20170126, 11:03  #1 
"Rashid Naimi"
Oct 2015
Remote to Here/There
1984_{10} Posts 
Regarding Squares
Can there be a square s such that 2 does not divide s and 3 does not divide s.
And s1 does not divide 6? For example s+1 divides 6? In other words are there any such squares which are not of the form 6n+1? If not, why not? Thank you in advance. 
20170126, 11:12  #2 
Jun 2003
2^{3}·607 Posts 
Let r be the root of s. If neither 2 nor 3 divides s, then neither 2 nor 3 divides r, Hence r=+/1 (mod 6) ==> r^2 = s == 1 (mod 6).
QED Last fiddled with by axn on 20170126 at 11:16 
20170126, 11:24  #3 
"Rashid Naimi"
Oct 2015
Remote to Here/There
1984_{10} Posts 
That does not prove that it can not be of the form 6n1. But it never is.
Why always of the form 6n+1 and never of the form 6n1? 
20170126, 12:37  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
quite obvious:
if we don't allow divisibility by 2 or 3 we get a number of form 6n+1 or 6n1 as axn says let's think what happens when we multiply them together if we allow them to be different numbers we get that: (6n+1)*(6v+1) =36nv+6n+6v+1 which is of form 6n+1 (6n+1)*(6v1) =36nv6n+6v1 which is of form 6n1 (6n1)*(6v1) =36nv6n6v+1 which is of form 6n+1 so the only way to get a number of form 6n1 is by multiplying two values that aren't the same remainder on division by 6 and hence their product is not a square. Last fiddled with by science_man_88 on 20170126 at 13:02 
20170126, 12:44  #5 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{6}×31 Posts 
Yes I think I figured it out too.
The reminder of such numbers over 6 can only be 1 or 5 And 1 squared and 5 squared will always have a reminder 1 over 6. Now the ingesting implication is that 6n1 will always be a prime or multiples of singular primes. It will never factor into squares, cubes or the likes. Isn't that correct? 
20170126, 13:00  #6  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
for example (6*n+1)^2*(6v1)^3 works out to be 6j1 edit: as does (6v1)^(2x+1) for any value x>=0 and really the correct implication is that it can never be an even power. Last fiddled with by science_man_88 on 20170126 at 13:30 

20170126, 14:12  #7 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{6}×31 Posts 
Do you have a counter example where 6n1 factors into odd powers?

20170126, 14:15  #8  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
11^3 = 1331 is 1 mod 6 17^3 = 4913 is 1 mod 6 23^3 = 12167 is 1 mod 6 29^3 = 24389 is 1 mod 6 35^3=5^3*7^3=42875 is 1 mod 6 etc. Last fiddled with by science_man_88 on 20170126 at 14:24 

20170126, 16:48  #9  
"Sam"
Nov 2016
2×163 Posts 
Quote:
The way you look at it squares take one of the 4 forms: 6n+0, 6n+1, 6n+3, 6n+4 So if you were asking if any squares of the form 6n1 exist then the answer to your question is no. 

20170126, 17:52  #10 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{6}×31 Posts 
Thank you for all the replies.
Are there any counter examples for even powers greater than squares? Say powers 6, 10 or the likes? Thank you SM 
20170126, 18:02  #11 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
there are no even powers greater than squares x^4= (x^2)^2 is the square of another value. x^6= (x^3)^2 is the square of another value they can all be turned into squares.

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