If $$2^6k+2\equiv4\pmod9$$ then this implies that $$2^6k+2\equiv4\pmod{18}$$since both sides are even. Can anyone explain to me why this inference is true?
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$9\mid 2n\Rightarrow,9\mid n\Rightarrow 18\mid 2n,$ by Euclid (or directly $,9\mid n = 9n!-!4(2n)).\ $ In OP $, 2n = 2^6k-2\ \ $ – Bill Dubuque Oct 04 '20 at 23:07
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Or $ $ let $,x = 2^6k!+!2,,$ so $,x\equiv 4\pmod{!9\ &\ 2}\Rightarrow x\equiv 4\pmod{!18!=!{\rm lcm}(9,2)}\ $ by CCRT = Constant-case Chinese Remainder Theorem or LCM Universal Property $\ \ $ – Bill Dubuque Oct 04 '20 at 23:27
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Or $ $ $ 9\mid 2n\Rightarrow 2n\bmod 18 = 9(2n/9 \bmod 2) = 9(0/1) = 0\ $ by the $!\bmod$ Distributive Law $\ \ $ – Bill Dubuque Oct 04 '20 at 23:36
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so $2^6k +2\equiv 4 \pmod 9$.
So there is an integer $m$ so that $2^6k + 2 = 4 + 9m$.
But the LHS is even so the RHS must be even which (because $9$ is odd) means that $m$ must be even.
SO there is an integer $n$ so that $k = 2n$ and ....
$2^6k + 2 = 4 + 9\cdot 2n = 4 + 18n$.
But that means $2^6k + 2 \equiv 4 \pmod {18}$.
......
ALternative explanation:
$2^6k + 2 \equiv 4\pmod 9 \implies$
$9|2^6k + 2 - 4$. But $2^6k +2 -4$ is even so $2|2^6k+2-4$. So both $2$ and $9$ divide $2^6+2 -4$ so then $\operatorname{lcm}(2,9) = 18$ also divides $2^6k +2-4$.
And as $18|2^6k + 2-4$ that is the very definiton of $2^6k + 2\equiv 4 \pmod {18}$.
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Lemma (for future use now that we did the above): If $a\equiv b\pmod n$ and $k$ is a common divisor of $a$ and $b$ then $a\equiv b \pmod{\operatorname{lcm}(k,n)}$
Pf: Same way we did the above.
fleablood
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i.e. $,a\equiv b\pmod{!n\ &\ k}\iff a\equiv b\pmod{{\rm lcm}(n,k)}\ $ by CCRT = Constant-case Chinese Remainder Theorem or LCM Universal Property $\ \ $ – Bill Dubuque Oct 04 '20 at 23:20