Let $r(n)$ denote the number of different ways in which a natural number $n$ can be expressed as the sum of two squares. Is there any function $f(n)$ such that $r(n) \geq n^{f(n)}?$
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No, because there are many $n$ for which $r(n)=0$ but $n^{f(n)} \gt 0$ for any real $f(n)$
Ross Millikan
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@Frpzzd: I don't think so because it answers the question as currently posed. – Ross Millikan Apr 10 '18 at 23:46
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Right! Thanks! Is there any condition that we can assume for $n$ and then find $f(n)$? – Megan Apr 10 '18 at 23:48
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@Megan: This is given in the answer here. It depends on the factorization of $n$ – Ross Millikan Apr 10 '18 at 23:51
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Got it! Thank you! – Megan Apr 10 '18 at 23:52