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?
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).)
??? 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.
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.