Why is $(k L - y) \mod L = (L - y) \mod L$? What's the rule that allows the constant $k$ to be ignored?
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$ a\bmod L = b\bmod L \iff a\equiv b\pmod{!L}$, i.e. $,a,$ and $,b,$ have equal remainders when divided by $L$ iff $,a-b,$ is divisible by $L$ (true here where $,a-b = kL-y - (L-y) = (k!-!1)L)\ \ $ – Bill Dubuque Nov 09 '20 at 22:11
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In your notation $a\pmod b = c\pmod b$ means that $a-c \pmod p =0 \pmod p$ this is just a consequence of the division algorithm. By doing that, notice that the $-y$ cancel and you end up having a $(k-1)\cdot L \pmod L,$ what is the remainder by dividing a number by itself? hint: $L = 1\cdot L +0.$
Phicar
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It doesn't "mean that". Rather, it can be proved to be equivalent to that - see my comment on the question. – Bill Dubuque Nov 09 '20 at 22:15
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@BillDubuque I know, but throwing equivalences to a coder was an overkill from my perspective. – Phicar Nov 09 '20 at 22:25
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But "means" is so imprecise that it often leads to errors, e.g. its denotation could be "by definition" or it could denote a (uni or bi)directional implication, etc. Here it is easy to be precise so there is no reason not to be. – Bill Dubuque Nov 09 '20 at 22:38
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