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What is the proof that for any given element $c$ of $F_q$, there exist two elements $a$ and $b$ of $F_q$ such that $a^2 + b^2 = c$. i know that $q$ is the characteristic of this field, but i don't see how this leads, in any way, to a solution to the proof. Thanks a lot.

Michael Hardy
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  • do you really mean that $q$ is the characteristic of $F_q$? Usually people write $q = p^k$, where $p$ is prime, so $F_q$ is the finite field of $q$ elements, and has characteristic $p$. – oxeimon May 06 '15 at 19:15
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    It is tangential but often one also writes $F_q$ when $q$ is a prime power, not a prime, in which case it is not the characteristic. – quid May 06 '15 at 19:16
  • ahh, that is very true. q, generally is the overall number of elements in $F$. –  May 06 '15 at 19:30
  • but then, having put it that way, what then is the general relationship between a characteristic and the primitive elements of $F_q$ am a little confused now :( –  May 06 '15 at 19:35
  • The question I declared this to be a duplicate of nominally covers the case of the prime field only. But if you check out the arguments, they work in any finite field. – Jyrki Lahtonen May 06 '15 at 21:00
  • @JyrkiLahtonen why not close the other way round? (also one cannot just say it is odd and deal with $2$). – quid May 06 '15 at 21:11
  • @quid: Basically I'm still looking for the first time this question came up. There was an identical occurrence 2 days ago. And another one. I have a recollection of this having showed up a couple years earlier today, but my search-fu is weak. The result is used as a step in many other answers, and unfortunately the search hits those... – Jyrki Lahtonen May 06 '15 at 21:19
  • @JyrkiLahtonen I had not seen that one. But honestly if you cannot find easily an argument for the finite fields case I wonder why you dupe-close it in a somewhat forced way. – quid May 06 '15 at 21:23
  • Fair enough, @quid. Reopening. I cannot rule out the possibility that I was simply immensely irritated by seeing the exact same question for the third time in a span of two days. Others and myself decided only to give a hint in a comment + links to the first of these. Eager answering of duplicates annoys me, but that's no excuse to use my superpowers. – Jyrki Lahtonen May 06 '15 at 21:29
  • @JyrkiLahtonen, i didnt see that. actually i tried searching for it before asking but nothing of the source came up. i wonder why. Thanks though. –  May 06 '15 at 21:31
  • @JyrkiLahtonen that this had shown up twice yesterday is unfortunate. Of course I could somehow have though it was asked at some point. Anyway the point is not that it is closed as a duplicate or that you closed it alone, but that I do not see why the general version should be closed into the special case. Actually Jack's version is quite thorough. Maybe that one should actually be the final target. – quid May 06 '15 at 21:38

1 Answers1

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Observe two things:

  • The set $S = \{ a^2 \colon a \in F\}$ has at least $(q+1)/2$ elements (note that at must two elements can have the same square and $0$ has a unique square).

  • $c= a^2 +b^2$ with $a,b \in F$ is the same as $S \cap (c-S)$ is non-empty.

quid
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  • I have tried that on $F_9$ and it seems true for this case, i just cant prove it for the general case. –  May 06 '15 at 19:52
  • Which of the two points is not clear? – quid May 06 '15 at 19:53
  • the second comment. The one with the intersection. Where do the $q - (q+1)/2$ elements belong? –  May 06 '15 at 20:04
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    The cardinality of $S$ is at least $(q+1)/2$ and the cardinality of $c-S$ is the same as the cardinality of $S$ thus you have two sets of cardinality at least $(q+1)/2$ is a set of cardinality $q$. The two sets thus cannot be disjoint. – quid May 06 '15 at 21:09
  • thanks, that makes it better :) –  May 06 '15 at 21:33