Solution Points to the Multinomial Lucas Congruence Mod p
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Joshua Crisafi (Advisor: Eric Rowland) Hofstra University
Introduction
The Family of Multinomial Functions
Visualizing the Results
In a previous paper, Dr. Rowland looked at special pairs of numbers that cause the n n! (this would be the (k + 1)-th number from the left binomial function k = k!·(n−k)! in the (n + 1)-th row of Pascal’s Triangle below) to satisfy a special property. This property has to do with modular arithmetic, also known as clock arithmetic, in which we take the remainder of a division. For instance, 41 divided by 7 gives remainder 6 and 48 divided by 7 also gives remainder 6, so we say that 41 ≡ 48 mod 7. However, dividing 43 by 7 gives remainder 1, so 43 ̸≡ 41 mod 7. Dr. Rowland found these aforementioned special pairs of numbers experimentally and noticed that they exhibited a 3-way symmetry but could not fully crack the code as to where they came from or how many there were. In this research, we aimed not only to solve that problem but also achieve a more general solution. Since the binomial function only takes 2 variables, we considered the multinomial function, which is like a generalization of binomial for any number of inputs, under these same properties. We experimentally determined the tuples (or lists) of points that satisfied our special property and analyzed them until we cracked the pattern.
The k-argument (or k-input) multinomial function is defined by m(n1, n2, . . . , nk ) = Pk ( i=1 ni)! (n1+n2+···+nk )! = Qk ((n )!) . To see that it is a k-argument extension of the binomial n1!n2!...nk ! i i=1 n function, notice that k = m(k, n − k). Therefore, results about the multinomial functions should give us insight into the binomial function as well. The more general problem which now needed to be solved was as follows:
Let’s visualize the special points of the 3-argument multinomial function. Since each point must be between 0 and p − 1 inclusive, they have to fit within a cube going from (0, 0, 0) to (p − 1, p − 1, p − 1). If we plot these triplets of numbers as points in 3-space, we can get a nice graph for each odd prime. For instance, if p = 11, we get the following diagram:
Pascal’s Triangle 1 1 1 1 1 1 1
2 3
4 5
6
1 3 6
10 15
1 1 4 10
20
1 5
15
1 6
Given an odd prime p and a positive integer k, which values of r1, r2, . . . , rk < p cause m(n1 · p + r1, n2 · p + r2, . . . , nk · p + rk ) ≡ m(n1, n2, . . . , nk ) · m(r1, r2, . . . , rk ) mod p2 for all n1, n2, . . . , nk , where r1, r2, . . . , rk and n1, n2, . . . , nk are natural numbers?
The Harmonic Numbers To understand the coming result, the harmonic numbers be introduced. The Pn 1 must definition of the n-th harmonic number is Hn = i=1 i = 11 + 12 + · · · + n1 . Because they are rational numbers and not integers, a new technique must be developed to reduce them in modular arithmetic. What we will do is reduce the numerator normally and then multiply by a multiplicative inverse of the denominator modp. For instance, suppose we want to reduce 115 mod 7. The numerator becomes 4 and 5 · 3 ≡ 1 mod 7, so our final answer will be 4 · 3 = 12 ≡ 5 mod 7. Another way to think about this is writing 115 as 11 · 51 and reducing 11 to 4 and 51 to 3 because 5 · 3 ≡ 1 mod 7. For one last example, notice that H3 ≡ 11 + 12 + 13 ≡ 116 ≡ 4 · 6 ≡ 24 ≡ 3 mod 7. Lastly, notice that, for any prime number p, p1 is undefined modp because p ≡ 0 mod p, so it is equivalent to dividing by 0. But if q ≥ p, then Hq will include a p1 term, so Hq mod p will be undefined. This result will be important soon.
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The Main Result Previous Research Summary In 1878, Édouard Lucas proved for all odd primes p and natural numbers n1, n2, r1, r2 p·n1+r1 n1 r1 that p·n2+r2 ≡ n2 · r2 mod p so long as r1, r2 < p. Dr. Rowland wanted to know if this congruence also held mod p2 (making it a stronger result), and it generally did not; however, fixing certain values of r1 and r2 made the congruence hold modp2 for all natural numbers n1 and n2. These pairs of natural numbers (r1, r2) where n1 r1 2 1 +r1 0 ≤ r1, r2 ≤ p − 1 such that p·n ≡ · mod p for all natural numbers n1 p·n2+r2 n2 r2 and n2 and for some odd prime p were the subject of Dr. Rowland’s previous paper. It was then shown that these pairs satisfy a 3-way symmetry; that is, if you know 1 pair, then you can always generate 2 more that must also satisfy the desired property. He could not, however, decipher the exact structure behind the points.
Let p be an odd prime and let r1, r2, . . . , rk < p be natural numbers. Then m(n1 · p + r1, n2 · p + r2, . . . , nk · p + rk ) ≡ m(n1, n2, . . . , nk ) · m(r1, r2, . . . , rk ) mod p2 for all natural numbers n1, n2, . . . , nk if and only if exactly one of the following holds:
In the above diagram, notice that there is a tetrahedron of points around the point (10, 10, 10), and remember that 10 = 11 − 1 = p − 1. These points come from condition 2 of the Main Result. The points closer to the bottom left corner come from condition 1 of the Main Result, and we can visually now see that conditions 1 and 2 of the Main Result are disjoint because their respective points lie in opposite ends of the cube.
1. Hr1 ≡ Hr2 ≡ · · · ≡ Hrk ≡ Hr1+r2+···+rk mod p
References 2. r1 + r2 + · · · + rk ≥ 2 · p
Discussion The proof of this result is very lengthy, but notice that the first condition above has the expression Hr1+r2+···+rk , which we known is undefined modp if r1 + r2 + · · · + rk ≥ p. This shows that conditions 1 and 2 are mutually exclusive.
[1] Édouard Lucas, Sur les congruences des nombres eulériens et des coefficients différentiels des functions trigonométriques, suivant un module premier, Bulletin de la Société Mathématique de France 6 (1878) 49–54. [2] Eric Rowland, Lucas’ theorem modulo p2, arXiv, https://arxiv.org/abs/ 2006.11701, (2020).