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::
Ken Denmead
Ken is a husband and father from the San Francisco Bay Area, where he works as civil engineer. He became the Publisher of GeekDad in 2007, and the owner in 2010. He also wrote the NYT bestselling GeekDad series of project books for parents and kids to share.
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.