So, I was having some fun today reading through the amazing-but-true thread at Reddit, when I had the idea to mash some crazy things together. Follow me carefully on this (and feel free to correct my math if you notice mistakes):
There are about 7 billion (7×10^9) people on Earth.
If you replaced every person with another Earth full of people (49 billion-billion, or 4.9×10^19), you’d still have fewer people than there are molecules in a drop of water (divide the number of people by Avogadro’s number to get about 8.1368×10^-5 moles; there are 18mL of water in a mole of water, so a we get a little over .0014 mL of water; a drop of water has been defined in some places as 0.05mL).
Extrapolating back out, we can figure out that there are 3.346×10^28 molecules of water in a cubic meter of water (1×10^6 gms of water per m3, 18 grams per mole, Avogadro’s number again; everything is assuming standard temperature and pressure). The recognized volume of our sun is 1.4×10^27 cubic meters. Multiply those together to find out that there would be 4.684×10^55 molecules in a volume of water equal to that of our sun. That’s a big freakin’ number.
But…
It would take enough water to fill 1.7 TRILLION suns to have as many molecules of water as there are ways to shuffle a standard deck of cards (52! = 8.067×10^67). That’s more than 4 times the number of stars in the Milky Way galaxy.
::head asplode::
Everybody.
(N).
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
With two any integers (a) and ( b)::
[a(a+1)/2]^2 – [b(b+1)/2]^2=(b+1)^3+(b+2)^3+……..+a^3.
Therefore:::
z(z+1)/2]^2 =[6(6+1)/2]^2+(6+1)^3+(6+2)^3+……..+z^3
[x(x+1)/2]^2=[5(5+1)/2]^2+(5+1)^3+(5+2)^3+……..+x^3.
[y(y+1)/2]^2=[2(2+1)/2]^2+(2+1)^3+(2+2)^3+……..+y^3.
And:
:[z(z-1)/2]^2=[6(6+1)/2]^2+(6+1)^3+(6+2)^3+……..+(z-1)^3.
[x(x-1)/2]^2=[5(5+1)/2]^2+(5+1)^3+(5+2)^3+……..+(x-1)^3
[y(y-1)/2]^2=[2(2+1)/2]^2+(2+1)^3+(2+2)^3+……..+(y-1)^3
Therefore, first equation:
[z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2=[6(6+1)/2]^2+(6+1)^3+(6+2)^3+……..+z^3 – [5(5+1)/2]^2-(5+1)^3-(5+2)^3-……..-x^3 – [2(2+1)/2]^2-(2+1)^3-(2+2)^3-……..-y^3.
And second equation:
[z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2=[6(6+1)/2]^2+(6+1)^3+(6+2)^3+……..+(z-1)^3 – [5(5+1)/2]^2-(5+1)^3-(5+2)^3-……..-(x-1)^3 – [2(2+1)/2]^2-(2+1)^3-(2+2)^3-……..-(y-1)^3
Because have an extra equation :
[6(6+1)/2]^2 – [5(5+1)/2]^2 – [2(2+1)/2]^2.=[5(5+1)/2]^2 – 2[2(2+1)/2]^2
Therefore, first equation become:
[z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2=[5(5+1)/2]^2 – 2[2(2+1)/2]^2+(6+1)^3+(6+2)^3+……..+z^3 -(5+1)^3-(5+2)^3-……..-x^3 – (2+1)^3-(2+2)^3-……..-y^3.
And second equation become:
[z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2=[5(5+1)/2]^2 – 2[2(2+1)/2]^2+(6+1)^3+(6+2)^3+……..+(z-1)^3 -(5+1)^3-(5+2)^3-……..-(x-1)^3 – (2+1)^3-(2+2)^3-……..-(y-1)^3
Having:
[5(5+1)/2]^2 – 2[2(2+1)/2]^2=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3.
Therefore:
First equation:
[z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3+(6+1)^3+(6+2)^3+……..+z^3 -(5+1)^3-(5+2)^3-……..-x^3 – (2+1)^3-(2+2)^3-……..-y^3.
And second equation
[z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3+(6+1)^3+(6+2)^3+……..+(z-1)^3 -(5+1)^3-(5+2)^3-……..-(x-1)^3 – (2+1)^3-(2+2)^3-……..-(y-1)^3
Define the function:
f(z,x,y)=[z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2
So:
f(z-1,x-1,y-1)=[z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2.
