- #1

- 144

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r^3 + 2(r^2)s + (s^3) = 0

I don't understand the next part of the solution. Why introduce this prime number that divides s.

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Suppose a prime number p divides s.

This implies p divides 2(r^2)s + (s^2)

Or, this implies also the above equals 2(r^2)pa + (p^3)(a^3) = p(2(r^2)a + (p^2)(a^3)) which implies p divides (r^3) which also means p divides r which is contradiction because (r,s) = 1

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So I follow the above steps, but what does prime number p dividing s or r have anything to do with this?

Next part of solution.

--> If s = 1, (r^3) + 2(r^2) + 1 = 0 means r((r^2) + 2r) = -1 implies r = 1 or -1, but 1 and -1 are not roots, seen by evaluating.

Same argument for s = -1.

This means that r/s is not a root, so not irreducible in Q[x]

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So I'm lost about this step too. Only thing that I think is that if you let s = 1 or -1, it's as if you are trying to find something about a root in Z...