a = 1+a^2. Since a can't equal zero, (then 0 = 1), I can divide by a, yielding

1=1/a + a. Substituting for 1 in the original equation,

a=1/a + a + a^2

-1/a = a^2

-1=a^3

a=-1

Checking into the original equation,

-1 = 1+(-1)^2

-1=2

Clearly, -1 doesn't equal two- however, can you find the error?

Here's an interesting polynomial pattern I picked up.

Sum[n, k=1](1) = n

Sum[n, k=1](2k-1) = n^2

Sum[n, k=1](3k^2-3k+1) = n^3

Sum[n, k=1](4k^3-6k^2+4k-1) = n^4

Sum[n, k=1](5k^4-10k^3+10k^2-5k+1) = n^5

Sum[n, k=1](6k^5-15k^4+20k^3-15k^2+6k-1) = n^6

Can you see the pattern?

.....

Well, although I'm having a big fat load of trouble nesting this polynomial pattern, it can be understood this way: n^x = Sum[n, k=1](Natural(n) - Triangle(n) + Tetrahedral(n) -... +/- 1), where Natural(n) is the nth natural number, Triangle(n) is the nth triangle number, and so on. (I'll explain what triangle numbers and tetrahedral numbers are in a second. {Not to offend those who already know what they are.})

The "..." in this formula represents an extension of the natural, triangle, tetrahedral... sequence. Natural numbers can be written in this way:

The nth natural number = n

Triangle numbers can be written in this way:

The nth triangle number = (n(n+1))/2

Tetrahedral numbers can be written in this way:

The nth tetrahedral number = (n(n+1)(n+2))/6

The relationship between these sequences is such- the nth number in the kth kind of number sequence (where natural numbers correspond to k=0, triangles correspond to k=1, and so on) is equal to Product[k, i=0](n+i)/[(k+1)!]

I'm sure I can use combinations in the nesting process instead of this series, (since combinations are closely related to Pascal's Triangle, where these numbers are directly taken from), but I haven't yet figured out how.

Any form of assistance is appreciated.

Logs of Negative Numbers