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The problem is "Can you find a value $n$ such that $n^2+1$ is divisible by $3$?"

My analysis: For the divisibility of $n^2+1$ by $3$, we need $n^2 \equiv 2 \pmod{3}$ in other words we need to show that $2$ is quadratic residue of $3$, but $2 \equiv -1 \pmod 3$ which imply that $2$ is quadratic non residue of $3$.Hence, no such $n$ is possible.

I recently learned about quadratic residue and this is probably my first application, so please check if I committed an error?

Thanks,

Srivatsan
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Quixotic
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2 Answers2

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Why does pointing out that $2\equiv -1\bmod 3$ show that $2$ is a quadratic non-residue? You need to fill in your argument. Here is a simple proof that $2$ is a non-quadratic residue mod $3$: $$0^2\equiv 0\bmod 3$$ $$1^2\equiv 1\bmod 3$$ $$2^2\equiv 1\bmod 3$$ The rest of your proof is fine.

Zev Chonoles
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  • Reciprocity was actually a typo since I was thinking if we could apply that while typing the last part, then but I guess quadratic reciprocity is not applied here because of 2.. right? Also, $2 \equiv -1 \pmod 3$ ain't this sufficient to prove the quadratic non-residue? – Quixotic Dec 04 '11 at 14:20
  • Where is your proof that $-1$ is not a quadratic residue modulo $3$? Why would pointing out that $2\equiv -1\bmod 3$ show that $2$ is a quadratic non-residue? – Zev Chonoles Dec 04 '11 at 14:20
  • Look at supplement law 1; for an odd prime $p$, $-1$ is a quadratic residue mod $p$ iff $p\equiv 1\bmod 4$. That's what I thought you meant when you said you were using quadratic reciprocity. You're correct that you can't flip because of the $2$. – Zev Chonoles Dec 04 '11 at 14:22
  • I thought it is implied that if a is non-residue to b then $a \equiv -1 \pmod b$ and $a \equiv 1 \pmod b$ otherwise ... just thiking on the lines of Fermat's little theorem ain't it ? – Quixotic Dec 04 '11 at 14:23
  • No. Consider that $3$ and $4$ are quadratic non-residues modulo $7$. – Zev Chonoles Dec 04 '11 at 14:24
  • Aha got it .. From Femat's little theorem we can only imply that if a and p are quadratic residue then $a^{(p-1)/2} \equiv 1 \pmod{p} $ right ? – Quixotic Dec 04 '11 at 14:27
  • As long as $a$ is relative prime to $p$, that's right (otherwise it is $0\bmod p$, of course). See Euler's criterion. – Zev Chonoles Dec 04 '11 at 14:28
  • Cool,but again $(a,p) \neq 1$ then the Fermat's theorem doesn't apply ... anyways, the only correct method to prove non-residue is to show all numbers $0,1,2,... (p-1)$ .. like you did in your answer ?! – Quixotic Dec 04 '11 at 14:29
  • Well, that's why quadratic reciprocity is so great, is that you don't have to do that :) Quadratic reciprocity is a major theorem though. Euler's criterion is simple to prove, but calculating $a^{(p-1)/2}\bmod p$ a lot can get tedious if $p$ is large. – Zev Chonoles Dec 04 '11 at 14:31
  • Um but actually,what I read about quadratic reciprocity form my book which states that: "If p and q are two positive odd prime numbers then q is quadratic residue p iff p is quadratic residue p, provided that the product $\large(\normalsize\frac{(p-1)}{2})\times (\frac{(q-1)}{2})$is even.In case this product is odd, the situation is reveresed, so that the p is a residue of q, iff q is a non-residue of p." but actually in this page, there is whole lot of different things!Do you know why is this ? – Quixotic Dec 04 '11 at 14:37
  • If you are not sure why the various formulations of quadratic reciprocity are equivalent, you could ask a separate question about that if you want to. Here is an example of how to use quadratic reciprocity to figure out whether anything is a quadratic residue modulo a prime. – Zev Chonoles Dec 04 '11 at 14:40
  • Cool!! Quadratic reciproicity seems to be a major theorem, I was actually reading it from "what is Mathematics?" although this book is very good but it doesn't provide much insight in this topic, can you suggest me any book/reference which provides sufficient information and yet not complicated... – Quixotic Dec 04 '11 at 14:52
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HINT $\rm\quad\ \ mod\ 3\!\!:\ n\not\equiv 0\ \Rightarrow\ n \equiv \pm1 \ \Rightarrow\ n^2 \equiv 1 \not\equiv -1$

Similarly $\rm\ mod\ 8\!\!:\ odd\ n\: \Rightarrow\ n \equiv\pm1,\pm3\ \Rightarrow\ n^2 \equiv 1,\:$ a result often of use in number theory.

Combining both we deduce $\rm\: n^2 \equiv 1\pmod{24}$ for odd $\rm\:n\:$ coprime to $3\:.\:$ More generally see here.

Note how work is halved using the balanced residue system $\rm\: 0,\pm1,\pm2,...,\pm\lfloor m/2\rfloor\pmod m$

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Bill Dubuque
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