Question

Is there anyway we can show that 2[2*k(k+1)]+1 generates all odd squares?

Answer 1

so like 1^2, 3^2, 5^2, and so on?

Answer 2

yes

Answer 3

i wonder if induction would work

Answer 4

actually it simplifies rapidly

Answer 5

It is true by definition.

Answer 6

Hint : Any odd square is of form 8k+1

Answer 7

oh wait k(k+1) is a product of consecutive integers
2|k(k+1)

Answer 8

so that is where you get your 2*2*2i+1

Answer 9

yes! that ends the proof unless you want to prove that any odd square is of form 8k+1

Answer 10

I'm going to post a visual proof of this as well. :)

Answer 11

there was a question asked by @sourwing
he wanted to prove that if n^2 is odd then n is odd
and he wanted a direct proof
so we can suppose then that n^2=2(2k(k+1))+1
so n^2=(2k+1)^2
and n=2k+1
do you think this is a good direct proof?

Answer 12

Looks legit to me !
also division algorithm is always there for proving these directly

Answer 13

Interesting... that looks pretty clever actually...

Answer 14

division algorithm
can you show me how to apply that

Answer 15

By division algorithm any number can be represented as 2k or 2k+1

if n is of form 2k then n^2 = (2k)^2 = 2(A) = even
if n is of form 2k+1 then n^2 = (2k+1)^2 = 2(B)+1 = odd
\(\blacksquare\)

Answer 16

I think thats not same as proving \( n^2 ~\text{is odd } \implies n ~\text{is odd}\)
but you can use the contrapositive to use division algorithm

its bit dumb compared to your trick

Answer 17

Here's that visual proof I mentioned: