RE: [PrimeNumbers] Digest Number 2748
Continuing to think out of the box consider the following,
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))
Conversely, for any real a,b we can find x,y to get x^2 + y^2 = z^2.
To: email@example.com; plesala@...
Date: Thu, 16 Jul 2009 00:00:06 -0400
Subject: Re: [PrimeNumbers] Digest Number 2748
> __________________________________________________________Hello Peter.
> 1a. Pythagoras theorem
> Posted by: "Peter Lesala" plesala@...
> Date: Tue Jul 14, 2009 9:13 am ((PDT))
> 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.
> 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
Cino Hilliard proved that your formula does not capture all the possible
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,
and we derive your form.
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.
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