Some number theory problems: tutorial set II: RMO and INMO

1. A simplified form of Fermat’s theorem: If x, y, z, n are natural numbers, and $n \geq z$, prove that the relation $x^{n} + y^{n} = z^{n}$ does not hold.

2. Distribution of numbers: Find ten numbers $x_{1}, x_{2}, \ldots, x_{10}$ such that (a) the number $x_{1}$ is contained in the closed interval $[0,1]$ (b) the numbers $x_{1}$ and $x_{2}$ lie in different halves of the closed interval $[0,1]$ (c) the numbers $x_{1}$, $x_{2}$, $x_{3}$ lie in different thirds of the closed interval $[0,1]$ (d) the numbers $x_{1}$, $x_{2}$, $x_{3}$ and $x_{4}$ lie in different quarters of the closed interval $[0,1]$,  etc., and finally, (e) the numbers $x_{1}$, $x_{2}$, $x_{3}, \ldots, x_{10}$ lie in different tenths of the closed interval $[0,1]$

3. Is generalization of the above possible?

4. Proportions: Consider numbers A, B, C, p, q, r such that: $A:B =p$, $B:C=q$, $C:A=r$, write the proportion $A:B:C = \Box : \Box : \Box$ in such a way that in the empty squares, there will appear expressions containing p, q, r only; these expressions being obtained by cyclic permutation of one another expressions.

5. Give an elementary proof of the fact that the positive root of $x^{5} + x = 10$ is irrational.

I will give you sufficient time to try these. Later, I will post the solutions.

Cheers,

Nalin Pithwa.

Some number theory training questions: RMO and INMO

Question 1:

Let us write an arbitrary natural number (for example, 2583), and then add the squares of its digits. ($2^{2}+5^{2}+8^{2}+3^{2}=102$). Next, we do the same thing to the number obtained. Namely, $1^{2}+0^{2}+2^{2}=5$. Now proceed further in the same way:

$5^{2}=25$, $2^{2}+5^{2}=29$, $2^{2}+9^{2}=85, \ldots$.

Prove that unless this procedure leads to number 1 (in which case, the number 1 will, of course, recur indefinitely), it must lead to the number 145, and the following cycle will repeat again and again:

145, 42, 20, 4, 16, 37, 58, 89.

Question 2:

Prove that the number $5^{5k+1} + 4^{5k+2} + 3^{5k}$ is divisible by 11 for every natural k.

Question 3:

The number $3^{105} + 4^{105}$ is divisible by 13, 49, 181 and 379, and is not divisible by either 5 or by 11. How can this result be confirmed?

Cheers,

Nalin Pithwa.

Some basics of Number Theory for RMO: part III: Fermat’s Little Theorem

Fermat’s Little Theorem:

The fact that there are only a finite number of essentially different numbers in arithmetic to a modulus m means that there are algebraic relations which are satisfied by every number in that arithmetic. There is nothing analogous to these relations in ordinary arithmetic.

Suppose we take any number x and consider its powers $x, x^{2}, x^{3}, \ldots$. Since there are only a finite number of possibilities of these to the modulus m, we must eventually come to one which we have met before, say $x^{h} \equiv x^{k} {\pmod m}$, where $k . If x is relatively prime to m, the factor $x^{k}$ can be cancelled, and it follows that $x^{l} \equiv 1 {\pmod m}$, where $l \equiv {h-k}$. Hence, every number x which is relatively prime to m satisfies some congruence of this form. The least exponent l for which $x^{l} \equiv 1 {\pmod m}$ will be called the order of x to the modulus m. If x is 1, its order is obviously 1. To illustrate the definition, let us calculate the orders of a few numbers to the modulus 11. The powers of 2, taken to the modulus 11, are

2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, $\ldots$

Each one is twice the preceding one, with 11 or a multiple of 11 subtracted where necessary to make the result less than 11. The first power of 2 which is $\equiv 1$ is $2^{10}$, and so the order of $2 \pmod {11}$ is 10. As another example, take the powers of 3:

3, 9, 5, 4, 1, 3, 9, $\ldots$

The first power of 3 which is equivalent to 1 is $3^{5}$, so the order of $3 \pmod {11}$ is 5. It will be found that the order of 4 is again 5, and so also is that of 5.