And define another function:
g(z,x,y)=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3+(6+1)^3+(6+2)^3+……..+z^3 -(5+1)^3-(5+2)^3-……..-x^3 – (2+1)^3-(2+2)^3-……..-y^3.
So:
g(z-1,x-1,y-1)=(-1)^3+(-2)^3+(2+1)^3+(2+2)^3+……..+5^3+(6+1)^3+(6+2)^3+……..+(z-1)^3 -(5+1)^3-(5+2)^3-……..-(x-1)^3 – (2+1)^3-(2+2)^3-……..-(y-1)^3
Suppose:
z^n=x^n+y^n.
Special case:
z^3=x^3+y^3.
So:
[z(z+1)/2]^2 – [x(x+1)/2]^2 – [y(y+1)/2]^2=[z(z-1)/2]^2 – [x(x-1)/2]^2 – [y(y-1)/2]^2.
So:
f(z,x,y)=f(z-1,x-1,y-1)=g(z,x,y)=g(z-1,x-1,y-1)!!!!!!!!
And more.
z^3=[z(z+1)/2]^2 – [a(a-1)/2]^2-(a+1)^3-(a+2)^3-……..-(z-1)^3 – a^3..
So;
with infinity value of integer (a) have infinity forms of z^3 that equal series cube integers.
How found two integers x and y that satisfies infinity equations.
ADIEU.
Dear everybody.
(N).
With two any integers (a) > ( b)::
a^3=[a(a+1)/2]^2 – [b(b+1)/2]^2 – [(b+1)^3+(b+2)^3+……..+(a-1)^3].
So:::
x^3=[x(x+1)/2]^2 – [3(4)/2]^2 – [4^3+5^3+……..+(x-1)^3].
y^3=[y(y+1)/2]^2 – [4(5)/2]^2 – [5^3+6^3+……..+(y-1)^3].
z^3=[z(z+1))/2]^2 – [5(6)/2]^2 – [6^3+7^3+……..+(z-1)^3].
So
4^3=[x(x+1)/2]^2 – [3(4)/2]^2 – [5^3+6^3+……..+(x-1)^3]. – x^3
5^3=[y(y+1)/2]^2 – [4(5)/2]^2 – [6^3+7^3+……..+(y-1)^3]. – y^3.
6^3=[z(z+1))/2]^2 – [5(6)/2]^2 – [7^3+8^3+……..+(z-1)^3]. – z^3.
Because:
6^3 – 3^3 – 4^3 – 5^3=0.
So:
[z(z+1))/2]^2 – [5(6)/2]^2 – [7^3+8^3+……..+(z-1)^3]. – z^3. – 3^3 – [x(x+1)/2]^2 + [3(4)/2]^2 |+ [5^3+6^3+……..+(x-1)^3]+x^3 – [y(y+1)/2]^2 + [4(5)/2]^2 + [6^3+7^3+……..+(y-1)^3]. +y^3=0.
So;
z^3 – x^3 – y^3.=[z(z+1))/2]^2 – [5(6)/2]^2 – [7^3+8^3+……..+(z-1)^3. – 3^3 – [x(x+1)/2]^2 + [3(4)/2]^2 |+ [5^3+6^3+……..+(x-1)^3]- [y(y+1)/2]^2 + [4(5)/2]^2 + [6^3+7^3+……..+(y-1)^3]..
We had had this equation for any trio integers (z,x and y) , if had an extra equation z^3 – x^3 – y^3=0.
Then:
z,x and y can not be integers.
ADIEU.
(N+).
Two formulas
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
And :
x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a)^3+(x+a+1)^3+……..+(y-1)^3].
Because:
(x+a)^3=[(x+a)(x+a+1)/2]^2 – [(x+a)(x+a-1)/2]^2.
So:
x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]
Suppose:
z^3=x^3+y^3.
So:
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]
Attention!
This equation have four invariant square integers . The rest will change over integers (a).
Corresponding to a value of (a) will have a corresponding equation.
Impossible found integers to satisfy too many equations.
