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Let $Q_n$ denote the quadratic residues in $U_n=\mathbb Z_n^\times$.Our instructor gave us a characterisation of $Q_n$ as follows:

$1.$ If $\gcd(m,n)=1$.then $a\in Q_{mn}\iff a\in Q_m $ and $a\in Q_n$.(Why?)

$2.$ For $p$ odd prime,$a\in Q_{p^k} \iff a\in Q_p$.

$3.$ $Q_2= Q_4=\{1\}$ and $Q_{2^k}=\{a:a\equiv 1(\mod 8)\}$ for $k\geq 3$.

Using these results,we are to find $Q_{160}$. I am not sure how to proceed.I know that $a\in Q_{160}$ iff $a\in Q_{2^5}$ and $a\in Q_5$ and $Q_{2^5}=\{a\in U_{32}:a\equiv 1(\mod 8)\}$ and $Q_5=\{1,4\}$. Then how to find $Q_{160}$ .Someone please help me with this. Also I want a proof of the result $(1)$.

Kishalay Sarkar
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1 Answers1

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The elements in $Q_{160}$ is an obvious one. Just apply the first relation to what you’ve got.

For the proof of the first equivalence, the $\implies$ direction is easy. LHS indicates $\exists x, k\in\mathbb{Z}$, $(x,mn)=1$ and $a=x^2+kmn$. Just group up $km$ and $kn$ respectively to conclude.

For the converse, RHS indicates $a=x^2[m]$ and $a=y^2[n]$ for some integers $x$ and $y$. If $x=y$, then the conclusion is obvious. If not, since $(m,n)=1$, there is $g$, $h$ integers such that $gm+hn=1$, i.e. $(x-y)gm+(x-y)hn=(x-y)$. By rearranging the order of the terms, we have $x-(x-y)gm=y+(x-y)hn$. In this last equation, square of LHS is congruent to $a$ modulo $m$ while that of RHS modulo $n$. Thus, the square of LHS is congruent to $a$ modulo $mn$.

Clément Dato
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  • Not an answer, since the question does not ask for a proof of (1), but rather how to find all quadratic residues when given (1),(2),(3). The proof you sketch of (1) is a standard application of CRT that works for roots of any polynomial with integer coef's, e.g. see here – Bill Dubuque Nov 09 '21 at 21:59
  • @BillDubuque: From the question: "Also, I want a proof of result (1)." – Jacob Manaker Nov 09 '21 at 22:33
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    @Jacob Thanks, I missed that. But that additional question at the end is a dupe as I indicated, and that's the only thing this answer addresses. – Bill Dubuque Nov 09 '21 at 23:01
  • @ClementDato This does not answer my question – Kishalay Sarkar Nov 10 '21 at 03:50
  • @Kishalay Sarkar You mean the first question? Elements in $Q_{32}$ are of form $1+8k$ in $\mathbb{Z}$, and those in $Q_5$, $1+5k$ or $4+5k$. We can enumerate all these elements in $Q_{160}$. The intersection would be the answer, by the first relation. – Clément Dato Nov 10 '21 at 06:59