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

1. ## 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?

2. 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.)

3. 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).)

4. 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

5. 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.

6. 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.

7. 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. =)

8. 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.

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# odd number raised to integral power is odd

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