Example:
62^2 -32^2=25^2-14^2 – c^2+d^2 – [e^3+f^3+g^3+h^3+t^3+r^3]
62^2 -32^2=25^2-14^2 – p^2+q^2 – [e^3+f^3+g^3+h^3+k^3+r^3]
62^2 -32^2=25^2-14^2 – t^2+n^2 – [e^3+f^3+g^3+h^3+t^3+s^3]
62^2 -32^2=25^2-14^2 – h^2+k^2 – [v^3+f^3+g^3+h^3+t^3+r^3]
Too many equations.
How satisfied them by only integers !!!!!!!!
So:
z^3=/x^3+y^3
Similar.
z^n=/x^n+y^n.
ADIEU.
Attention:
Only (N+).
Fermat’s last Theorem z ^ 3 = x ^ 3 + y ^ 3 is capable exists a solution if fully meet the following conditions:
First step:
(1+2+3+4+……..+a)^2+(1+2+3+4+……..+b)^2=v^2.
In fact, using the computer, this equation has the ability to survive.
Second step:
(1+2+3+4+……..+a+1)^2+(1+2+3+4+……..+b+1)^2=s^2.
Third step:
v=1+2+3+4+……..+c.
In fact, using the computer, this equation has the ability to survive.
Fourth step:
s=1+2+3+4+……..d.
Fifth step:
d=c+1
If all five steps are satisfied.This equation is capable of existence.
[z(z+1)/2]^2 – [z(z-1)/2]^2=[x(x+1)/2]^2- [x(x-1)/2]^2+[y(y+1)/2]^2 – [y(y-1)/2]^2.
Because:
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2.
Mean this equation is capable of existence.
z^3=x^3+y^3.
However, too hard to satify all five equations in same time..
And more:
Attention about series of number:
1,3,6,10,15,21,28,36,45……..
Recognize:
10 and 15 are two number consecutive which belong this string.
Having:
15^2 – 10^2=5^3.
Or:
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2.
Impossible in same time exist both:
[z(z+1)/2]^2=[x(x+1)/2]^2+[y(y+1)/2]^2
And
[z(z-1)/2]^2=[x(x-1)/2]^2+[y(y-1)/2]^2
Attention:
All numbers as z(z+1)/2 and x(x+1)/2 and y(y+1)/2 and z(z-1)/2 and x(x-1)/2 and y(y-1)/2 are belong this string and they are Pythagorean
This is main proof:
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
Define:
x<x+a<y.
x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a)^3+(x+a+1)^3+……..+(y-1)^3]
Suppose:
z^3=x^3+y^3.
Because also:
(x+a)^3= [(x+a)(x+a+1)/2]^2 – [(x+a)(x+a-1)/2]^2.
Therefore a system of equations is generated.
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+b)(x+b+1)/2]^2 + [(x+b)(x+b-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+b-1)^3+(x+b+1)^3+……..+(y-1)^3]
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+c)(x+c+1)/2]^2 + [(x+c)(x+c-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+c-1)^3+(x+c+1)^3+……..+(y-1)^3]
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+d)(x+d+1)/2]^2 + [(x+d)(x+d-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+d-1)^3+(x+d+1)^3+……..+(y-1)^3].
……..
Can not count the number of equations
And more;
Using the two formulas by rotation affect each other:
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
And
[z(z+1)/2]^2=1^3+2^3+……..+z^3.
This method makes root equation z^3=x^3+y^3 is structured as a Transfiguration equation which having unlimited formats as Robot Bumblebee Transformer.
General case:
z ^ n = x ^ n + y ^ n.
Mean:
z ^ (n-3) * z ^ 3 = x ^ (n-3) * x ^ 3 + y ^ (n-3) * y^ 3.
You can structure this equation according to your own discretion when use z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2 then use [z(z+1)/2]^2=1^3+2^3+……..+z^3 then continue by same to become your own format.
ADIEU.
Attention:
Only (N+).
Fermat’s last Theorem z ^ 3 = x ^ 3 + y ^ 3 is capable exists a solution if fully meet the following conditions:
First step:
(1+2+3+4+……..+a)^2+(1+2+3+4+……..+b)^2=v^2.
In fact, using the computer, this equation has the ability to survive.
Second step:
(1+2+3+4+……..+a+1)^2+(1+2+3+4+……..+b+1)^2=s^2.
Third step:
v=1+2+3+4+……..+c.
