More fun in 2015, Part 8
How many polynomials p ( x ) with integer coefficients (PICs) divide x 2 0 1 5 − 1 , in the sense that x 2 0 1 5 − 1 = p ( x ) q ( x ) for some PIC?
Here is some background info for those who have not studied this kind of number theory yet:
For any positive integer n , we define the cyclotomic polynomial Φ n ( x ) = ∏ ( x − w ) , where the product is taken over all primitive n th roots of unity, w . We are told that Φ n ( x ) is a PIC and that it cannot be factored into two PICs of positive degree ( Φ n ( x ) is "irreducible" ) .
Extra-credit question: Explain why the cyclotomic polynomials have integer coefficients.
The answer is 512.
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1 solution
Exactly! Thanks! (+1)
Small correction: 2015 has 8 positive integer divisors (and 16 integer divisors)
For the benefit of other readers, would you mind explaining, in layman's terms, why the cyclotomic polynomials have integer coefficients.
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I fail to understand. Isn't it that x^2015-1 is the 2015th cyclotomic polynomial???????????? these 2 shares all 2015 roots and moreover have same coefficient of x^2015(=1). Please explain to me.
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Please anyone explain
The cyclotomic polynomial involves only the primitive roots of unity, e 2 k π i / 2 0 1 5 , where k is co-prime with 2 0 1 5 . Thus the degree of Φ 2 0 1 5 ( x ) is ϕ ( 2 0 1 5 ) = 1 4 4 0 . Each 2015-th roots of unity is a primitive root of unity for some divisor d of 2015, so that x 2 0 1 5 − 1 = ∏ d ∣ 2 0 1 5 Φ d ( x ) , as Patrick states. Since 2015 has eight positive divisors d , there will be eight such (irreducible) factors Φ d ( x ) .
Edited, thanks.
Almost got caught out by the "negative" divisors.
Loving these probems by Otto. They are so beautiful and always leave me wanting to learn more about them.
I too am confused. I challenge you to list the 8 divisors. The prime factorization is (5)(13)(31)
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The 8 positive integer divisors of 2015? Sure: 1, 5, 13, 31, 65, 155, 403, 2015.
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@Patrick Corn OK, I see each of the unique divisors d creates a partition over the integers n less than or equal to 2015 because n with gcd(n,2015) falls in one class d, and each polynomial degree i for any integer will have a gcd(i, d)=1 and thus be a primitive root of the non reducible cyclotomic polynomial. I still don't see why multiplying these primitive roots in each factor would always yield integer coefficients (Part b of the question) but I can probably find that in one of my text books. @Debjani Banerjee I recommend that you read the second half of math.bard.edu/belk/math318/CyclotomicPolynomials.pdf and Jones Elementary number theory chapters 5 and 6, especially proofs of theorems 5.8 and 6.5 before this makes sense.
i'm still confused it's prime factorisation is (5).(13).(31) and so , by Euler's phi function it should have 1440 cyclotomic(PIC) divisors .....help me out
But why do cyclotomic polynomials have integer coefficients? And is it coincidence that the 105th is the first with a coefficient that is not 0, 1, or -1?
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The answer to your first question is in the wiki . The answer to your second question is that it turns out that Φ n ( x ) has coefficients 0 , ± 1 for any n of the form 2 a p b q c , where p , q are odd primes. (This is due to Migotti, from the late 19th century. I am not sure how elementary the proof is.)
The second paragraph of the question implies that the factorization of x n − 1 into irreducibles is x n − 1 = d ∣ n ∏ Φ d ( x ) and 2 0 1 5 has 8 positive integer divisors.
Since x 2 0 1 5 − 1 factors as a product of 8 distinct irreducible integer monic polynomials, the answer is 2 8 ⋅ 2 = 5 1 2 , where that last factor of 2 is due to multiplying each monic factor by the units ± 1 .