Question

The square of any odd number is 1 more than a multiple of 8

true or false

Answer 1

is this case, can zero be considered as a multiple of eight ?,

or is this the only exception , which would make the original statement false

Answer 2

@TuringTest

Answer 3

\[5^2=25\]\[\implies (8 \times 3)+1=25\]so true.

Answer 4

\[1^2-1=0\times8\],

but is zero a multiple of eight ?

Answer 5

i am assuming 1 is an odd number ,

Answer 6

i think yes because after multiplying with zero the result is coming 0 so not surely but i think that 0 is the multiple of 8.

Answer 7

According to Wikipedia and several other sites Zero (0) is in fact a multiple of everything.

Answer 8

so m i correct @PhoenixFire ???

Answer 9

I would say the statement is true. Including zero.

Answer 10

but that definition is inconsistent with other definitions,
that would imply that
the lowest common multiple of any numbers would be zero

Answer 11

yes i'm also with ur opinion.

Answer 12

@UnkleRhaukus No, because the LCM is defined as a positive integer.

Answer 13

so what is LCM of 10 and 0

Answer 14

If either of the two integers are 0 then the LCM 0.

Answer 15

From Wikipedia: "usually denoted by LCM(a, b), is the smallest positive integer that is divisible by both a and b.[1] If either a or b is 0, LCM(a, b) is defined to be zero."

Answer 16

well that is clearly a contradiction ,

Answer 17

Hmmm... I don't think that's right. I think the LCM of 10 and 0 would be 10. Majority of the definitions I find agree with one thing: the LCM cannot be 0. However, 0 is still a multiple of everything and is included in the explanation of finding LCMs.

Answer 18

@Zarkon

Answer 19

\[\rm \large (2n + 1)^2 =4n^2 + 4n + 1 = 4(n^2 +n ) + 1 \]

Answer 20

even?

Answer 21

Dude, \(\rm \large n \) can be even.

Answer 22

\[\large \rm 4(4k^2+2k)+1=8(2k^2+k)+1\]Proved if n=2k=even

Answer 23

^what did you prove?

Answer 24

If n = 2k + 1 = odd,
\[\rm \large 4(4k^2+4k+1+2k+1)+1=4(4k^2+6k+2)+1=8(2k^2+3k+1)+1\]

Answer 25

and yes @UnkleRhaukus 0 is a multiple of 8

Answer 26

Multiple of 8 + 1 ^^

Answer 27

im not convinced ,

Answer 28

Why not?

Answer 29

mathematicians never are

Answer 30

unless things become too complicated to understand....only then do they agree (when they can't understand it anymore)

Answer 31

i think it's a mathematician's psychological disorder

Answer 32

The square of any odd number is given by (2n + 1)^2 where n can be of any parity.
= 4n^2 + 4n + 1

Because n can be of any parity, we'd first prove it for odd, then even which I did above.

Answer 33

masochism

Answer 34

because im not convinced i know the definition of '' is a multiple of '' that is intended in the question

Answer 35

but i do agree with @UnkleRhaukus though... @ParthKohli 's proof is incomplete

Answer 36

How so? Do you want one full post of this proof?

Answer 37

1) that's impossible

2) refer to the definition

Answer 38

you'll have to let 2k^2 + 3k = x

and then rewrite 8x + 1 <--only then will that proof be complete

Answer 39

it's a formality in proving...you won't understand....

Answer 40

The square of any odd number is given by (2n + 1)^2 where n can be of any parity.
= 4n^2 + 4n + 1
= 4(n^2 + n) + 1

Because n can be of any parity, we'd first prove it for odd, then even.

Case 1 - EVEN: if n is even, then n = 2k where k is another integer.
We have 4(n^2 + n) + 1
sub n = 2k,
4(4k^2 + 2k) + 1
= 8(2k^2 + k) + 1

That's a multiple of 8 + 1.

Case 2 - ODD: if n is odd, then n = 2k + 1 where k is another integer.
We have 4(n^2 + n) + 1
= 4((2k + 1)^2 + 2k + 1) + 1
= 4(4k^2 + 4k + 1 + 2k + 1) + 1
= 4(4k^2 + 6k + 2) + 1
factoring 2 out
= 8(2k^2 + 3k + 1) + 1

Multiple of 8 + 1 ^^

Answer 41

QED

Answer 42

@lgbasallote I've manipulated and rewritten it as 8x + 1 in both cases.

Answer 43

@UnkleRhaukus

Answer 44

you still haven't

Answer 45

OK.

EVEN: let x = 2k^2 + k
then we are left with 8x + 1

ODD: let x = 2k^2 + 3k + 1
then we are left with 8x + 1

Answer 46

Wow, this is the first time I successfully proved anything by myself.

Answer 47

i still believe false

Answer 48

What else?

Answer 49

ParthKohli's proof is correct.

Answer 50

i am not convinced that can zero be considered as a multiple of eight

Answer 51

lol

Answer 52

All multiples of 8 are in form 8k where k is an integer.
8k = 0
k = 0

and k is an integer

QED #2

Answer 53

Here's an interesting fact :
1^2 = 8*0 +1
3^2 = 8*1 +1
5^2 = 8*3 +1
7^2 = 8*6 +1
9^2 = 8*10 +1
11^2 = 8*15 +1
...
1,3,6,10... are triangular numbers of the form (1/2)n(n+1), but zero is not a triangular number .

the triangle numbers can be related to the square numbers by
\[(2n+1)^{2}=8T _{n}+1\]= \[T _{n-1}+6T _{n}+T _{n+1}\]
\[T _{n}\] is a triangular number