Prove that if $a$ and $b$ are odd positive integers, then there are integers s and t such that $a = bs + t$ where |t| < b, and t is either zero or odd Prove that if a and b are odd positive integers, then there are integers s and t such that $$a = bs + t,$$ where $|t| \lt b$, and $t$ is either zero or odd.
 A: Since you know the Division Algorithm, there are $s,t$ such that $a = bs+t$ with $\,t \in [0,b\!-\!1]$. If $\,t\ne0\,$ is even then $a = b(s\!+\!1)+t-b,\,$ for $\,t\!-\!b \in [-(b\!-\!1),0],\,$ and $t$ even, $b$ odd $\Rightarrow$ $t-b$ odd.
i.e. if the remainder $\, t\ = a\ {\rm mod}\ b = 2n \ne  0,\,$ toggle its parity by replacing it by $\,t\!-\!b\equiv t\!\!\ \pmod b$.
A: Consider
$$S=\{a-nb:n\in\mathbb Z\}$$
there exists a unique $N\in\mathbb Z$  such that 
$$0 \leq a-Nb \lt b$$
If $ a-Nb$  is odd or zero, let $s=N,t=a-Nb$. otherwise, let $s=N+1, t=a-(N+1)b$
A: I am also looking at this problem right now. I am a bit confused. We know that both $a$ and $b$ are odd and by the division algorithm we have that $a=b\cdot s + t$ where $b>0$ and $0\le t< b$. Consider $a=b\cdot s + t$. Since $a$ and $b$ are both odd, then we may consider two cases:

*

*$s$ is even

*$s$ is odd

If 1., then clearly $t$ has to be either odd or $=0$. If 2., then either $t=0$ or $t$ is even!? Since if we assume that $s$, $a$ and $b$ are odd then we may write $$a=b\cdot s + t=(2n+1)(2j+1)+t=\bigl[ 2\cdot (2nj+n+j)+1 \bigr]+t,$$ so $t$ must be either $=0$ or even whenever $a$ is odd. Is this correct? Because if it is, then it is not true that $t=0$ or $t$ is odd (?).
