Determine the number of factors for the polynomials x^1000000 - y^1000000

Question

What exactly does it mean to find the number of factors? and how would you do that?

Answer 1

ps This was the answer my teacher gave. I just don't understand exactly why or how she got it though.

(x^500000 + y^500000)(x^500000 - y^500000)
(x^500000 + y^500000)(x^250000 + y^250000)(x^250000 - y^250000)
...
(x^500000 + y^500000)(x^250000 + y^250000)(x^125000 + y125000)
this is the part I don't know how she got to:
(x^62500 + y^62500)(x^31250 + y^31250)(x^15625 + y^15625)(x^15625 - y^15625)
⬆️
7 factors

Answer 2

a^2 - b^2 = (a-b)(a+b)
as long as a and b are also squares, you can repeat the process.
Note that in the last step, the exponents are no longer even, so the terms ar no longer perfect squares.

Answer 3

a^2 - b^2 = ( a + b ) ∙ ( a - b )

x^1000000 - y^1000000 = (x^500000)^2 - (y^500000)^2 =

( x^500000 + y^500000 ) ∙ ( x^500000 - y^500000 )

x^500000 - y^500000 = ( x^250000 )^2 - ( y^250000 )^2 =

( x^250000 + y^250000 ) ∙ ( x^250000 - y^250000 )

x^250000 - y^250000 = ( x^125000 )^2 - ( y^125000 )^2 =

( x^125000 + y^125000 ) ∙ ( x^125000 - y^125000 )

x^125000 - y^125000 = ( x^62500 )^2 - ( y^62500 )^2 =

( x^62500 + y^62500 ) ∙ ( x^62500 - y^62500 )

x^62500 - y^62500 = ( x^31250 )^2 - ( y^31250 )^2 =

( x^31250 + y^31250 ) ∙ ( x^31250 - y^31250 )

x^31250 - y^31250 = ( x^15625 )^2 - ( y^15625 )^2 =

( x^15625 + y^15625 ) ∙ ( x^15625 - y^15625 )

15625 can be further factorized because 15625 = 5^6