If $ a \equiv 3\pmod 4$ , $ b \equiv 3\pmod 4 $ Then $ab \equiv 1\pmod4$

Question

If $ a \equiv 3\pmod 4$ , $ b \equiv 3\pmod 4 $ Then $ab \equiv 1\pmod4$.

How can I prove that? If I multiply $a \equiv 3\pmod 4 $ with $ b \equiv 3\pmod 4$

then will it look like? $ab \equiv 9\pmod4 \rightarrow ab \equiv 1\pmod4$ ?

Yes, that's correct. Alternatively, you could use the naive approach that we have $a=4m+3$ and $b=4n+3$ for some integers $m,n$. So, we have,$$ab=(4m+3)(4n+3)=4(4mn+3m+3n+2)+1\implies ab\equiv 1\pmod 4$$ — Prasun Biswas, Dec 12 '16 at 18:13
http://math.stackexchange.com/questions/1603652/show-that-if-c-mid-a-b-and-c-mid-a-b-then-c-mid-aa-bb/1603658#1603658 — user236182, Dec 12 '16 at 18:14

score 1 · Answer 1 · answered Dec 12 '16 at 18:14

1

You know $a=4m+3,\ b=4n+3$ for some integers $m,n$ Now muptiply these, and try to factor out $4$ from as much as possible.

answered Dec 12 '16 at 18:14

coffeemath

score 0 · Answer 2 · answered Dec 12 '16 at 18:15

0

we define $$a=3+4m$$ and $$b=3+4n$$ with $$m,n\in\mathbb{Z}$$ then we have $$ab=1+(8+12m+12n+16mn)$$ thus we can write $$ab \equiv 1 \mod 4$$

answered Dec 12 '16 at 18:15

score 0 · Answer 3 · answered Dec 12 '16 at 18:23

$[a\equiv3\pmod4]\wedge[b\equiv3\pmod4]\implies$

$[\exists{m}\in\mathbb{Z}:a=4m+3]\wedge[\exists{n}\in\mathbb{Z}:b=4n+3]\implies$

$[\exists{m,n}\in\mathbb{Z}:ab=(4m+3)(4n+3)]\implies$

$[\exists{m,n}\in\mathbb{Z}:ab=16mn+12m+12n+9]\implies$

$[\exists{m,n}\in\mathbb{Z}:ab=16mn+12m+12n+8+1]\implies$

$[\exists{m,n}\in\mathbb{Z}:ab=4(4mn+3m+3n+2)+1]\implies$

$[\exists{k}\in\mathbb{Z}:ab=4k+1]\implies$

$[ab\equiv1\pmod4]$

3 Answers3