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I am trying to find all of the answers to $r_2(n^2) = 420$, where $N < 10^{11}$. It is for finding lattice points on a circle with points $(0,0), (N,0), (0,N)$, and $(N,N)$. I am (pretty) sure that all of the following answers work:

359125, 469625, 612625, 718250, 781625, 866125, 933725, 939250, 1047625, 1119625, 1225250, 1288625, 1336625, 1366625

where $r_k(n)$ is the number of different squares $k$ (in this case, the sum of 2 different squares) that add up to $n$. (Source)

I have noticed that all of these are divisible by $5^2$, but I know that the answer to this problem ends in 309.

What kinds of numbers would get me 420 for this?

  • 0
    Why do you know the last digits of the answer, where is the problem from?2012-02-11
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    This is for a math/programming problem that was given to me. I was given the answer and asked to make a program that gives me that answer for the question.2012-02-11
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    @Awk34: Please put the information into the body of the post. Already others have asked as well. People shouldn't have to go look up the mathematica manual, or wade deep into the comments, to find out what your question is.2012-02-11
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    @AndréNicolas I am not understanding what you mean. None of my $n$s are perfect squares or twice perfect squares. Also, 125 does not seem to be a factor of all answers, but so far 25 does.2012-02-12
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    @Awk: I checked a few, they were OK. The smallest not divisible by $25$ will be $(5)(13^3)(17^2)$. You will have to make sure not to forget, for example, $(5^7)(13^3)$.2012-02-12
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    So are all answers divisible by 5 and (13 or 17)? I still don't think this will give me enough optimization to find all answers N under $10^{11}$2012-02-12
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    For those interested in the source for this problem, it is Euler Project Problem 233.2012-03-05
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    If, as @user26300 indicated, this question came from Project Euler, [this meta discussion](http://meta.math.stackexchange.com/questions/1090/re-project-euler-questions) would be relevant.2012-03-05
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    It did come from PE. I hope what I have asked isn't considered cheating.2012-03-06

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