Question

Every odd integer is the difference of 2 squares. Direct Proof.

Let m and n be any integers such that;

2m^2 – (2n+1)^2 = 2k +-1

4m^2 – 4n^2 -4n -1 = 2k +- 1

2(2m^2 -2n^2 -2n) – 1 = 2k +- 1

Does this work?

Answer 1

(2m)^2 – (2n)^2 = 2k +-1

4m^2 – 4n^2 = 2k +- 1

2(m^2 -n^2) = 2k +- 1

Doesnt seem to fit

Answer 2

every odd integer is the difference of 2 consecutive squares

Answer 3

do you mean 2 consecutive squares
or
do you mean the square of 2 consecutive numbers
?

Answer 4

1 - 0 = 1
2 - 1 = 3
3 - 2 = 5
4 - 3 = 7
5 - 4 = 9.

Answer 5

thoseare lazy squares :)

Answer 6

3-2=5?

Answer 7

3^2 - 2^2 = 5 :)

Answer 8

oh lazy squares i get it

Answer 9

ok oh thats neat never realized that i don't think maybe i forgot
let n be the fist integer and let n+1 be the next integer
(n+1)^2-n^2=n^2+2n+1-n^2=2n+1
2n+1 is odd

Answer 10

so (n+1)^2-n^2 is odd

Answer 11

go ahead satellite rain on my parade

Answer 12

your last line is the only one needed right?

Answer 13

yes

Answer 14

yes

Answer 15

maybe

Answer 16

i was wondering if the proof had to be that every odd number is the difference of any 2 squares; but i might be reading to much into it

Answer 17

of course. general odd integer
\[2k+1=(k+1)^2-k^2\]

Answer 18

and of course the teacher said that k would indicate the same integer

Answer 19

would have to include subs

Answer 20

as in the proof that
\[\sum_{k=0}^n 2k+1=n^2\]

Answer 21

start with
\[\sum_{k=0}^n(k+1)^2-k^2=n^2\] and then do the algebra to see that you get
\[\sum_{k=0}^n 2k+1\]

Answer 22

we havent learnted no sum notation so its not allowed ... lol

Answer 23

lol. what you wrote above is exactly the same thing right? like the telescoping sum. one term minus the previous one

Answer 24

it is... but I gots the antiGuass teacher ...

Answer 25

that aint how I showed it to you !!! YOU FAIL!!!

Answer 26

antiguasse? lol

Answer 27

think photoshop comes with an anti gausse feature

Answer 28

i believe it does, but you have to read it in korean