contrapositive of $a^2 + b^2$ is divisible by $8$ iff $a$ and $b$ are both even

Question

I'm working on creating a contrapositive statement to the one above and this is what I've come up with:

$a^2 + b^2$ is not divisible by $8$ iff $a$ or $b$ are not even.

I know that typically a contrapositive setup follows something like If $A$ then $B$ Contrapositive: Not $B$ then not $A$

I'm just not sure if I have the contrapositive statement right or it should be saying $a^2 + b^2$ is not divisible by $8$ iff $a$ and $b$ are not even.

Hint: $m$ divides $n$ iff $n=qm$ for some $q\in\Bbb Z$; in particular, $c$ is even iff $c=2d$ for some $d\in \Bbb Z$. — Shaun, Apr 14 '19 at 22:19
You can simplfy saying $a^2+b^2$ is not divisible by $8$ iff at least one of $a$, $b$ is odd. — Bernard, Apr 14 '19 at 22:22
@Bernard I thought that it implied at least one of but I wasn't sure. Thank you! — Tunifish17, Apr 14 '19 at 22:33
https://math.stackexchange.com/questions/3186105/sum-of-square-of-integers-when-divided-by-8-cant-leave-a-remainder-of-7 — lab bhattacharjee, Apr 15 '19 at 03:17

William Elliot · Accepted Answer · 2019-04-15T00:02:59.050

0

The proposition is false.
$2^2 + 4^2 = 20$ is not divisible by 8.

The contrapositive to "P implies Q" is "not Q implies not P".
The notion of a contrapositive of an equivalence is bizarre.
It is simply negating both sides.

Thus, 8 does not divide $a^2 + b^2$ iff a or b is odd.
That of course, is false by the above counterexample.

Is there an odd a and an integer b for which 8 divides $a^2 + b^2$?

edited Apr 15 '19 at 00:02

answered Apr 14 '19 at 23:44

William Elliot

18,025

I can't tell, you may have meant bizarre. I don't know what baraaz means. – Will Jagy Apr 14 '19 at 23:50

contrapositive of $a^2 + b^2$ is divisible by $8$ iff $a$ and $b$ are both even

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