# Thread: 2, 5, 8, 11, 14...

1. ## 2, 5, 8, 11, 14...

How can I prove that no square number can be in the sequence?

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...

Is it to do with mod 5? And if so how would I do it

2. Originally Posted by Natasha
How can I prove that no square number can be in the sequence?

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...

Is it to do with mod 5? And if so how would I do it
First you need find the general term of your sequence. Now it looks to me
as though the sequence is:

$\displaystyle a_n=2+3n, n=0,1,2,..$

Which suggests you consider the squares modulo 3. Now any natural number
may be written:

$\displaystyle N=3k+\rho$,

for some natural number $\displaystyle k$, and $\displaystyle \rho= 0, 1\ \mbox{or}\ 2$. So

$\displaystyle N^2=9k^2+6k \rho + \rho^2$.

Therefore $\displaystyle N^2(mod\ 3)$ is $\displaystyle \rho^2(mod\ 3)$, but

$\displaystyle \rho^2(mod\ 3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2$.

So as $\displaystyle a_n, n=0,1,2,..$ is congruent to $\displaystyle 2(mod\ 3)$ it cannot be a square for any
natural number $\displaystyle n$

RonL

3. Originally Posted by CaptainBlack
First you need find the general term of your sequence. Now it looks to me
as though the sequence is:

$\displaystyle a_n=2+3n, n=0,1,2,..$

Which suggests you consider the squares modulo 3. Now any natural number
may be written:

$\displaystyle N=3k+\rho$,

for some natural number $\displaystyle k$, and $\displaystyle \rho= 0, 1\ \mbox{or}\ 2$. So

$\displaystyle N^2=9k^2+6k \rho + \rho^2$.

Therefore $\displaystyle N^2(mod\ 3)$ is $\displaystyle \rho^2(mod\ 3)$, but

$\displaystyle \rho^2(mod\ 3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2$.

So as $\displaystyle a_n, n=1,2,..$ is congruent to $\displaystyle 2(mod\ 3)$ it cannot be a square for any
natural number $\displaystyle n$

RonL
Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??

4. Originally Posted by Natasha
Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??
Yes, but the +2 form is slightly more convienient when dealing with modular
arithmetic.

RonL

5. Originally Posted by CaptainBlack
Yes, but the +2 form is slightly more convienient when dealing with modular
arithmetic.

RonL
I see. Thanks!