I need to prove: $a$ is a square in $\mathbb{Z}$ iff $a\mod p$ is a square in $\mathbb{F_p}$ for every $p$ prime. First side ($\rightarrow$) is trivial and is derived directly from multiplicativity of modulo. Now I need to prove the other side ($\leftarrow$). I tried taking $p>a$ so $a\equiv a\mod p$, and I know that $a$ is a square so there exists $b<p$ s.t $\space$$b^2\mod p \equiv a$. But, $b$ is not necessarily the root of $a$ in $\mathbb{Z}$, so I'm stuck. I also know that for each $p$, $(\frac{a}{p})=1$ but this doesn't get me anywhere. Any hint would help.
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@StevenStadnicki fixed it for every $p$ prime, thanks – Math101 Nov 30 '20 at 21:59
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$b$ is not necessarily the root of $a$ in $\mathbb{Z}$ ? Why – M. Di Nov 30 '20 at 22:03
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Not that it seems to help an awful lot, but we do have http://oeis.org/A144294. There might be better and more accessible primes to go for than "the lowest", though. – Arthur Nov 30 '20 at 22:05
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if $a\ne0$, there actually exist two $b<p$ s.t. $b^2\equiv a\pmod p$, because if $b$ satisfies that, then so does $p-b$ – J. W. Tanner Nov 30 '20 at 22:06
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1@M.Di The argument is a little awkwardly written there but consider e.g. $a=10$, $p=13$. $a$ is a square in $\mathbb{F}_p$, and $a\lt p$ but $a$ isn't a square in $\mathbb{Z}$. – Steven Stadnicki Nov 30 '20 at 22:06
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@StevenStadnicki: I think OP is saying $a$ is a square in $\mathbb F_p$ for all $p$, not just $13$ – J. W. Tanner Nov 30 '20 at 22:07
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@StevenStadnicki But also $10$ isn't a square modulo $7$. Which is the point of this problem: show that there always is at least one prime that makes $a$ non-square modulo that prime, as long as $a$ itself is non-square. – Arthur Nov 30 '20 at 22:07
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@J.W.Tanner I understand that; I was just saying what '$b$ is not necessarily the root of $a$ in $\mathbb{Z}$' means, as I understand it, trying to clarify that point. – Steven Stadnicki Nov 30 '20 at 22:08
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Working with just a 'sufficiently large' $p$ won't be enough to show what you're after; for instance, 2 is not a square in $\mathbb{Z}$ but it is a square modulo infinitely many primes (by reciprocity). So the argument is going to have to be more thorough than just that. – Steven Stadnicki Nov 30 '20 at 22:09
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1@StevenStadnicki I'll clarify what I mean. If for every $p$ prime, $a \mod p $ is a square in $\mathbb{F_p}$, then it is a square in $\mathbb{Z}$. It has to be a square in every finite field for the argument to work, obviously, but I don't know how to prove it. – Math101 Nov 30 '20 at 22:15
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1Cf. this unanswered question and this answered question – J. W. Tanner Nov 30 '20 at 22:23
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@J.W.Tanner Just for my own understanding, is 27 a square in $\mathbb{F_3}$? – user2661923 Nov 30 '20 at 22:24
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@user2661923: I would say yes, because $27\equiv 0^2\bmod \mathbb F_3$ – J. W. Tanner Nov 30 '20 at 22:26
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Note that $7$ is a square modulo $19, 29, 31, 37, 47, 53, $ and $59$ – J. W. Tanner Nov 30 '20 at 22:27
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@J.W.Tanner I was afraid that you were going to say that. If that were not the case, then the $\leftarrow$ proof becomes trivial, since $a$ is a square $\iff$ all of the exponents in its prime factorization are even. Oh well. – user2661923 Nov 30 '20 at 22:27
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If think the easiest way is to use quadratic reciprocity, that if $a$ is not a square then for $p\ne 2$, $(\frac{a}p) =\chi(p) $ with $\chi$ a non-trivial Dirichlet character modulo $4a$,
there is $m>0$ such that $\chi(m)=-1$, some prime $p|m$ is such that $-1=\chi(p)=(\frac{a}p)$
reuns
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While I am familiar with QR, I'm not familiar with Dirichlet characters. Is there another way to approach the contra-positive? – Math101 Dec 01 '20 at 08:31
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$n\to (\frac{n}q)$ is the quadratic Dirichlet character modulo $q$ (completely multiplicative and $q$ periodic); quadratic reciprocity is saying that $\chi$ is completely multiplicative and periodic and not always $1$ or $0$. Completely multiplicative implies that if all the $p|m$ are such that $\chi(p)=1$ then $\chi(m)=1$. – reuns Dec 01 '20 at 16:39