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I'm trying to prove this statement:

For all integers $a$ and $b$, $4|(a+3b)$ if and only if $a ≡ b \pmod{4}$.

It's straightforward to prove that if $a ≡ b \pmod{4}$ then $4|(a+3b)$ will be valid. I'm unsure how to prove if $4|(a+3b)$ then $a ≡ b \pmod{4}$.

I've already proved that the relation defined by $4|(a+3b)$ is an equivalence relation (reflexive, symmetric, and transitive). I've used the definition of congruence to re-frame the question as this:

For all integers $a$ and $b$, if $4|(a+3b)$ then $4|(a-b)$.

I've tried thoroughly to get $a-b$ as a multiple of $4$, but I keep running into a wall where I can only prove it is a multiple of $2$.

Thank you for any help anyone can provide!



EDIT: Thank you so much for the answers I've gotten back! They have really helped and I understand congruence much better now. Thanks for taking the time to help!!

bluerockrq
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  • I think you're pretty much done: $a-b = a+3b - 4b$. – Matthew Leingang Nov 26 '19 at 03:33
  • If $a+3b=4k$, add $-4b$ to both sides. Then $a-b=4(k-b)$. – Randy Marsh Nov 26 '19 at 03:33
  • How to brainstorm. If two numbers are divisible by $4$ then their difference is divisible by $4$. So If $4|a+3b$ then we would have $4|(a+3b)-(a-b) = 4b$. We can make this a prove if $4|a+3b$ then $4|a-b\iff 4|(a+3b)-(a-b)=4b$. So $4|a-b$. – fleablood Nov 28 '19 at 02:04

4 Answers4

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Hint: Write $a+3b=(a-b)+4b$.

Pythagoras
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$4|a+3b$

$\iff a+3b\equiv0\pmod4$

$\iff a\equiv-3b\pmod4$

$\iff a\equiv b\pmod4$

$\iff 4|a-b$

J. W. Tanner
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Hint $\, $ Notice that we have: $\, $ if $\ \color{#c00}{m\equiv n}\ $ then $\ m\equiv 0\iff n\equiv 0,\ $ via an obvious proof.

$\!\!\!\begin{align}{\rm In\ OP}\qquad m &= a+\color{#0a0}3b\\ n &= a-b_{\phantom |}\\ \hline\ {\rm since}\ \ \ m-n &= 4b\equiv 0^{\phantom |}\end{align}\!\!\Rightarrow\, \color{#c00}{m\equiv n}\ $

Simpler: $ $ note that $\ \color{#0a0}{3}\equiv -1\,\Rightarrow\, m\equiv n\ $ via Congruence Product & Sum Rules.

Bill Dubuque
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Well $4|a+3b$ means $a+3b \equiv 0\pmod 4$ so $a \equiv -3b \equiv(-3)b \equiv 1b\equiv b \pmod 4$.

Alternatively. There is $k$ so that $4k =a+3b = a+ 4b -b$ so $4(k-b) = a-b$ so $4|a-b$ and $a\equiv b \pmod 4$.

Or let $a\equiv a'$ and $b \equiv b'\pmod 4$ where $a'=0,1,2,3$ and $b'=0,1,2,3$ then $a = 4k+a'$ and $b = 4j + b'$ and $a +3b = 4k + a' + 12j + 3b'= 4k+12j + a'+3b'$ and so $4|a'+3b'$ And $3b' = 0, 3, 6, 9$ if $b' = 0,1,2,3$ which mean $a'+3b'$ must be then only multiple of $4$ equal are greater than by at must $3$. i.e. $a'+3b' = 0, 4, 8, 12$ and so $a' = 0,1,2,3$ if $b' = 0,1,2,3$. So $a' = b'$ and $a\equiv b\pmod 4$.

Or $4|a+3b$ means $4|a + 3b -4b$ so $4|a-b$ and $a\equiv b \pmod 4$.

fleablood
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