.

Giving grand demonstrations without common factors like

91004468168113^2 + 28515500892816^2 = 5^40

You might think this is all ridiculous.

But if you can prove that any of the series 4, 24, 117, 527, 3116... is not perfect power then you've won a million $US!!

(Not including 4)

But I'm completely in the dark. Can this be done for all numbers? Is 5 special? 13 had a taste?

Hmm?

Help me antoncan, and I'll throw in 5g of Picamilon if you can come up with a general formula

[Thelema]

actually, I'm completely convinced that the only numbers 'ere involved are 0, 1, 2, 5, 13

but maybe that's every 2nd number of the Fibonacacci Series, so 34 should be the next bet.

something like

a^2 +b^2 = 34^2n

[woof woof woof]

some real genius stuff going on in here. I wish I was better equiped to understand this. Interesting prizes! hahaha

Edited July 7, 2013 by woof woof woof

[DELETED ACCOUNT]

Alrighty, *takes a deep breath*, here we go...

Nice pickup about A067312. From this, it's actually really straightforward to determine a method for generating the values!

• Take any primitive Pythagorean triple. For example, say, (8,15,17).
• To enumerate all values for (a,b,c) where a^2 + b^2 = c^2 where c = d^n, we can use any Pythagorean Triple as a starting point in the sequence generator for A067312, which is (x + iy)^n, d=|(x + iy)^n| (the absolute magnitude, corresponding to the hypotenuse). For the initial triple of (8,15,17), the start of the sequence is (as the triple <(x + yi), c, exponent>, where a=|x|, b=|y|):

(8 + 15i), 17, 1

(-161 + 240i), 289, 2

(-4888 - 495i), 4913, 3

(-31679 - 77280i), 83521, 4

(905768 - 1093425i), 1419857, 5

So, for example, we find that 905768^2 + 1093425^2 = 1419857^2 = (17^5)^2.

All this takes is the method for determining powers of complex numbers, or simply repeated multiplication of a complex number with itself, n times.

Bingo!

Edited July 8, 2013 by antoncan

[Thelema]

Wowsers! I'll have to give this some thought. My main questions right now are:

1) Does this guarantee that a,b,c of the final solution are all relatively co-prime?

2) Does this mean that a solution exists for all c^2n? for some a^2 + b^2?

3) Is there a method for determining which, of any, of a,b are powers of some d?

Nice one man....if 1) is true, and you can clarify me on the point of 2) then the 5g picamilon is yours too!

PM me!

[Thelema]

I think I got it
only for "prime centred square numbers" 5,13,41,61,113,181,313..." general formula

giving so far a


Right Hand Side of General Theorem 1

= [n² + (n+1)²]^2k

where [n² + (n+ 1)²] is prime .

Edited July 9, 2013 by Thelema

[DELETED ACCOUNT]

Wowsers! I'll have to give this some thought. My main questions right now are:

1) Does this guarantee that a,b,c of the final solution are all relatively co-prime?

2) Does this mean that a solution exists for all c^2n? for some a^2 + b^2?

3) Is there a method for determining which, of any, of a,b are powers of some d?

Nice one man....if 1) is true, and you can clarify me on the point of 2) then the 5g picamilon is yours too!

PM me!

1) Well, if you enumerate all primitive Pythagorean Triples (using, say, Euclid's method), this guarantees that each (a,b,c) are relatively co-prime. Also, since any powers of relatively co-prime values (a^k, b^j, c^i) are also relatively co-prime, it follows that powers of primitive Pythagorean Triples are also co-prime, so the method ought to guarantee this.

2) No, this means that: for any c in a Pythagorean Triple (a0,b0,c), there exists some a and b such that a^2 + b^2 = c^2n for all non-negative integers n >= 1, as found by the relationship (x + yi)^n, c=|(x + yi)^n|, where initially x = a0, y = b0, and a = |x|, b = |y|. Note that this doesn't actually show that there isn't some c which is not the 3rd value of some Pythagorean triple which satisfies the equation, so I don't know for sure whether or not it covers all solutions (see below)!

3) The above method can compute solutions, however as I said above, I don't know if it's "complete" in the sense it can generate all possible solutions where they may exist. My intuition is that since Euclid's method can generate all primitive triples, that the complex number method will also find all solutions which are powers of c. Any ideas?

I've yet to check your formula, but it looks interesting! How did you get to it?

Edited July 9, 2013 by antoncan

[Thelema]

note slight change to formula

my formula essentially is based on the idea that all c^2n collapse into s^2n where s are prime-centred square numbers. It was a bit of late-early morning mathematical work that followed simply from the formula from prime centred squares and the relationship between the first and last numbers in unique Pythagorean triples.

