Natural Blogarithms Ramblings of a Christian Mathematician and Bioinformaticist

There are a couple of problems above.

#1) A=ab/2 WOULD be an integer.

#2) If the greatest common factor of a set of numbers is 1, then it does have a GCF. You cannot say “(In other words, if they have no common factors it is a primitive pythagorean triple”

In response to Chris (above): No, I don’t believe such a pair of triples exists, but I don’t have a proof for you. I have an idea for one, though.

If {a, b, c} and {2a, d, c} were two triples, then by number theoretic arguments involving parity and the fact that squares are 0 or 1 mod 4, we could conclude, I think, that a must be even. Since I got a contradiction (that a is both even and odd) by assuming a to be odd.

Assuming a is even, I think we can also then assume these triples to be PPTs and eventually that c – a, c + a, c – 2a, and c + 2a are all perfect squares. And if x = c – 2a, then we can get that x, x + a, x + 3a, and x + 4a are all perfect squares. I’m pretty sure it can be proven without too much trouble that the squares simply can’t be spread out so linearly like that, no matter how you choose them.

Sorry. Best I can do for now.

Yes, I am mistaken. That’s not sufficient, or at least I can’t prove that it is. But I’m so close to certain about squares not spreading linearly. And if v^2 + u^2 + 4uv is a square, then 2uv must equal 2u + 2v + 1, or 4u + 4v + 4, or 6u + 6v + 9, or 8u + 8v + 16, . . . Eliminating the odd numbers, factoring minorly and dividing by 2, we get that uv = 2(u + v + 1), or 4(u + v + 2), or 6(u + v + 3), , , , uv = 2n(u + v + n) for some natural number n. We can assume gcd(u,v)= 1 and that they are of opposite parity. Something’s gotta give, but I don’t know what.

NEVERMIND!!!! A funny thing happens when you square 2a, you get 4a^2 not 2a^2! Back to square one, so to speak.

Ah, no. The arcs are not in the same ratio as their corresponding chords. So, the angle I called O/2 would actually be smaller than that, less than half of angle O, I think. But this general route still may well give a genuine contradiction.

get a life Keith!!!!!!!!!!!!!!!

Pythagorean Triples Summary and Clarification:

Regarding the question about whether or not two Pythagorean triples of the form {a,b,c} and {2a,d,c} exist. I don’t think so — still don’t have a proof — but the question leads to lots of interesting questions about squares and rationals.

Let’s assume such triples exist. We can also assume without loss of generality that they are both primitive triples. If we now assume that a is odd, we get a contradiction. Observe:

{a,b,c} = {u^2 – v^2, 2uv, u^2 + v^2}
{2a,d,c} = {2xy, x^2 – y^2, x^2 + y^2}

for some u,v and x,y such that each pair is relatively prime and of opposite parity.

That means 2xy = 2a ⇒ xy = a, which means a is even, since either x or y is even.

Therefore a must be even and

{a,b,c} = {2uv, u^2 – v^2, 2uv, u^2 + v^2}
{2a,d,c} = {2xy, x^2 – y^2, x^2 + y^2}

where 2xy = 4uv and x^2 + y^2 = u^2 + v^2.

From here we get that c – 2a, c – a , c + a, and c + 2a are squares, because these are really

x^2 – 2xy + y^2, u^2 – 2uv + v^2, u^2 + 2uv + v^2, and x^2 + 2xy + y^2

OR (x – y)^2, (u – v)^2, (u + v)^2, and (x + y)^2.

I suspect that no set of squares are spaced out like that. In other words, I seriously doubt that there exist four squares of the form n^2, n^2 + k, n^2 + 3k, and n^2 + 4k. However, that’s just my mathematical intuition. I just don’t think that a parabola or a square root function will bend through those four linearly spaced points without hitting an irrational.

If such triples do exist, though, we would get the following bonus set of triples:

{4m, 4(u^2 – 4uv + v^2) – 1, 4(u^2 – 4uv + v^2) + 1}
{4(u – v), 4(u – v)^2 – 1, 4(u – v)^2 + 1}
{4(u + v), 4(u + v)^2 – 1, 4(u + v)^2 + 1}
{4w, 4(u^2 + 4uv + v^2) – 1, 4(u^2 + 4uv + v^2) + 1}

for m, w that equal the square roots of c – 2a, c + 2a, respectively.

Interesting.

But while it does seem true that triples with prime hypotenuses are singular, we don’t have the situation that p = a^2 + b^2. We have p^2 = a^2 + b^2.

I also don’t know how we would be guaranteed that both k^2 and l^2 are the sums of squares (that both k and l are hypotenuses). Yet, again, it does seem to be true that hypotenuses factor into other hypotenuses.

I’m looking into infinite descent proofs, but I’m very skeptical of them. There exist infinite descent proofs that arithmetic progressions of four squares cannot occur. This is similar to our situation.

Anyway, thanks for the tip. Will look into it.

