What is the Pattern for Solving X^2 and X^3 Equations?

[Russell E. Rierson]

X^2 :

3^2 = 2*(1+2)+3

4^2 = 2*(1+2+3)+4

5^2 = 2*(1+2+3+4)+5

6^2 = 2*(1+2+3+4+5)+6

X^3 :

3^3 = 3*(1*2 + 2*3)+3

4^3 = 3*(1*2+2*3+3*4)+4

5^3 = 3*(1*2+2*3+3*4+4*5)+5

6^3 = 3*(1*2+2*3+3*4+4*5+5*6)+6

 

[matt grime]

you keep writing these as if there is something deep going on. A little elementary algebra should show you it's not.

 

[honestrosewater]

Russell,

for x and n in N, you can write the sequence x^n using partial sums and n!s. (n factorials)

For n=1, think sigma[y+1], y is nonegative integer
For n=2, think sigma[2y+1]
For n=3, think sigma[6([y^2+y]/2)+1]
...

You can see the connection by expanding (x+1)^n.
Now look at n! and the partial sums of (1, 1, 1...), i.e. (1, 2, 3...) (1, 3, 6...) (1, 4, 10...)...

I didn't see the connection at first either :)

Happy thougths
Rachel

 

[Russell E. Rierson]

you keep writing these as if there is something deep going on. A little elementary algebra should show you it's not.

It's called brainstorming. Everything does not necessarily start out with ...deep thoughts.


3^2 = (1+1)*(1+2) + 3

4^2 = (1+1)*(1+2+3) + 4

5^2 = (1+1)*(1+2+3+4) + 5

[...]

3^2 = (1+1)*(1+2) +3

3^3 = (1+2+3)*(1+3) + 3

3^5 = (1+2+3)*(1+2+3+4)*(1+3) + 3

3^7 = (1+2+3+4+5+6+7+8+9+10+11+12+13)*(1+2+3)*(1+3) +3

 

[Russell E. Rierson]

5^2 = 4*5/2 + 5*6/2

13^2 = 12*13/2 + 13*14/2

17^2 = 16*17/2 + 17*18/2

25^2 = 24*25/2 + 25*26/2

5^2 = 10+15

13^2 = 78+91

17^2 = 136+153

25^2 = 300+325

1+2+3+...+N = N*[N+1]/2

X^2 = X*[X-1]/2 + X*[X+1]/2

X^3 = X*[X^2-1]/2 + X*[X^2+1]/2

X^4 = X*[X^3-1]/2 + X*[X^3+1]/2

X^n = X*[X^(n-1) -1]/2 + X*[X^(n-1) + 1]/2

 

[matt grime]

It's called brainstorming. Everything does not necessarily start out with ...deep thoughts.

But why do you feel the need to just write them out repeatedly without explanation?

 

[Russell E. Rierson]

5*2-1 = 3^2

5*3+1 = 4^2

[5*2-1] + [5*3+1] = 5*2 + 5*3 = 5*[2+3] = 5^2


[13*2-1] + [13*11+1] = 5^2 + 12^2 = 13^2

[17*4 - 4] + [17*13 + 4] = 17^2


Z*U + K = X^2

Z*V - K = Y^2



[Z*U + K] + [Z*V - K] = Z*U + Z*V = Z*[U + V] = Z*Z = Z^2

= X^2 + Y^2


Z = U + V

 

[Russell E. Rierson]

The General Equation?

3*(3+4) + 4*|4-3| = 5^2

3*(3+4) + 4*1 = 5^2

3*(3+4) + 4*(1 + 5^2) = 5^3

3*(3+4) + 4*(1 + 5^2 + 5^3) = 5^4

3*(3+4) + 4*(1 + 5^2 + 5^3 + 5^4) = 5^5

etc...


5*(5+12) + 12*|12 - 5| = 13^2

5*(5+12) + 12*(7 + 13^2) = 13^3

5*(5+12) + 12*(7 + 13^2 + 13^3) = 13^4

etc...

The equation? :

p is a prime number > 2.

z^p = x*(x+y) + y*( |y-x| +...+ z^(p-1) )

 

[Russell E. Rierson]

More random thoughts:


a^n + b^n = c^n

[a+b] > c

[a+b] - d = c


[a+b] = [c+d]


[a+b]^2 = [c+d]^2


a^2 + 2ab + b^2 = c^2 + 2cd + d^2

iff

a^2 + b^2 = c^2

then

2cd + d^2 = 2ab

[...]

