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Let $p$ and $q$ be distinct prime numbers. The integers $a_1, \ldots, a_p$ and $b_1,\ldots,b_q$ are such that the sums $a_i + b_j$ form a complete residue system modulo $pq$ (that is, there is precisely one sum which is $0$ mod $pq$, precisely one which is $1$ mod $pq$, etc.). Prove that $a_1,\ldots,a_p$ form a complete residue system mod $p$ and $b_1,\ldots,b_q$ form a complete residue system mod $q$.

Any idea on how to approach this? Any help appreciated!

DesmondMiles
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    Hints: (1) show it's impossible for some $a_{i_1}\equiv a_{i_2}\pmod p$ and $b_{j_1}\equiv b_{j_2}\pmod q$ simultaneously. (2) If some $a_{i_1}\equiv a_{i_2}\pmod p$ but all the $b_j$ are distinct modulo $q$, can the $a_i+b_j$ all be distinct modulo $pq$? – Greg Martin Mar 04 '21 at 23:34
  • Yeah I also thought of this strategy, but how do we show (1)? – DesmondMiles Mar 05 '21 at 09:29
  • You say "the other direction is obvious," but it seems to only be obvious in light of the exact fact you need to prove (1) – jackson Mar 05 '21 at 12:15
  • By "the other direction" I mean that it's obvious that if $a_i$ form a CRS and $b_j$ form a CRS, then $a_i + b_j$ form a CRS. – DesmondMiles Mar 05 '21 at 15:04
  • @DesmondMiles you're right, it's not as immediate as I had thought. I keep wanting to leverage the fact that these functions are injective if and only if surjective—so that we can look at whether residues are repeated, rather than whether they cover everything. Thus I hope to establish the contrapositive: if the $a_i$ aren't a CRS (mod $p$) or the $b_j$ aren't a CRS (mod $q$), then the $a_i+b_j$ aren't a CRS (mod $pq$). – Greg Martin Mar 05 '21 at 18:21
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    @DesmondMiles The other direction is false (let alone obviously true). Consider $p=2,q=3, a_1 = 0, a_2 = 1, b_1 = 0, b_2 = 1, b_3 = 2$. – mathworker21 Mar 11 '21 at 21:32
  • @mathworker21 In view of your comment, one may wonder if more can be said on the $a_i$ and $b_j$ under the hypothesis that tha $a_i+b_j$ are a CRS ; in particular, are $a_i$ and $b_j$ equivalent modulo $pq$ to arithmetic progressions ? I have asked a related question. – Ewan Delanoy Mar 12 '21 at 10:03

2 Answers2

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Here's an idea.

Suppose $\{a_i+b_j\}_{i,j}$ forms a complete residue system mod $pq$. Then $$0 = \sum_{k=0}^{pq-1} k = \sum_{i,j} (a_i+b_j) = q(a_1+\dots+a_p)+p(b_1+\dots+b_q) \pmod{pq}.$$ This implies $$a_1+\dots+a_p \equiv 0 \pmod{p}$$ and $$b_1+\dots+b_q \equiv 0 \pmod{q}.$$ Using these, and that $$\sum_{i,j} (a_i+b_j)^2 = q(a_1^2+\dots+a_p^2)+p(b_1^2+\dots+b_q^2)+2(a_1+\dots+a_p)(b_1+\dots+b_q),$$ we obtain $$a_1^2+\dots+a_p^2 \equiv 0 \pmod{p}$$ and $$b_1^2+\dots+b_q^2 \equiv 0 \pmod{q}.$$ We can continue like this to get higher and higher moments (it stops at $p-1$ for $a_i$'s and $q-1$ for $b_i$'s, I think). I wonder if this implies $\{a_i\}_i$ is a complete residue system or is all $0$s. Same with $\{b_j\}_j$.

mathworker21
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Let's assume $p<q$. You can only conclude that the the $a_i$'s form a complete system of residues modulo $p$. Let's note that $\displaystyle \sum_{k=0}^{pq-1} k^t\equiv 0 \pmod{p}$ if and only if $p-1\nmid t$ and $\displaystyle \sum_{k=0}^{pq-1} k^t\equiv -q\pmod{p}$ for $p-1|t$. Same goes for switching $p$ and $q$.

Inductively one can show that $\displaystyle \sum_{i=1}^{p} a_i^{n}\equiv 0 \pmod{p}$ for $n\leq p-2$;this formalizes mathworkers post . More precisely if we denote with $\displaystyle A_n=\sum_{i=1}^p a_i^n$ and $\displaystyle B_n=\sum_{i=1}^q b_i^n$ we have $$\sum_{i,j} (a_i+b_j)^n=\sum_{k=1}^{n-1} A_k B_{n-k} \dbinom{n}{k}+qA_n+pB_n$$

Now for $n=p-1$ we have $\displaystyle \sum_{k=0}^{pq-1} k^t\equiv -q \pmod{pq}$ and thus $A_{p-1}\equiv -1\pmod{p}$. Now Fermat's little theorem gives $a_i^{p-1} \equiv \{0,1\} \pmod{p}$ and thus we must have one number divisible by $p$ and $p-1$ coprime with $p$ amongst the $a_i$. Wlog $p|a_1$.

Now considering the polynomial $\displaystyle f(X)=\prod_{i=2}^{p} (X-a_i)$ in $\mathbb{F}_p[X]$ we can show that all the coefficients are zero; these are symmetric sums in the $a_i$ and using Newton's relations between power sums and symmetric sums. Thus $f(X)=X^{p-1}-a$ and since $a_{i}^{p-1}\equiv 1 \pmod{p}$ we have $a=1$. Since $\displaystyle X^{p-1}-1=\prod_{j=1}^{p-1} (X-j) $ in $\mathbb{F}_p[X]$ we have shown that $\{a_i\}_{i=1}^p$ form a complete system of residues modulo $p$.

Note that the same strategy guarantees that $q|B_n$ for $n<q-1$ using the same recurrence relation. By the same argument as above one gets that $\{b_i\}$ form a complete system of residues mod $q$ also.

Vlad Matei
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