I add

I originally had the left side of the equation but now I realise it was wrong. back to the drawing board on that.

Can you find the left side of the general equation?

[Thelema]

OK so I've been away for a while thinking a bit about reducing this whole thing down algebraically:

I was a *little" bit wrong above:

To be complete:

for n=2

FIRST GENERAL EQUATION FOR IRREDUCIBLE CO-PRIME TRIPLETS IS

1) (2n+1)^2 + (n^2+(n+1)^2 -1)^2 = (n^2 + (n+1)^2)^2 for odd numbers

OR

2) 4n^2 + (4n^2-1)^2 = (4n^2 +1)^2 for all even numbers divisible by 4.

first observation being that for all a^2 + b^2 = c^2 then there is ALWAYS some a^2 that fits the bill, unless a^2 is even and not divisible by 4.

for n=4

SECOND GENERAL EQUATION FOR IRREDUCIBLE CO-PRIME TRIPLETS IS

1) 24^2 +7^2 = 5^4

120^2 + 119 ^2 = 13^4

485^2 + 220^2 = 25 ^4

1519^2 + 720^2 = 41^4

3479^2 + 1320^2 = 61^4

6936^2 + 2023^2 = 85^4

12319^2 + 3360^2 = 113^4

16820^2 + 12615^2 = 145^4

19873^2 +6864^2 = 181^4

43095^2 + 22984^2 = 221^4

40716^2 + 26975^2 + = 265^4

where the right hand side is simply (n^2 + (n+1)^2)^4

But I am at a complete loss to describe the algebraic relationship between the right and left hand sides here.

HELP! It seems all over the place!

2) the right hand side is simply (4n^2 +1)^4 which reduces into only one irreducible set of coprime triplets, the first few of which are:

240^2 + 161^2 = 17^4

1081^2 + 840^2 = 37^4

4199^2 + 468^2 = 65^4

So at the moment a lot of my work is going into working out the algebraic notation for this crazy n=4 notation for a when c^4. If I can get that nailed then I'm pretty sure I can go on to extrapolate to a general theorem, or at least it will make the work I might easier. I have an idea that logarithms or deviations from square roots or cross products with other reducible c^4 triplets might be involved. Man, Antoncan, you gotta help me on this one. Does your Euclidian multiplication of imaginaries shed any light on the algebraic terms of "a" involved in this "c^4" case?

I realise all of a sudden that we have not actually PROVED that c^n can only take the form of a^2+b^2. Thus we are not able to deduce reliably that c^2n = a^2 + b^2 without loss of generality. I will try to address this in in future posts, but this is a thorny issue indeed.

Edited July 27, 2013 by Thelema

[Thelema]

2OK and it seems totally false that in these situations a^2 +b^2 must = c^2n, in fact they can equal c^n, provided c is still of the form (4n^2+1) or (n^2 + (n+1)^2).
For example, staring right at you in the face, is 11^2 + 2^2 = 5^3 the next simple example is 13^3 = 9^2 + 46^2

Hence these 2 "magic " squares:

for (n^2 + (n+1)^2): (5,13,25,41,61,85,113,145 e.t.c across;n down for n>=0)
.
1 1 1 1 1 1
2 3 4 5 6 7
4 12 24 40 60 84
11 46 117 236 415 621
24 120 278 1519 3429 5544
41 597 3116 x x x


for 4n^2 +1 (17,37, 65, 101...)

1 1 1 1
15 35 63 99
52 198 524 970
240 1081 4199 9401




That might help ye some in derivative-land.

Edited August 4, 2013 by Thelema

[Thelema]

oops

Edited August 4, 2013 by Thelema

[Thelema]

Boyle Conjecture.

Edited August 4, 2013 by Thelema

[Thelema]

SUMMARY

1) There are 2 series that complete all instances of known additions of powers A^i + B^j = C^k. The (4n^2 +1)^k series and the (n^2 + (n+1)^2)^k series are the 2 variants of C.

2) So far I have found 3 complete algebraic formulations for sums of squares (i = j = 2)

a) for k=2 and C=4n^2 +1: 4n^2 + (4n^2-1)^2 = (4n^2 +1)^2

for k=2 and C=n^2 + (n+1)^2: (2n+1)^2 + (n^2+(n+1)^2 -1)^2 = (n^2 + (n+1)^2)^2

c) for k=3 and C=n^2 + (n+1)^2: (2n^3 - 3n - 1)^2 + (n(2n^2 + 6n + 3))^2 = (n^2 + (n+1)^2)^3

I hope to add more in the future. For now I would like to find the complement for k=3 for C= 4n^2 +1, and also the algebraic formulations for k=4.