Question

Let m and n be integers. Then prove that m and n have different parity iff m^2-n^2 is odd.

Answer 1

You can do this through contradiction, by first assuming m and n are both even, then assuming they're both odd, and showing neither can be true.

Here's a hint: If m is odd, then m^2 is odd. If m is even, then m^2 is even.

Answer 2

All squared even numbers are even and all squared odd numbers are odd.

when you subtract an odd and a even the result is an odd number

Answer 3

noo i need to use the if and only if format

Answer 4

You'll have to prove it backwards eventually, but I'd focus on proving it forwards first.

Answer 5

wait let me copy and paste the format

Answer 6

\[
(2 p+1)^2-(2 q)^2=4 p^2+4 p-4 q^2+1
\]

Answer 7

It is one line proof

Answer 8

wiat let me show u an example of the way it needs to be proved

Answer 9

\[
(2 p)^2-(2 q+1)^2=4 p^2-4 q^2-4 q-1
\]

Answer 10

\[
(2 p + 1)^2 - (2 q)^2 =
4 p^2 + 4 p - 4 q^2 + 1 = 2 (2 p^2 + 2 p - 2 q^2) + 1= 2 k +1
\]

where \[ k=2 p^2 + 2 p - 2 q^2

\]
So it is odd.

Answer 11

@swissgirl did you get it?

Answer 12

yaaaa ok i was just looking at the format i gotta use

Answer 13

i gotta show that if p then q
and if q then p

Answer 14

I thought i had to use a more complicating format

Answer 15

ok i gotcha then

Answer 16

No, it is easy.
Take one even 2p+1 and one even 2q and expand you get it in the from
2k +1 so it isodd.

Answer 17

that was great. thanks