Is the lowest common multiple of two odd numbers, an odd number? I was really just wondering, if this holds true. I've tried to find a counterexample, but I couldn't. I thought that maybe I could prove it with: $$
lcm(a,b) = \prod_{\text{p prime}}p^{\max\{ord_p(a), ord_{p}(b)\}}.
$$
Maybe the statement isn't even actually true?
 A: It is true, and it follows easily from the formula you listed in your post. If we consider 2 odd numbers $a,b$, then their prime decomposition does not contain the number $2$. As $lcm(a,b)$ can be expressed as a product of prime powers (see your formula), that do not contain $2$ because it is not in the prime decomposition of either $a$ or $b$, it follows that $lcm(a,b)$ is odd, because a product of odd numbers remains odd.
A: Recall: $\gcd(a,b) \cdot \operatorname{lcm}(a,b) = ab$. Thus, if $a = p_{1}^{a_1} \ldots p_k^{a_k}$ and $b = p_{1}^{b_1} \ldots p_k^{b_k}$ are the prime-power factorizations of $a$ and $b$, where each $a_i$ and $b_j$ is a nonnegative integer, then $$\gcd(a,b) = p_1^{\min\{ a_1, b_1\}} \ldots  p_k^{\min\{ a_k, b_k\}}.$$ Since $\min\{x,y\} + \max \{ x,y\} = x + y$ for all $x$, $y \in \mathbb{R}$, it follows that $$ \begin{aligned}[t] \operatorname{lcm}(a,b) = \dfrac{ab}{\gcd(a,b)} &= \dfrac{p_{1}^{a_1} \ldots p_k^{a_k} \cdot p_{1}^{b_1} \ldots p_k^{b_k}}{p_1^{\min\{ a_1, b_1\}} \ldots  p_k^{\min\{ a_k, b_k\}}} \\ &= p_{1}^{a_1 + b_1 - \min\{a_1,b_1\}} \ldots p_{k}^{a_k + b_k - \min\{a_k,b_k\}} = p_{1}^{\max\{a_1,b_1\}} \ldots p_k^{\max\{a_k,b_k\}} .\end{aligned}$$
A: Shorter:
Let $m=\operatorname{lcm}(a,b)$, $d=\gcd(a,b)$. We have
$$md=ab\equiv 1\cdot 1\mod 2,\enspace\text{hence}\enspace m,d\equiv 1\mod2.$$
A: Hint $ $ odd $ a,b\mid 2n\,\Rightarrow\, a,b\mid n,\,$ so an even natural is not a least common multiple of odds.
A: If $a,b$ are odd, then their product is odd.
Since their product is a common multiple of $a,b$, it is a multiple of $lcm(a,b)$, which means that $lcm(a,b)$ is odd.
