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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(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^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+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.