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## Main Question or Discussion Point

I managed to get through all of the problems in chapter three of

I noticed that when

For example, 7 = 2

Furthermore, if the powers in the polynomial are successive multiples of

Lastly, the same is true for an arithmetic progression of powers if the lowest power can be factored out to leave the appropriate form, for example:

A proof of this lemma would be immensely useful, because if it is true then it provides extremely clean solutions to a number of the problems in the chapter, but I'm at a loss as to how to prove it.

*Problem Solving through Problems,*and I am now on to chapter 4 in the section on polynomials. A few problems I have encountered so far involve polynomials of the form:*P(x) = 1 + x + x*^{2}+ x^{3}+....+x^{n}I noticed that when

*n*, the degree of the polynomial, is a sum of the first*k*powers of two (starting from 0), then*P(x)*can be factored as such:*P(x) = (x+1)(x*

^{2}+1)(x^{4}+ 1)....(x^{2k}+ 1)For example, 7 = 2

^{0}+ 2^{1}+ 2^{2}so:*1 + x*

^{2}+ x^{3}+ .... + x^{7 }= (x + 1)(x^{2}+1)(x^{4}+ 1)Furthermore, if the powers in the polynomial are successive multiples of

*m*(starting from 0) then the powers in the factorization are the first*k*powers of two times*m*, for example:

1 + x

1 + x

^{4}+ x^{8}+ x^{12}= (x^{4}+1)(x^{8}+1)Lastly, the same is true for an arithmetic progression of powers if the lowest power can be factored out to leave the appropriate form, for example:

*x*^{2}+ x^{3}+ x^{4}+ x^{5}= x^{2}(*1 + x + x*^{2}+ x^{3}) = x^{2}*(x + 1)(x*

^{2}+1)A proof of this lemma would be immensely useful, because if it is true then it provides extremely clean solutions to a number of the problems in the chapter, but I'm at a loss as to how to prove it.