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Let $r \in \mathbb{N}$ be a natural number. Let $$L \geq 2(r-1)²$$ A paper (on quantum information theory, I'm not an expert in number theory or so...) I'm recently reading now says

"One can easily check that $L$ can be represented as $$L = a(r-1)+br$$ where $a, b \in \mathbb{Z}$ are positive integers and $$K = a + b$$ is even."

My first attempt was of course to apply something like the euclidean algorithm, which gives us $$L = k(r-1)+m$$ where $k, m$ are positive integers, $m<r-1$ (I tried to apply euclidean algorithm for $r$ instead of $r-1$ at first, but this seems to be more useful...)

As we have $$L = k(r-1) + m \geq 2(r-1)²$$ with $m \leq r-2$, we know $$k+1-m-2 \geq 2r(r-1) -(r-2)-2 \geq 2r² -3r -1 \geq (r-1)²$$ for all $r\geq 2$. (The case $r=1$ is not interesting.)

Therefore we can write $$L = (k+1)(r-1)+m-r+1 = (k+1-m-2)(r-1)+(m+1)r$$ where now $(k+1-m-2), (m+1)$ are positive integers. But their sum equals $$(k+1-m-2)+(m+1)=k$$ which of course does not have to be even.

Can somebody give me a hint how to manipulate this way of writing $$L=a(r-1)+br$$ where $a+b$ is even?

This whole thing is a little weird, if you consider e.g. $r=5, L=33\geq 32$, you could write $$33=5r+2(r-1)$$ where $5+2$ is not even, but you also can write $$33=r+7(r-1)$$ where $1+7$ is even (I don't know if an expert in stuff like this would consider this as weird, but I do...)

Thanks for any help :)

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While I'm not sure how to make the OP's idea work, the following is a solution to the problem he poses.

First suppose $(r-1) \nmid L$. Notice that the condition that $a,b$ exist so that $L = a(r-1) + br$ is equivalent to saying that $a,b$ exist so that $r-1 = \frac{L-b}{a + b}$. Choose $b$ so that $L-b = 2k(r-1)$, where $k$ is chosen so that $2k(r-1)$ is the largest multiple of $2(r-1)$ less than or equal to $L$. Then we must have $a + b = 2k$ and that $0 < b \leq r-1$. Because $k \geq r-1$, it follows that $a > 0$.

Now suppose $L = m(r-1)$ for some odd number $m$. Then take $b = r-1$ and $a = m-r$.

Now suppose $L = 2m(r-1)$ for some odd number $m$. Then take $b = 2(r-1)$ and $a = 2(m-r)$.

That should be all the cases.

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