Topics in Number Theory: Some problems involving congruence relations (Part I)

For this note I assume that you know the basics properties of divisibility including those of the congruence relation ab(modc)a\equiv b \pmod{c}. See for example Modular arithmetics

I will show some nice problems involving the congruence relation, most of them are very useful in problem solving, thus we can consider this problems as lemmas.

Problem 1. If gcd(a,n)=1\gcd(a,n)=1 then there exists an integer bb such that ab1(modn)ab \equiv 1 \pmod{n}. bb is called the multiplicative inverse of aa modulo nn.

Solution. Consider the numbers a1,a2,a3,,ana\cdot 1, a\cdot 2, a\cdot 3, \ldots, a\cdot n. If aiaj(modn)a\cdot i\equiv a\cdot j \pmod{n} for 1i,jn1\leq i,j\leq n, then a(ij)0(modn)a\cdot(i-j)\equiv 0\pmod{n}, since gcd(a,n)=1\gcd(a,n)=1 we conclude that (ij)0(modn)(i-j)\equiv 0 \pmod{n} and then i=ji=j. Thus the numbers a1,a2,a3,,ana\cdot 1, a\cdot 2, a\cdot 3, \ldots, a\cdot n are pairwise distinct modulo nn, since this set consists of nn numbers then it is a complete residue system modulo nn. In particular, for some 1bn1\leq b\leq n we have ab1(modn)a\cdot b\equiv 1\pmod{n}.

Problem 2. Let n>1n>1 be an integer:

  • If nn is prime then (n1)!1(modn)(n-1)!\equiv -1 \pmod{n}. [Wilson's Theorem]

  • If nn is composite and n4n\neq 4 then (n1)!0(modn)(n-1)!\equiv 0 \pmod{n}.

Hint for the first part. If n2n\geq 2 is an odd prime, prove that the set {2,3,,n2}\{2, 3, \ldots, n-2\} can be divided in pairs {a,a}\{a,a'\} where aa and aa' are inverse each other.

Problem 3. If pp is a prime and 1k<p1\leq k<p then (pk)0(modp){p \choose k}\equiv 0 \pmod{p}.

Solution. We know that (pk)=p!k!(pk)!{p \choose k}=\frac{p!}{k!(p-k)!} then (pk)k!(pk)!=p!{p \choose k}\cdot k!\cdot (p-k)!=p!. The number p!p! is divisible by pp while k!(pk)!k!\cdot (p-k)! is not, then (pk){p \choose k} is divisible by pp.

Problem 4. If pp is a prime then (a+b)pap+bp(modp)(a+b)^p\equiv a^p+b^p \pmod{p}.

Hint. Use the previous problem.

Now, you can practice with the following problems. To avoid confusion I will call these problems N1, N2, etc.

Problem N1. ¿Does there exist a positive integer nn for wich the set {n,n+1,,n+17}\{n, n+1, \ldots, n+17\} can be partitioned in two subsets A\mathcal{A} and B\mathcal{B} such that the product of the elements of A\mathcal{A} is equal to the product of the elements of B\mathcal{B}?

Problem N2. Let p3p\geq3 be a prime. For each i=1,2,,p1i = 1, 2, \ldots, p-1 denote by rir_i the remainder when ipi^p is divided by p2p^2, thus 0ri<p20\leq r_i<p^2. Calculate r1+r2++rp1r_1+r_2+\cdots+r_{p-1}.

Please post your comments and solutions!

Note by Jorge Tipe
5 years, 10 months ago

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Problem N1:

We claim the answer is no.

Lemma: No element in the set can be a multiple of 19.

Proof: Assume that an element in the set is divisible by 19. Then, since the set is composed of 18 consecutive integers, there can be no other element in the set that is divisible by 19. However, then the product of the elements in the two sets cannot be equal, as one has a factor of 19, while the other doesn't, a contradiction. \square

Since the set is composed of 18 consecutive integers, none of which is divisible by 19, the set must be equivalent to the set {1,2,,18}\{1,2,\cdots,18\} when reduced modulo 19. If this set can be partitioned into two subsets with the same product, the product of the elements must be a square. By Wilson's theorem, the product of the elements is congruent to 1-1 modulo 19. However, it is well-known that 1-1 is a quadratic residue modulo an odd prime pp if and only if that prime is 11 modulo 4. Since 193(mod4)19\equiv 3\pmod 4, we conclude that no such integer nn exists. \blacksquare

Problem N2:

We claim the answer is p2(p1)2\dfrac{p^2(p-1)}{2}.

Lemma: ip+(pi)p0(modp2)i^p+(p-i)^p\equiv 0\pmod{p^2}, and ri+rpi=p2r_i+r_{p-i}=p^2.

Proof: By the binomial theorem, ip+(pi)pip+pp(p1)pp1i+(pp2)p2ip2+(pp1)pip1ip(modp2)ip+00+0+(pp1)pip1ip(modp2)ip+p2ip1ip(modp2)0(modp2)\begin{aligned} i^p+(p-i)^p&\equiv i^p+p^p-\dbinom{p}{1}p^{p-1}i+\cdots-\dbinom{p}{p-2}p^2i^{p-2}+\dbinom{p}{p-1}pi^{p-1}-i^p\pmod{p^2} \\ &\equiv i^p+0-0+\cdots-0+\dbinom{p}{p-1}pi^{p-1}-i^p\pmod{p^2} \\ &\equiv i^p+p^2i^{p-1}-i^p\pmod{p^2} \\ &\equiv 0\pmod{p^2} \end{aligned} However, also note that, for 1ip11\le i\le p-1, p2ipp^2\nmid i^p. Since 0<ri<p20<r_i<p^2, 0<rpi<p20<r_{p-i}<p^2, and ri+rpi0(modp2)r_i+r_{p-i}\equiv 0\pmod{p^2}, we must have that ri+rpi=p2r_i+r_{p-i}=p^2, as desired. \square

By the lemma, and since p1p-1 is even, the set of integers r1,r2,,rp1r_1,r_2,\cdots,r_{p-1} can be paired, rir_i with rpir_{p-i}, such that the sum of each pair is p2p^2. The total number of pairs is p12\dfrac{p-1}{2}, and so we conclude r1+r2++rp1=p2(p1)2r_1+r_2+\cdots+r_{p-1}=\dfrac{p^2(p-1)}{2}. \blacksquare

Daniel Chiu - 5 years, 10 months ago

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Awesome! Really nice job

Eddie The Head - 5 years, 9 months ago

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Excellent!!

Kishan k - 5 years, 10 months ago

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Since Problem 4 has no proof, I will write one, though the hint really gives it away..

By the binomial theorem we have that, (a+b)p=k=0p(pk)akbpk(a + b)^p = \sum\limits^p_{k=0} \binom{p}{k} a^k b^{p - k}

By Problem 3, we know that 1k<p(pk)0(modp)1 \leq k < p \Rightarrow \binom{p}{k} \equiv 0 \pmod p, and so all of the summands except for the first and last ones are zeroed. We are then left with,

(a+b)p(p0)ap+(pp)bpap+bp(modp) (a + b)^p \equiv \binom{p}{0} a^p + \binom{p}{p} b^p \equiv a^p + b^p \pmod p\ \quad \square

Ben Frankel - 5 years, 9 months ago

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