So many roads to go down. For example, using the fact that (c – a)(c – 2a),
(c – a)(c + 2a), (c + a)(c – 2a), (c + a)(c + 2a) are all squares and the quadratic formula I can get the following triples

{x, a/2, c + 3a/2}, {y, a/2, c – 3a/2}, {r, 3a/2, c + a/2}, {s, 3a/2, c – a/2}

where x^2 = (c + a)(c + 2a), y^2 = (c – a)(c – 2a),

r^2 = (c – a)(c + 2a), s^2 = (c + a)(c – 2a).

All I need is one good contradiction!

BRAVO KEITH, WELL DONE.

The Beauty of Truth is that it is always simple.

You spoke about generalisation, i.e. comparing the
triples {a,b,c} and {na,d,c}.

I have another generalisation in mind, which I have
studied for more than 10 years.
It is the quadratic form ax^2+(2a-1)xy+ay^2=k, where a,k are integers.
When a=1 we get x^2+xy+y^2=k
When a=1/2 we get (1/2)x^2+(1/2)y^2=k
When a=-3 we get -3x^2-7xy-3y^2=k

My problem was:
Let ax1^2+(2a-1)x1x2+ax2^2=k
Let ay1^2+(2a-1)y1y2+ay2^2=l
Is there any z1 and z2 such that
az1^2+(2a-1)z1z2+az2^2=kl
and that z1 and z2 can be written as a function
of a,x1,x2,y1,y2 ?

Yes there is: z1=ax1y1+ax2y1+(a-1)x1y2+ax2y2
z2=-ax1y1-(a-1)x2y1-ax1y2-ax2y2

When a=1 we can multiply ellipses
When a=1/2 we can mutiply circles and form pyth. triples
When a=-3 we can solve Pell’s equations:
-3x^2-7xy-3y^2=1

I have mainly been working on the case a=1 as a basis for solving Fermats Last Theorem

I am planning to create a website and publicate the
results so far.

A NEW QUESTION

Before I come up with my new question, one last
remark about Keith’s proof of the fact that if
{a,b,c} is a triple, then {d,2a,c} is not.

It was shown that if they did exist, it was leading
to a contradiction: “a = 2yz = xw – yz. So, 3yz = xw. This is impossible, since x, y and w,z are relatively prime and x cannot be equal to w or z (nor can y be w or z). Yet if 3yz = xw, then x = 3z. But this forces y = w. Impossible!”

Maybe there are readers of this Blog who ask the
question: why impossible? (It is more a statement
then a proof).

We read in Keith’s comment of 16 aug:
x^2 = (c + a)(c + 2a), y^2 = (c – a)(c – 2a),
meaning: 9z^2=(c + a)(c + 2a), y^2 =(c – a)(c – 2a)

This means x is divisible by 3 and lets assume
(c+a) is divisible by 3. Then (c-2a) is divisible
by 3, which implies y is divisible by 3.
And both x and y cannot be divisible by 3.

And now my question. Or rather 2 questions:

In my last post I wrote:
“When a=-3 we can solve Pell’s equation:
-3x^2-7xy-3y^2=1″
Meaning I wanted to solve -3x^2-7xy-3y^2=1 in integers.
So, lazy as I was, I first asked advice from my
academic formed teachers,how to handle the problem.
They all advised to substitute x=p+q, and y=p-q
which gives q^2-13p^2=1

Why are mathematiciens so eager in their wish to
“simplify” things, even if that means that they
dramatically change the original question?

Question 1) Why is solving -3x^2-7xy-3y^2=1
totally different from solving q^2-13p^2=1 ?

Question 2) Let q^2-Dp^2=1
What can we say about D for q^2-Dp^2=-1 being
solvable. This question is about the negative
Pell’s equation.

DON’T GIVE UP YET

We have the formula’s:
x^2+y^2=k^2 x=3z
w^2+z^2=l^2 y=w

therefore 9z^2+w^2=k^2 9z^2=k^2-w^
and z^2=l^2-w^2
3^2=5^2-4^2

Further we have the identity formula’s:
(a^2-b^2)(c^2-d^2)=(ac+bd)^2-(bc+ad)^2

(see: http://www.geocities.com/fredlb37/node8.html)

So we have:
9z^2=k^2-w^2 what we can wtite as:

(5^2-4^2)(l^2-w^2)=(5l+4w)^2 – (4l+5w)^2= k^2-w^2
with l>w , k>w and (5l+4w)>(4l+5w)

You have something usefull here??

BASIC NUMBER THEORY AND THE NECESSITY OF ACCEPTING NEGATIVE INTEGERS AS OF EQUAL IMPORTANCE

In my post of 02 September 2009 I showed my discovery:
“My problem was:
Let ax1^2+(2a-1)x1×2+ax2^2=k
Let ay1^2+(2a-1)y1y2+ay2^2=l
Is there any z1 and z2 such that
az1^2+(2a-1)z1z2+az2^2=kl
and that z1 and z2 can be written as a function
of a,x1,x2,y1,y2 ?
Yes there is:

z1=ax1y1+ax2y1+(a-1)x1y2+ax2y2
z2=-ax1y1-(a-1)x2y1-ax1y2-ax2y2 ”

For which I claim copyright; and for most things that will follow, unless stated otherwise.

