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let $p=8k+1$ prime number, show that there exist two set $A=\{a_{1},a_{2},\cdots,a_{\frac{p-1}{2}}\},B=\{b_{1},b_{2},\cdots,b_{\frac{p-1}{2}}\}$ such that $A\bigcup B=\{1,2,3,\cdots,p-1\},A\bigcap B=\varnothing $,and $$\prod_{i=1}^{\frac{p-1}{2}}a_{i}\equiv\prod_{i=1}^{\frac{p-1}{2}}b_{i}\pmod {p^3}$$

This problem should be treated with the knowledge of quadratic residue.

JMP
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math110
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    What have you tried? E.g. Can you do this mod $p^2$? Or mod $p$? – Calvin Lin Nov 17 '20 at 04:32
  • How does one proof this if we don't put the restriction to the knowledge of quadratic residue? in any case reminds me a little bit of the first proof under "prime modulus" of the Wilson theorem on wikipedia – Dabed Dec 07 '20 at 19:24
  • Doesn't this imply that $\prod_{i=1}^\frac{p-1}{2} (a_i-b_i) \equiv 0 \pmod {p^3}$? – JMP Dec 26 '20 at 11:40
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    @JMP: Of course it doesn't. – metamorphy Dec 27 '20 at 07:25
  • The unusual modulus reminds me of Wolstenholme's theorem. – Adam Zalcman Dec 30 '20 at 06:49
  • Is this even possible? I think using discrete logarithms we can check if this is possible for $p=17$. However, taking a generator of $\mathbb Z_{p^3}$ and assuming that the discrete logarithm is random, the problem can be translated to given $p-2$ elements of $Z_{p^2(p-1)}$ can we partition them into two subsets of equal sum. This is probably false, I don't think that $p-2$ elements are enough. This does not prove anything and only hunch based according to my additive number theory knowledge. nothing concrete here. Certainly finding such a partition for some primes $17, 41, 57$ should help. – Elaqqad Dec 31 '20 at 16:41
  • For $p=17$ using base $3$ the discrete logarithms are $[0,734, 1 , 1468, 3765, 735, 3691, 2201,2, 4499, 295, 1469 , 195, 4425, 3766, 2926]$. It remains only to find a partition with equal sum modulo $p^2(p-1)=4624$. – Elaqqad Dec 31 '20 at 16:50
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    It's true for 17 with $A = {1, 6, 9, 10, 12, 13, 14, 15}$ and $B = {2, 3, 4, 5, 7, 8, 11, 16}$. – yotam maoz Jan 07 '21 at 13:12
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    Search based on @Elaqqad's method results $13$ such partitions for $p=17$ (note there are several typos above, the correct discrete logarithms / multiplicative orders for base 3 are $[0,734,1,1468,3765,735,3691,{\bf2202},2,4499,295,1469,{\bf196},4425,3766,{\bf2936}]$), and many more (in the order of $10^5$ when I stopped) solutions for $p=41$, e.g., $A={1, 3, 6, 7, 8, 9, 14, 15, 17, 21, 25, 27, 30, 31, 32, 34, 35, 36, 39, 40}$ and $B={2, 4, 5, 10, 11, 12, 13, 16, 18, 19, 20, 22, 23, 24, 26, 28, 29, 33, 37, 38}$. – Wiley Mar 10 '21 at 21:17

1 Answers1

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This problem appeared in Kvant (M2300). I guess it is more classical than that. I will quote the construction given in the solution.

Let $p-1=4k$. Let us note that the number of odd quadratic residues is equal to the number of even quadratic residues; same goes for non residues. This is because $x$ and $p-x$ have the same Legendre symbol; $-1$ is a square for $p\equiv 1 \pmod{4}$. There are $2k$ nonzero quadratic residues and $2k$ quadratic residues. We denote by $a_1, \ldots, a_k$ the odd quadratic residues, $b_1=p-a_1,\ldots, b_k=p-a_k$ the even quadratic residues, $c_1,\ldots, c_k$ the odd quadratic nonresidues, and finally $d_1=p-c_1,\ldots, d_k=p-c_k$ the even quadratic nonresidues. The claim is we can take $\mathcal{A}=\{a_1,\ldots, a_k, d_1,\ldots, d_k\}$ and $\mathcal{B}=\{b_1,\ldots, b_k, c_1,\ldots, c_k\}$. Note that this is consistent with the above numerical constructions given in the comments.

