Mathematics - 2015-03-19 (page 6 of 6)

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Chris's sis

23:00

It's not just about days, weeks, but months, years.

robjohn

@Chris'ssis Ah, the alternating version

Chris's sis

@robjohn Yeap.

robjohn

@Chris'ssis you mean $k$ and $k+3$ instead of $2011$ and $2014$?

Chris's sis

@robjohn For that particular case, yes. It can be generalized in terms of $n$ and $k$.

robjohn

@Chris'ssis do you have further generalization?

Chris's sis

23:03

@robjohn 2 generalizations, for the simple one and for the alternating one (I sent the first one to AMM).

I also plan to send the second one to some math magazine.

robjohn

@Chris'ssis Yes, you mentioned that this morning.

I have to work harder... I still have some answers without $0$ votes on my profile ;-)

Chris's sis

@robjohn I think the guys from AMM receive a lot of stuff to be published. I receive answers back with much difficulty. I mean it takes a lot of time to receive a feedback for a proposed problem.

The amount of work there must be insane.

robjohn

@Chris'ssis will they take just problems, or do they require an article to accompany it?

Chris's sis

@robjohn Well, my solution is a bit complex, but it doesn't require some mentioned articles by other authors or so.

If you referred to that.

I also thought initially to consider it as an article.

robjohn

23:31

@Chris'ssis I wasn't sure if they wanted more context than a proof. It has been a while since I've read AMM

Chris's sis

@robjohn It's about proposed problems section. The articles cleary need more context.

Committing to a challenge

Ideas for finding $a,b$ that satisfy $ax+b=a^{-1}x - a^{-1}b$ in $\Bbb Z/ 29 \Bbb Z$?

@Robjohn Is there some easy way to find these $a,b$?

(other than the trivial case)

(e.g. a=1, b=0)

robjohn

@Committingtoachallenge is this supposed to be true for all $x$?

Committing to a challenge

Yep

The motivation is trying to find all keys for the affine cipher $ax+b = X$ such that encyption is the same as decryption

so $ax+b = X$ means decryption is $x=\frac{X-b}{a}$

But if encryption and decryption are the same, then that big $X$ is the same as the little $x$, since either way we obtain the same thing

so $ax+b = \frac{x-b}{a}$

all keys (a,b) I mean

And this is in $\pmod {29}$ here

robjohn

23:49

@Committingtoachallenge if you use $x=0$, then you need $b=-a^{-1}b$ which means that $a=-1$ or $b=0$

@Committingtoachallenge if $b=0$ then we need $a^2=1$ and if $a=-1$, then $b$ can be anything.

@Committingtoachallenge Therefore, it looks as if you have two solutions: 1. $a=-1$ or 2. $a=1$ and $b=0$

Committing to a challenge

So the second is definitely the case, since it doesn't change anything in encryption or decryption

robjohn

@Committingtoachallenge The first is something like $b-x$

Committing to a challenge

And the first just inverses it and shifts it in both cases

evinda

Do you maybe have an idea?

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