The reason that I choose to publish my discoveries
is the fact that I believe that knowledge should be
public and accessible.
Another reason for publishing is to prevent others
to make mistakes I did.
I must confess that when I did my discovery described above, I realised that I had
defined things (such as multiplying quadratic forms) totally wrong for more than 20 years.

So let’s rock and roll…

Let’s start at the very beginning
A very good place to start:

We will examine the most special ellips:
x^2+xy+y^2=k

Until we finally arrive at Fermat’s Last Theorem.

(It will be a long, long journey…so stay by my
side…when we walk through the storm, maybe I
can be your gide..)

Links below to see the revised version of my paper on ordering the Pythagorean triples by Pellian sequences (based on Pell-type recursion preserving leg difference):

http://rationalargumentator.com/issue232/PPTs_Pellian.html

http://rationalargumentator.com/issue232/Pellian.pdf

HEY! How about a contest???

Who can either unearth or craft a proof that no pair of Pythagorean triples of the form {a, b, c} and {2a, d, c} exists?

Without loss of generality, I believe, we can safely assume that the triples are primitive and the doubled leg is the even one, ie a is even.

We could extend this to any or all n such that no pair of Pythagorean triples of the form {a, b, c} and {na, d, c} exists.

Whaddayou think???

We can show that no such pairs of triples exist using Dickson’s Theorem:

______________

THEOREM:

To find integer solutions to x^2 + y^2 = z^2 find positive integers r, s, t
such that r^2 = 2st is a perfect square.

Then: x = r + s, y = r + t, z = r + s + t .

______________

Since by assumption the hypotenuses are the same what you are asking for is:

1) [ x, y, z] where s = (z – x), t = (z – y), r = (z – s – t).
2) [2x, d, z] where s’ = (z – 2x), t’ = (z –d), r’= (z – s’– t’).

This requires that (s + t) = (s’ + t’) , that r = r’ , AND that r^2 = 2st = 2s’t’ .

But since st = s’t’ , we can substitute s for s’ or for t’.

But then s = (z–x) AND s= (z–2x) which is impossible because r , s , t must be positive integers. ( r can not be equal to zero).

see: Dickson’s Method http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples#VI.

PS The proof I alluded to above (July 1, 2011) using the Fibonacci Box method is
definitely invalid.

Corollaries and conjecture:
If a pair of triples of the form {a,b,c} and {2a,d,c} exist, then it is easy to show that
c – 2a, c -a, c + a and c + 2a are squares. Just look at their forms in Euclid’s algorithm,
given that a must be even (must equal 2mn, c – a = m^2 – 2mn + n^2,
c – 2a = r^2 – 2rs + s^2, etc.)
Let these squares be represented as M^2, N^2, O^2 and P^2. The first and last two
differ by a = N^2 – M^2, and the middle two differ by 2a. So, we can represent them
as M^2, N^2, 3N^2 – 2M^2 and 4N^2 – 3M^2, with M and N odd. It appears from
inspection of tables that when 3N^2 – 2M^2 or 4N^2 – 3M^2 is a square, the other
misses being an odd square by a factor of 8. In other words, either O or P must be irrational, it appears. In still other words, if you take an interval with perfect square
endpoints and advance it by double its length, one of the new endpoints is not a
perfect square; the image of the original interval under the square root function has
integral length, while the image of the new one has irrational length. If this conjecture is proven true, it would suffice in proving that the original triples could not
exist.
Also, if we represent O as N +q and P as N + p, then we get that q^2 + 2Nq over
p^2 + 2Np must equal 2/3. There are some inherent constraints: p and q are even
integers, with p bigger, and N must be greater than or equal to 1/2q^2 + Nq + 1.
If that quotient can never be exactly 2/3 for qualifying entries of p, q and N, then no
such original triples exist.

Further to my comment of 19 July 2012, the best way to prove Fermat’s Last Theorem is to recognise that with n an integer greater than 2, the binomial expansion of (p+q)^n- (p-q)^n will only have an nth root if p equals q, so that with p equalling q every term in the binomial expansion will contain p^n or q^n as a common factor, and the other factors will add up to 2^n. For example, for n=3, 6p^3+2p^3 equals 2^3 X p^3. This method could be applied to n=2 which are not Pythagorean Triples apparently overlooked by Andrew Wiles.

You copied incorrectly. The area is ALWAYS an EVEN integer.

In reply to Keith Raskin’s comment of 15 September 2012, the whole objective of mathematical proof is to reduce the problem to something which is totally obvious. Andrew Wiles and his associates seem to have forgotten this.