[a+b]^3 = [c+d]^3

a^3 + 3ba^2 + 3ab^2 + b^3 = c^3 + 3dc^2 + 3cd^2 + d^3


iff

a^3 + b^3 = c^3


then

3ba^2 + 3ab^2 = 3dc^2 + 3cd^2 + d^3

ba^2 + ab^2 = dc^2 + cd^2 + [d^3]/3

 

[Russell E. Rierson]

Fermat Algebra

a^3 + b^3 = (a+c)^3


a^3 + b^3 = (b+d)^3


(a+c) = (b+d)


a^3 = (b+d)^3 - b^3


b^3 = (a+c)^3 - a^3


a^3 + b^3 = (b+d)^3 - b^3 + (a+c)^3 - a^3


a^3 + b^3 = 3ca^2 + 3ac^2 + 3db^2 + 3bd^2


a^3 + b^3 = 3ac(a+c) + 3bd(b+d)


a+c = b+d


a^3 + b^3 = 3ac(a+c) + 3bd(a+c)


a^3 + b^3 = 3*(a+c)*(ac+bd)


(a+c)*(ac+bd) must be a certain multiple of 3 in order for a^3 + b^3 to be a "cube"


3*9 = 3^3


3*72 = 6^3


3*243 = 9^3


3*576 = 12^3


Interesting...

 

[Hurkyl]

Sheesh, the arithmetic's even wrong. (what happened to d^3 and c^3?)

 

[Russell E. Rierson]

Sheesh, the arithmetic's even wrong. (what happened to d^3 and c^3?)

Thanks for your excellent help Hurkyl

a^3 + b^3 = (a+c)^3

a^3 + b^3 = (b+d)^3

a^3 + b^3 = a^3 + 3ac^2 + 3ca^2 + c^3

a^3 + b^3 = b^3 + 3bd^2 + 3db^2 + d^3


b^3 = 3ac^2 + 3ca^2 + c^3

a^3 = 3db^2 + 3bd^2 + d^3

a^3 - d^3 = 3db^2 + 3bd^2

b^3 - c^3 = 3ca^2 + 3ac^2


same principle:


3*9 = 3^3


3*72 = 6^3


3*243 = 9^3


3*576 = 12^3


Interesting...

 

[Russell E. Rierson]

The abc conjecture:

http://www.math.unicaen.fr/~nitaj/abc.html#Consequences

The abc conjecture implies the asymptotic form of the Fermat Last Theorem, i.e. that there are only finitely many solutions to the equation x^n+y^n=z^n with gcd(x,y,z)=1 and n> 3.

Asymptotic Fermat using L'Hopital's rule:

B > A

A^x + B^x

[A^x + B^x]^[1/x]

L'Hopital's Rule:

Limit f(x)/g(x) = Limit f'(x)/g'(x)

Take the natural log

Ln[A^x + B^x]^[1/x] = Ln[A^x + B^x]/x


= f(x)/g(x)


L'Hopital's Rule...

Limit Ln[A^x + B^x]/x =

[(A^x)*Ln[A] + (B^x)*Ln]/[A^x + B^x] / 1

= LnA/[1 + [B/A]^x] + LnB/[1+[A/B]^x]

Take the limit

= 0 + Ln

e^LnB = B

Limit

for B > A

[A^x + B^x]^[1/x] = B

 

[Russell E. Rierson]

Interesting...

x+y = A

x-y = B

[A+B]/2 = x

[A-B]/2 = y


x^2 + y^2 = [A^2 + B^2]/2

x^3 + y^3 = [A^3 + 3AB^2]/4

x^5 + y^5 = [A^5 + 10A^3 B^2 +5AB^4]/16



So [A^3 + 3AB^2]/4 cannot be a cube...

It is a cube if A = B but x and y are integers greater than zero so A cannot equal B...

 

[Russell E. Rierson]

$x+y = A$

$x-y = B$

$[A+B]/2 = x$

$[A-B]/2 = y$


$x^2+y^2 = [A^2 + B^2]/2$

$x^3+y^3 = [A^3 + 3AB^2]/4$

$x^4+y^4 = [A^4 + 6A^2 B^2 + B^4]/8$

$x^5+y^5 = [A^5 + 10A^3 B^2 + 5AB^4]/16$

$x^6+y^6 = [A^6 + 15A^4 B^2 + 15A^2 B^4 + B^6]/32$

$x^7+y^7 = [A^7 + 21A^5 B^2 + 35A^3 B^4 + 7AB^6]/64$

etc...