All even squares are twice the sum of two square numbers. All odd squares are the twice of the sum o

[Charles Kusniec]

Dear administrator, please delete these post. Thank you.
The correct is in link: https://www.mersenneforum.org/showthread.php?t=27893
Thenk you.

[Charles Kusniec]

All even square numbers can be written in the form of two times the sum of two square numbers.

If \(x=2y^2+2k^2=square=s^2\), then the square is even.

We get solutions for integer \(s\) only when:
\(y=a_{n+1}=6*a_n-a_{n-1}\) for \(a_{-1}=k\) and \(a_0=7k\).

Then, for k=1 we get solutions for integer \(s\) only when:
\(y=a_{n+1}=6*a_n-a_{n-1}={1,7,41,239,…}=\) https://oeis.org/A002315
Where \(a_{-1}=k=1\) and \(a_0=7k=7\).

Then, for \(k=2\) we get solutions for integer \(s\) only when:
\(y=a_{n+1}=6*a_n-a_{n-1}={2,14,82,478,2786,16238,…}=\) https://oeis.org/A077444 ,
Where \(a_{-1}=k=2\) and \(a_0=7k=14\).

And so on…

See the figure [All even squares are twice the sum of two square numbers.]Attachment 27043

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All odd square numbers can be written in the form of the twice of the sum of two oblong numbers plus 1.

If \(x=2y^2-2y+2k^2-2k+1=square=s^2\), then the square is odd.

We get solutions for integer \(s\) only when:
\(y=a_{n+1}=6*a_n-a_{n-1}-2\) for \(a_{-1}=k\) and \(a_0=7k-3\).

Then, for \(k=1\) we get solutions for integer \(s\) only when:
\(y=a_{n+1}=6*a_n-a_{n-1}-2={1,4,21,120,697,…}=\) https://oeis.org/A046090
Where \(a_{-1}=k=1\) and \(a_0=7k-3=4\).

Then, for \(k=2\) we get solutions for integer \(s\) only when:
\(y=a_{n+1}=6*a_n-a_{n-1}-2={2,11,62,359,2090,…}=\) Axxxxxx
Where \(a_{-1}=k=2\) and \(a_0=7k-3=11\).

And so on…

See the figure [All odd squares are the twice of the sum of two oblong numbers plus 1.]Attachment 27042

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Map of colors for the figures:
Attachment 27041

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Some questions:
1. Why \(6a_{n}\) is present in both recurrence solutions?
2. Why the oblong recurrence solution has \(-2\)?
3. Why \(7k\) is present in both recurrence solution in the initial term \(a_{0}\)?
4. Why does the \(a_{0}\) initial term of the oblong recurrence solution has \(-3\)?

[kriesel]

Why not do it yourself? It's your blog, so you would have sufficient permissions there.

[Charles Kusniec]

Really? I do not know where is this command. Please, could you tell me where is? Thank you.

[Charles Kusniec]

LLTest, I cannot access this moderation menu. For me nothing happens when I click on the top right corner the way you showed. Thank you.

[EdH]

Instead of deleting for you in this case, we need to try to get you working as moderator within your blog. Let's troubleshoot:

It looks like you have moved threads, or was that done by another means?

I think you may have to do the steps kriesel suggested as 1 3 2, but let's see.

step 1 allows you to choose the posts on which you wish to perform the action. Check there. The little Go box (near the Moderation box) should count how many posts you have chosen.

step 3 chooses the action. Click within the dialog box just under Moderation, below the last post on the page (probably says Merge Posts). Click on preferred action.

step 2 says, "Go." do the task. You may need to provide another verification at this point, when deleting.

[Charles Kusniec]

Hi EdH,
Suddenly everything got confused in this post.
I had prepared everything to fall into the NUMBER THEORY group, but when I posted it fell here and still fell wrong.
Anyway
(1) I did not move and do not know how to move threads;
(2) I also don't know why I have a blog under my name (I didn't create it, don't know how to manage it and don't care to manage it)
I appreciate your explanation, but none of it is open for me.
Anyway, I prefer to keep posting my mathematical notes without managing anything.
The only thing I ask now is to clean up this post so that only one clean structure appears.
Thank you very much and have a good week,

[Viliam Furik]

Originally Posted by Charles Kusniec

All even square numbers can be written in the form of two times the sum of two square numbers.

This one is trivial. If the square number is even, it must be divisible by 2, and also 4, as all powers of prime divisors of squares must be even for it to be a square number. If you take a square s and divide it by 2, you get s/2, which is still divisible by 2. And because quarters of even square numbers are still square numbers, then s/2 = 2 * s_1, and s_1 is still a square number, equal to the quarter of the original square number s.

[Charles Kusniec]

@Viliam Furik I lost where is the sum of 2 square numbers in your reasoning...

If you are saying "then \(s/2 = 2 * s_1\), and \(s_1\) is still a square number" as the sum of two squares, it is true, but incomplete.

This is a trivial solution for \(square_1=2*(square_2+square_2)\) = 2 times the sum of two equal squares.

There are an infinite number of non-trivial solutions with the sum of 2 different squares.

Some non-trivial solutions for \(square_1=2*(square_2+square_3)\): 100=2*(49+1), 400=2*(196+4), 900=2*(441+9), ...etc.

See all the possible solutions at https://www.mersenneforum.org/attach...chmentid=27043 .

[Dr Sardonicus]

Originally Posted by Charles Kusniec

Hi EdH,
Suddenly everything got confused in this post.
I had prepared everything to fall into the NUMBER THEORY group, but when I posted it fell here and still fell wrong.
Anyway
(1) I did not move and do not know how to move threads;
(2) I also don't know why I have a blog under my name (I didn't create it, don't know how to manage it and don't care to manage it)
I appreciate your explanation, but none of it is open for me.
<snip>

It would appear that you have a blog under your own name to serve as a repository for those of your posts/threads the Powers that Be deem to be unsuitable wherever else you may post them on the Forum.

The reason you can not use Blogger powers is that you don't have Blogger powers.

It would thus seem your best option is to post exclusively to your blog.

[Charles Kusniec]

Originally Posted by Dr Sardonicus

It would appear that you have a blog under your own name to serve as a repository for those of your posts/threads the Powers that Be deem to be unsuitable wherever else you may post them on the Forum.

The reason you can not use Blogger powers is that you don't have Blogger powers.

It would thus seem your best option is to post exclusively to your blog.

Thank you Dr Sardonicus. If so, then would it be possible for me to be given the Blogger powers to delete everything that does not refer to math content from this thread? Thank you.

(...and before I forget, although I don't know who you are, I really enjoy your math observations. I would be very happy if you could study this thread. I believe there is an "expansion" of this thread that solves a problem that was very famous...)