# Proof that an odd number raised to an integral power is odd

• Sep 1st 2010, 12:25 PM
Discriminant
Proof 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 $(2k+1)^n$ will produce polynomials that would consist of terms with even coefficients and $1^n$, which is to say odd numbers. How can I prove this for all n?
• Sep 1st 2010, 12:27 PM
undefined
Quote:

Originally Posted by Discriminant
Sorry if this is too elementary for this sub-forum.

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

You can simply use that odd times odd equals odd.

(And for completeness you might want to address the negative integers.)
• Sep 1st 2010, 01:17 PM
HallsofIvy
Two ways:
1) Use the bionomial theorem:
$(2k+1)^n= \sum_{i=0}^n \begin{pmatrix}n \\i\end{pmatrix}2^ik^i= 2\sum_{i=1}^n\begin{pmatrix}n \\i\end{pmatrix}2^{i-1}k^i+ 1$

2) Proof by induction:
Certainly if n= 1, $(2k+ 1)^1= 2k+ 1$ is odd. Suppose that, for some i, [tex](2k+ 1)^i[tex] is odd. Then $(2k+1)^{i+1}= (2k+1)^i(2k+1)$ 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 PM
tonio
Quote:

Originally Posted by Discriminant
Sorry if this is too elementary for this sub-forum.

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

$\forall\,n\in\mathbb{N}\,,\,\,(2k+1)^n=\sum\limits ^n_{i=0}(2k)^i$ , according to Newton's binom, and this is a sum of

even numbers except for the first index $i=0$ , for which we have $(2k)^0=1$ , and thus the result is, by definition, an odd number.

Tonio
• Sep 1st 2010, 01:28 PM
Discriminant
Quote:

Originally Posted by undefined
You can simply use that odd times odd equals odd.

(And for completeness you might want to address the negative integers.)

(i) $(2n+1)(2n+1) = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1$
(ii) $(2n+1)^{-n} = \frac{1}{2n + 1}$. An odd number divided by an odd number will produce an odd number, for:
(iii) $\frac{2n+1}{2n+1} = 1$, which is odd.
• Sep 1st 2010, 01:36 PM
undefined
Quote:

Originally Posted by Discriminant
(i) $(2n+1)(2n+1) = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1$
(ii) $(2n+1)^{-n} = \frac{1}{2n + 1}$. An odd number divided by an odd number will produce an odd number, for:
(iii) $\frac{2n+1}{2n+1} = 1$, which is odd.

??? 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.

• Sep 1st 2010, 01:48 PM
Discriminant
Quote:

Originally Posted by undefined
??? 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.

Thanks! I got confused there with your previous note about completeness and the negative exponents and produced complete nonsense, which attributes parity to fractions. =)
• Sep 1st 2010, 06:28 PM
melese
Quote:

Originally Posted by Discriminant
Sorry if this is too elementary for this sub-forum.

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

I jsut want to show a different way. The result is clear for 1 because $1^n=1=2\cdot0+1$, odd. By the Fundamental Theorem of Arithmetic any integer $m>1$ has a unique factorizaton into primes (not considering the order), say $m=p_1^{m_1}p_2^{m_2}\cdots{p_r}^{m_r}$; the exponents are positive. Then $m^n=p_1^{nm_1}p_2^{nm_2}\cdots{p_r}^{nm_r}$.

It follows that $n^m$ and $n$ have precisely the same prime divisors. In particular, $2|m$ if and only if $2|m^n$, i.e., if $m$ is of the form $2k+1$ for some $k$ integer, then $m^n$ is of the same form.