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$n=2m-1\implies n+1=2m$.
So, $n\equiv 1\pmod 4, 2m=n+1\equiv 2\pmod 4\implies 2m=n+1\equiv 0\pmod 2$ $\implies n\equiv -1\pmod 2\implies n\equiv 1\pmod 2$.
But, how to find $m$ from this last line of equivalence relations is not clear.

jitender
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    If $n=1+4k$ then we have $4k+1=2m-1\implies 2m=4k+2\implies m=2k+1$. – lulu Feb 20 '18 at 01:52
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    $2m \equiv 2 \mod 4 \implies 4|2m - 2 \implies 2|m - 1\implies m \equiv 1 \mod 2$. Or $2m = 2 + 4k$ so $m = 1 + 2k$ so $m \equiv 1 \mod 2$ – fleablood Feb 20 '18 at 02:10

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$$n = 2m-1\equiv 1\pmod{4}\quad\Rightarrow\quad 2m \equiv 2\pmod{4}.$$ Now, note that in general, if $k\ne 0$ and $ak\equiv bk\mod{ck}$, this means that $ak$ is $bk$ plus a multiple of $ck$, say $rck$: $$ak = bk + rck.$$ Dividing through by $k$ gives $a = b+cr$, so that $a\equiv b\mod c$. Applying this to $2m\equiv 2\mod{(4 = 2\cdot 2)}$ gives $$m\equiv 1\pmod{2}.$$

rogerl
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  • But, then my approach is wrong as it must divide $2m$ by $2$, to get $m$, when dividing the modulus by $2$. – jitender Feb 20 '18 at 01:56
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    Remember $ak = bk \mod ck \implies a \equiv b \mod c$. Your approach isn't wrong so much but in going from $2m \equiv 2 \mod 4$ to $2m \equiv 0 \mod 2$ you are simply throwing away information. – fleablood Feb 20 '18 at 02:17
  • @fleablood Sorry, don't understand your comment. I never went to $2m\equiv 0\mod{2}$... – rogerl Feb 20 '18 at 13:56
  • Oh, sorry rogerl. That comment was meant to be a direct response to jitender's comment "But, then my approach is wrong as it must divide 2m by 2, to get m, when dividing the modulus by 2. " Your method is just fine. Although for the sake of the inexperience of the OP, it might be worth noting that $2m \equiv 2\mod 2*2 \implies m \equiv 2 \mod 2$ is valid and why. – fleablood Feb 20 '18 at 16:48