In fact, using the computer, this equation has the ability to survive.
Fourth step:
s=1+2+3+4+……..d.
Fifth step:
d=c+1
If all five steps are satisfied.This equation is capable of existence.
[z(z+1)/2]^2 – [z(z-1)/2]^2=[x(x+1)/2]^2- [x(x-1)/2]^2+[y(y+1)/2]^2 – [y(y-1)/2]^2.
Because:
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2.
Mean this equation is capable of existence.
z^3=x^3+y^3.
However, too hard to satify all five equations in same time..
And more:
Attention about series of number:
1,3,6,10,15,21,28,36,45……..
Recognize:
10 and 15 are two number consecutive which belong this string.
Having:
15^2 – 10^2=5^3.
Or:
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2.
Impossible in same time exist both:
[z(z+1)/2]^2=[x(x+1)/2]^2+[y(y+1)/2]^2
And
[z(z-1)/2]^2=[x(x-1)/2]^2+[y(y-1)/2]^2
Attention:
All numbers as z(z+1)/2 and x(x+1)/2 and y(y+1)/2 and z(z-1)/2 and x(x-1)/2 and y(y-1)/2 are belong this string and they are Pythagorean
This is main proof:
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
Define:
x<x+a<y.
x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a)^3+(x+a+1)^3+……..+(y-1)^3]
Suppose:
z^3=x^3+y^3.
Because also:
(x+a)^3= [(x+a)(x+a+1)/2]^2 – [(x+a)(x+a-1)/2]^2.
Therefore a system of equations is generated.
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+b)(x+b+1)/2]^2 + [(x+b)(x+b-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+b-1)^3+(x+b+1)^3+……..+(y-1)^3]
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+c)(x+c+1)/2]^2 + [(x+c)(x+c-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+c-1)^3+(x+c+1)^3+……..+(y-1)^3]
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+d)(x+d+1)/2]^2 + [(x+d)(x+d-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+d-1)^3+(x+d+1)^3+……..+(y-1)^3].
……..
Can not count the number of equations
And more;
Using the two formulas by rotation affect each other:
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
And
[z(z+1)/2]^2=1^3+2^3+……..+z^3.
This method makes root equation z^3=x^3+y^3 is structured as a Transfiguration equation which having unlimited formats as Robot Bumblebee Transformer.
Similar,this method is used for general case:
z ^ n = x ^ n + y ^ n.
Mean:
z ^ (n-3) * z ^ 3 = x ^ (n-3) * x ^ 3 + y ^ (n-3) * y^ 3.
You can structure this equation according to your own discretion when use z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2 then use [z(z+1)/2]^2=1^3+2^3+……..+z^3 then continue by same to become your own format.
ADIEU.
Fermat’s last Theorem is the transformer bumblebee Robot.
Original equation:
z^n=x^n+y^n.
Mean:
z^(n-3)*z^3=x^(n-3)*x^3+y^(n-3)*y^3.
Using the formula z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2 to convert z^3 become the exponent 2. Then using the formula [z(z+1)/2]^2=1^3+2^3+……..+z^3 to convert the exponent 2 become the exponent 3.
Repeated several times with the same method.
The transformer bumblebee Robot was created according your own structure.
Simplest format about Transformer Bumblebee Robot.
Using two formulas:
z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
And define x<x+a<y.
x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a)^3+(x+a+1)^3+……..+(y-1)^3]
Because:
(x+a)^3= [(x+a)(x+a+1)/2]^2 – [(x+a)(x+a-1)/2]^2
So also:
x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]
Original equation:
z^3=x^3+y^3.
According to above method, the transformer bumblebee Robot system was created:
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+b)(x+b+1)/2]^2 + [(x+b)(x+b-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+b-1)^3+(x+b+1)^3+……..+(y-1)^3]
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+c)(x+c+1)/2]^2 + [(x+c)(x+c-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+c-1)^3+(x+c+1)^3+……..+(y-1)^3]
[z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+d)(x+d+1)/2]^2 + [(x+d)(x+d-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+d-1)^3+(x+d+1)^3+……..+(y-1)^3].
……..
Flood robot on the planet Orion galaxy looking for the integers to eat.
Certainly not enough the integer for the large number of transformer bumblebee Robot
ADIEU.