Regarding Squares

[a1call]

Can there be a square s such that 2 does not divide s and 3 does not divide s.
And
s-1 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.

[axn]

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

[a1call]

That does not prove that it can not be of the form 6n-1. But it never is.
Why always of the form 6n+1 and never of the form 6n-1?

[science_man_88]

Originally Posted by a1call

Why always of the form 6n+1 and never of the form 6n-1?

quite obvious:

if we don't allow divisibility by 2 or 3 we get a number of form 6n+1 or 6n-1 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)*(6v-1) =36nv-6n+6v-1 which is of form 6n-1
(6n-1)*(6v-1) =36nv-6n-6v+1 which is of form 6n+1

so the only way to get a number of form 6n-1 is by multiplying two values that aren't the same remainder on division by 6 and hence their product is not a square.

[a1call]

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 6n-1 will always be a prime or multiples of singular primes. It will never factor into squares, cubes or the likes.

Isn't that correct?

[science_man_88]

Originally Posted by a1call

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 6n-1 will always be a prime or multiples of singular primes. It will never factor into squares, cubes or the likes.

Isn't that correct?

my example was too simple. Technically, it can be divisible by any odd number of 6v-1 for some value v and any number of 6n+1.
for example (6*n+1)^2*(6v-1)^3 works out to be 6j-1 edit: as does (6v-1)^(2x+1) for any value x>=0 and really the correct implication is that it can never be an even power.

[a1call]

Do you have a counter example where 6n-1 factors into odd powers?

[science_man_88]

Originally Posted by a1call

Do you have a counter example where 6n-1 factors into odd powers?

5^3 = 125 is -1 mod 6
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.

[carpetpool]

Originally Posted by a1call

Can there be a square s such that 2 does not divide s and 3 does not divide s.
And
s-1 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.

x^2 = (0, 1, 3, 4) mod 6. for any integer x. Meaning 0, 1, 3, 4 are quadratic residues mod 6 and 2, 5 are non residues.

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 6n-1 exist then the answer to your question is no.

[a1call]

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

[science_man_88]

Originally Posted by a1call

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

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.