## RE: [PrimeNumbers] Digest Number 2748

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• Hi, Continuing to think out of the box consider the following, Theorem 1. For non zero real numbers a,b,c,x,y,z if x = 2ab, y=a^2-b^2,z=a^2+b^2 then x^2+b^2 =
Message 1 of 2 , Jul 20, 2009
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Hi,
Continuing to think out of the box consider the following,

Theorem 1.
For non zero real numbers a,b,c,x,y,z
if x = 2ab, y=a^2-b^2,z=a^2+b^2 then
x^2+b^2 = z^2.

Clearly, Theorem 1 describes a primitive solution.
A general solution would be x = 2abk, y=(a^2-b^2)k,z=(a^2+b^2)k
for any k.

The Proof comes directly from evaluating the expression.

For any real x,y > 0, there exists real a,b such that
x = 2ab and y = a^2-b^2. Then
a = x/(2b) and
y = x^2/(4b^2) - b^2. Multiplying by 4b^2 we get
4yb^2 = x^2 - 4b^4. Rearranging and dividing by 4
b^4 + yb^2 = x^2/4. Adding (1/2y)^2 (completing the square) we have
b^4 + yb^2 + y^2/4 = (x^2+y^4)/4
(b^2+y/2)^2 = (x^2+y^4)/4
b^2 = sqrt(x^2+y^2)/2 - y/2

Now this is always positive since
sqrt(x^2+y^2)/2 > y/2 and squaring we get
(x^2+y^2)/4 > y^2/4 or x^2 > 0 which is true.

As a result we can solve for real b and a as we desired.

b = sqrt(sqrt(x^2+y^2)/2 - y/2)
a = x/(2b)

Thus, we have shown for any x,y there are real a,b such that
x = 2ab
y = a^2-b^2

If x,y,z are the legs of a right triangle XYZ, the right angle at Z and
vertex at A, with sides x,y,z we construct line ZD perpendicular
to z at point D. Then let XD = t and DY = s so s+t = z.

From similar triangles,
(a^2-b^2)/t = z/(a^2-b^2) or (a^2-b^2)^2 = tz and
2ab/s = z/(2ab) or 4a^2b^2 = sz. Multiplying and adding we have
a^2+b^2 = (s+t)z = z^2. Therefore, z=a^2+b^2 and by
Theorem 1, the sides x,y,z of a right triangle obey the rule
x^2+y^2 = z^2. :-)

Similarly, for any real x,y we can find a and b such that x=2ab,y=a^2-b^2
and z=a^2+b^2 to get x^2+b^2 = z^2.

Example, a=Pi, b=exp(1).

gp > g(x,y) = b=sqrt(sqrt(x^2+y^2)/2-y/2);a=x/2/b;print(a);print(b);print(a^2+b^2);

gp > g(Pi,exp(1))

1.853730863795006085660173257
0.8473702183385572400889453922
4.154354402313313572948121467

Conversely, for any real a,b we can find x,y to get x^2 + y^2 = z^2.

unutterably,

Cino Hilliard.

From: kermit@...
Date: Thu, 16 Jul 2009 00:00:06 -0400
Subject: Re: [PrimeNumbers] Digest Number 2748

> __________________________________________________________
> 1a. Pythagoras theorem
> Posted by: "Peter Lesala" plesala@...
> Date: Tue Jul 14, 2009 9:13 am ((PDT))
>
> Hi,
>
> I figured out that there is a formula for sets of integers for which Pythagoras theorem is true.
>
>
>
> The formula is
>
> 2n + 1 2n(n + 1) 2n(n + 1) + 1
>
> The question I am having difficulty with is whether or not there are sets of integers that do not belong to the set which is defined by the above relation;
>
> Thank you.
>
> Peter.
>
>
>

> Messages in this topic (3)
> __________________________________________________________
> 1c. Re: Pythagoras theorem
> Posted by: "cino hilliard" hillcino368@... hillcino368
> Date: Tue Jul 14, 2009 3:21 pm ((PDT))
>
>
> Hi Peter,
>
>
>
> Not 100% on topic but close enough for a response from me.
>
>
>
> Chances are number fiddlers, hypothesizers and number enthusiasts make this
>
> discovery every day. I pounced upon it messing around with x^n+y^n = z^n
>
> in my yout back in the late 50's. At that time I assumed your result were the
> only solutions.
>
>
>
>
> if x^2+y^2=z^2 then
>
> x = a^2-b^2
>
> y = 2ab
>
> z = a^2+b^2

Hello Peter.

Cino Hilliard proved that your formula does not capture all the possible
triples.

Cino derived the general form

(a^2 - b^2) ^2 + ( 2 a b )^2 = ( a^2 + b^2)^2

If a - b = 1, and b = n, then a = n + 1,

What about if a - b = 3.

(4^2 - 1^2)^2 + (2 * 4 * 1)^2 = ( 4^2 + 1^2).

15^2 + 8^2 = 17^2.

225 + 64 = 289

(15,8,17) is a triple that is not generated by your form.

Kermit Rose

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