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

• Feb 12th 2006, 11:43 AM
Natasha
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 :confused:
• Feb 12th 2006, 12:15 PM
CaptainBlack
Quote:

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 :confused:

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

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

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

$N=3k+\rho$,

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

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

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

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

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

RonL
• Feb 12th 2006, 12:27 PM
Natasha
Quote:

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

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

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

$N=3k+\rho$,

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

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

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

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

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

RonL

Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??
• Feb 12th 2006, 12:35 PM
CaptainBlack
Quote:

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
• Feb 12th 2006, 12:38 PM
Natasha
Quote:

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

RonL

I see. Thanks! :)