a=2, b=3, 2ab=12

f(a,b) = a

^{2} + 2ab + b

^{2} = 4 + 12 + 9

g(a,b) = 2(i2)(i3) =

= 2(-1)2*3 = -12

f((a,b) - g(a,b) = a

^{2} + b

^{2} + 12 - 12 = a

^{2} + b

^{2} = 13 which is a positive integer.

QED

NOW try it with n > 3 for the

**binomial theorem** for (a + b)

^{3} and compare with terms in a

^{3} + b

^{3} - g

_{3}(a,b) , any positive integers a and b

Hint: in the expansion, try sticking an i =

in front of the a's and b's in g

_{3}(a,b) and see if you can come up with integers....

(If you're confused about what g

_{3}(a,b) is, let me know...)

Ok, I'll make it easier: (a + b)

^{3} = a

^{3} + 3a

^{2}b + 3 ab

^{2} + b

^{3}
g

_{3}(a,b) = 3a

^{2}b + 3 ab

^{2} = 12 + 12 = 24 and cannot be eliminated by subtraction - that is, multiplication of a positive integer by (i

^{2})

g'(ia,ib) = 3(ia)

^{2}(b) + 3(ia)(ib)

^{2} = -12i -12i = -24i

so g

_{3}(a,b) cannot be eliminated by subtracting -g

_{3}(a,b), and therefore (a + b)

^{3} cannot be a positive integer.

That is, there will always be odd terms in the expansion for higher orders n > 2, which means that there will always be factors of i (as opposed to i

^{2}= (

)

^{2} = -1 in the inner terms in the expansion, so the expansion cannot be eliminated by subtraction for n > 2, a and b positive integers.

(Sigh) Oh Lord, but it's hard to be humble when you're perfect in every way....

(Does anyone know the toll-free number of the Abel committee?)

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