A question regarding multiples of 3

[guifb99]

Why doesn't any odd multiple of 3 can give "24^-1(x²-1)=y" as a result (y) a natural number? Obviously no even number will make "y" a natural number, but all of the odd numbers do, but odd multiples of 3 (3, 9, 15, 21, 27...).

x=1⇒y=0
x=3⇒y=1/3
x=5⇒y=1
x=7⇒y=2
x=9⇒y=10/3
x=11⇒y=5
x=13⇒y=7
x=15⇒y=28/3
x=17⇒y=12
x=19⇒y=15
x=21⇒y=55/3
.
.
.

 

[jedishrfu]

Can you explain this better perhaps by using an example?

 

[guifb99]

Can you explain this better perhaps by using an example?

I'm sorry, I typed it wrongly, it's not "1/2", it's "1/24".

 

[jedishrfu]

Okay, but can you explain this better perhaps by using an example?

Is this a homework assignment?

 

[guifb99]

Okay, but can you explain this better perhaps by using an example?

Is this a homework assignment?

There, now I think it's WAY better understandable. I'm sorry. Lol

 

[TeethWhitener]

Why doesn't any odd multiple of 3 can give "24-1(x2-1)=y" as a result (y) a natural number? Obviously no even number will make "y" a natural number, but all of the odd numbers do, but odd multiples of 3 (3, 9, 15, 21, 27...).

You mean, why does the expression:
$$\frac{1}{24}(x^2-1)$$
not return an integer when $x$ is an odd multiple of 3? Think about what "odd multiple of 3" means mathematically and substitute that for $x$ in the expression above to see what you get.

 

[mfb]

Can x²-1 be divisible by 3 if x is divisible by 3?