Sorry if this is too elementary for this sub-forum.

With n = 1, 2, 3, 4 I can see that will produce polynomials that would consist of terms with even coefficients and , which is to say odd numbers. How can I prove this for all n?

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- Sep 1st 2010, 12:25 PMDiscriminantProof that an odd number raised to an integral power is odd
Sorry if this is too elementary for this sub-forum.

With n = 1, 2, 3, 4 I can see that will produce polynomials that would consist of terms with even coefficients and , which is to say odd numbers. How can I prove this for all n? - Sep 1st 2010, 12:27 PMundefined
- Sep 1st 2010, 01:17 PMHallsofIvy
Two ways:

1) Use the bionomial theorem:

2) Proof by induction:

Certainly if n= 1, is odd. Suppose that, for some i, [tex](2k+ 1)^i[tex] is odd. Then is the product of two odd numbers and so odd, as undefined says.

(To prove "the product of two odd number is odd" just look at (2k+ 1)(2j+1).) - Sep 1st 2010, 01:19 PMtonio
- Sep 1st 2010, 01:28 PMDiscriminant
- Sep 1st 2010, 01:36 PMundefined
??? You should recheck what you wrote

Odd times odd is odd, we learn this in primary school, proof without modular arithmetic (2k+1)(2m+1) = 4km+2k+2m+1

Use weak induction on exponent n in {0,1,2,...} and for negative exponents note that the result is only an integer if absolute value of base is 1.

Edit: HallsofIvy already did it. - Sep 1st 2010, 01:48 PMDiscriminant
- Sep 1st 2010, 06:28 PMmelese
I jsut want to show a different way. The result is clear for 1 because , odd. By the Fundamental Theorem of Arithmetic any integer has a unique factorizaton into primes (not considering the order), say ; the exponents are positive. Then .

It follows that and have precisely the same prime divisors. In particular, if and only if , i.e., if is of the form for some integer, then is of the same form.