Let $A=a_1\ldots a_k$ and similarly we define $B,C,D$. Let us obtain first a relation between $A$ and $B$. Note that $2^{2k}AB$ is a product of all even numbers which are quadratic residues, but this time between $1$ and $2p-1$. These can only be $b_1, b_2,\ldots, b_k, p+a_1, \ldots, p+a_k$ in some order . Thus $$2^{2k} AB=\prod_{i=1}^k (p-a_i)(p+a_i)=\prod_{i=1}^k (p^2-a_i^2)\equiv A^2\left(1-p^2\sum_{i=1}^k \frac{1}{a_i^2}\right) \pmod{p^3}.$$ Now note that $2\displaystyle \sum_{i=1}^k \frac{1}{a_i^2}\equiv \sum_{i=1}^{k} \frac{1}{a_i^2}+\sum_{i=1}^{k} \frac{1}{b_i^2}\pmod{p}$. If we invert a quadratic residues it is still a quadratic residue thus $\displaystyle \sum_{i=1}^{k} \frac{1}{a_i^2}+\sum_{i=1}^{k} \frac{1}{b_i^2}\equiv \sum_{i=1}^k a_i^2+\sum_{i=1}^k b_i^2\pmod{p}$. If we take a primitive root $g$ mod $p$ the quadratic residues are $g_i^{2j}$ for $j=0,\ldots, 2k-1$. Thus $$\sum_{i=1}^k a_i^2+\sum_{i=1}^k b_i^2\equiv \sum_{j=0}^{2k-1} (g^4)^{j}=\cfrac{g^{8k}-1}{g^4-1}\equiv 0\pmod{p} $$ since $p>5$. Putting it all together $2^{2k} AB\equiv A^2 \pmod{p^3}$ and thus $B\equiv 2^{-2k} A\pmod{p^3}$. A similar argument shows that $C\equiv 2^{2k} D \pmod{p^3}$ and we are done.

Vlad Matei
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  • Very nice. A couple nitpicks though: you're using the symbol $A$ first as a set, then as a product. This is a little confusing. Also, I feel it would be good to be more explicit about the fact that the residues need to be divided into four equi- and even-numerous groups, which explains why this only works for primes $p$ of the form $8k+1$. – user3733558 Mar 12 '21 at 09:24
  • Thanks,can you tell me How to find the Kvant problem and solution? – math110 Mar 12 '21 at 12:09
  • The solution is the one I wrote – Vlad Matei Mar 12 '21 at 12:13
  • Would this still work with $p=8k+5$? and if I'm not wrong I think you use $\sum\frac{1}{a_i^2}\equiv\cfrac{g^{8k}-1}{g^4-1}\equiv 0\pmod{p}\Rightarrow\sum_{i=1}^k \frac{1}{a_i^2}\equiv \cfrac{g^{8k}-1}{g^4-1}\equiv 0\pmod{p^3}$ why is that? – Dabed Mar 12 '21 at 15:02
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    @DanielD., the proof only requires the 2nd term in $2^{2k}AB$, which is $p^2\sum_{i=1}^ka_i^{-2}\equiv0\pmod{p^3}$. That's why $\sum_{i=1}^k a_i^{-2}\equiv0\pmod{p}$ suffices. But you are right that it wouldn't work with $p=8k+5$. (See a separate comment) – Wiley Mar 12 '21 at 15:25
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    @VladMatei, very nice find, though there are a few places in the proof that you implicitly use the fact that $k$ is an even number (which is why the statement only holds when $p$ is a prime number of the form $8k+1$): first of all, $2^kAB$ *is a product of all even numbers which are quadratic residues*. For this to be true, $2$ also needs to be a quadratic residue modulo $p$, which requires $p\equiv\pm1\pmod8$; secondly, $\prod_{i=1}^k (p^2-a_i^2)\equiv (-1)^kA^2-p^2A^2\sum_{i=1}^ka_i^{-2} \pmod{p^3}$, so $k$ needs to be even for the rest to follow. – Wiley Mar 12 '21 at 15:37
  • I used that if $x$ is a quadratic residue then $2x$ is also is a quadratic residues when I was considering $2^{2k} AB$. $2$ is a quadratic residue mod $p$ only for $p\equiv 1,7 \pmod{8}$. Since you also use $-1$ is a square you need $p\equiv 1 \pmod{4}$ thus the condition of the problem. – Vlad Matei Mar 12 '21 at 15:38
  • @Wiley oh I see my mistake thanks, about $p=8k+5$ I was actually more inclined to think it would work but now I see it can't be. – Dabed Mar 12 '21 at 16:28