NeverSayNever's blog

Jacobi's Two Square Theorem
By NeverSayNever, 10 years ago, In English

Jacobi's Two Square Theorem: The number of representations of a positive integer as the sum of two squares is equal to four times the difference of the numbers of divisors congruent to 1 and 3 modulo 4.

I find this on Google , but not able to find any proof . I have brute force it and find that this is correct. Can anyone tell me any proof or whether any test case on which this theorem fails. Really needed please help.

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10 years ago, # |
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Google gives me this on the first page of the results.

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    10 years ago, # ^ |
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    Thank you so much

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      10 years ago, # ^ |
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      You could google it yourself.

      But I still don't see all the four ways of representing 1 as a sum of two squares. 1 = 02 + 12 = 12 + 02. How else?

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        10 years ago, # ^ |
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        02 + ( - 1)2 = ( - 1)2 + 02 = 1

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          10 years ago, # ^ |
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          Oh, the statement is so ambiguous. Two important facts are omitted: negative numbers' squares are allowed and negative divisors are forbidden. It's a shame :).

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        10 years ago, # ^ |
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        for this theorem , We consider both order as well as sign.

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    10 years ago, # ^ |
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    I would be thankful if you explain what is going on in this paper.

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      10 years ago, # ^ |
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      I'd be glad to do it, but my knowledge of formal power series approaches to zero.

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        10 years ago, # ^ |
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        So why you write it as an answer for the question?

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          10 years ago, # ^ |
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          Why not to do that? I googled for "Jacobi's Two Square Theorem" and found this article on the first page of the search result.

          I find this on Google , but not able to find any proof .

          If the author couldn't understand the proof, he must say that he didn't understand the proof, but not what I've just cited.

          BTW, such mathematics-heavy questions should be asked on specialized forums, not here.

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10 years ago, # |
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There is much more easier proof; let p be a prime in form 4k + 1. Then exists a number x such that x2 =  - 1 (p). For example, satisfies this condition because of Wilson Theorem.

Lets consider numbers of type a + bx, where are a, b integers in range . Obviously there are more than p such numbers, so, by Pigeonhole principle, we can get four numbers a, b, c, d({a, b} ≠ {c, d}) such that a + bx = c + dx (p), so, a - c = x(d - b) (p), then, (a - c)2 =  - (b - d)2 (p), so, (a - c)2 + (b - d)2 = 0 (p).

a, b, c, d are in range , so (a - c)2 + (b - d)2 obviously is between 0 and 2p. It's divisible on p, so, it is exactly equal to p. Theorem is proved.

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    10 years ago, # ^ |
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    It is not theorem topicstarter is talking about.

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      10 years ago, # ^ |
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      Oh, I so blind:(

      But, 5 users found it interesting and even put the pluses, so it wasn't very harmful:)

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