It will be seen that the successive powers of x are periodic; when we have reached the first number l for which $x^{l} \equiv 1$, then $x^{l+1} \equiv x$ and the previous cycle is repeated. It is plain that $x^{n} \equiv 1 {\pmod m}$ if and only if n is a multiple of the order of x. In the last example, $3^{n} \equiv 1 {\pmod 11}$ if and only if n is a multiple of 5. This remains valid if n is 0 (since 3^{0} = 1), and it remains valid also for negative exponents, provided $3^{-n}$, is interpreted as a fraction (mod 11) in the way explained earlier (an earlier blog article).

In fact, the negative powers of 3 (mod 11) are obtained by prolonging the series backwards, and the table of powers of 3 to the modulus 11 is:

$\begin{array}{cccccccccccccc} n & = & \ldots & -3 & -2 & -1 & 0 & 1 &2 & 3 & 4 & 5 & 6 & \ldots \\ 3^{n} & \equiv & \ldots & 9 & 5 & 4 & 1 & 3 & 9 & 5 & 4 & 1 & 3 & \ldots \end{array}$

Fermat discovered that if the modulus is a prime, say p, then every integer x not congruent to 0 satisfies

$x^{p-1} \equiv 1 {\pmod p}$….call this as equation A.

In view of what we have seen above, this is equivalent to saying that the order of any number is a divisor of $p-1$. The result A was mentioned by Fermat in a letter to Frenicle de Bessy of 18 October 1640, in which he also stated that he had a proof. But, as with most of Fermat’s discoveries, the proof was not published or preserved. The first known proof seems to have been given by Leibniz (1646-1716). He proved that $x^{p} \equiv x {\pmod p}$, which is equivalent to A, by writing x as a sum $1+ 1 + 1 + \ldots + 1$ of x units (assuming x positive), and then expanding $(1+1+ \ldots + 1)^{p}$ by the multinomial theorem. The terms $1^{p} + 1^{p} + \ldots + 1^{p}$ give x, and the coefficients of all the other terms are easily proved to be divisible by p.

Quite a different proof was given by Ivory in 1806. If $x \not\equiv 0 {\pmod p}$, the integers

$x, 2x, 3x, \ldots, (p-1)x$

are congruent in some order to the numbers

$1, 2, 3, \ldots, p-1$.

In fact, each of these sets constitutes a complete set of residues except that 0 (zero) has been omitted from each. Since the two sets are congruent, their products are congruent, and so

$(x)(2x)(3x) \ldots ((p-1)x) \equiv (1)(2)(3)\ldots (p-1){(\pmod p)}$

Cancelling the factors 2, 3, ….(p-1), as is permissible we obtain the above relation A.

One merit of this proof is that it can be extended so as to apply to the more general case when the modulus is no longer a prime.

The generalization of the result A to any modulus was first given by Euler in 1760. To formulate it, we must begin by considering how many numbers in the set 0, 1, 2, …, (m-1) are relatively prime to m. Denote this number by $\phi(m)$. When m is a prime, all the numbers in the set except 0 (zero) are relatively prime to m, so that $\phi(p) = p-1$ for any prime p. Euler’s generalization of Fermat’s theorem is that for any modulus m,

$x^{\phi(m)} = 1 {\pmod m}$…relation B

provided only that x is relatively prime to m.

To prove this, it is only necessary to modify Ivory’s method by omitting from the numbers $0, 1, 2, \ldots, (m-1)$ not only the number 0, but all numbers which are not relatively prime to m. These remain $\phi(m)$ numbers, say

$a_{1}, a_{2}, \ldots, a_{\mu}$, where $\mu = \phi(m)$.

Then, the numbers

$a_{1}x, a_{2}x, \ldots, a_{\mu}x$

are congruent, in some order, to the previous numbers, and on multiplying and cancelling $a_{1}, a_{2}, \ldots, a_{\mu}$ (as is permissible) we obtain $x^{p} \equiv 1 {\pmod m}$, which is relation B.

To illustrate this proof, take $m=20$. The numbers less than 20 and relatively prime to 20 are :

1, 3, 7, 9, 11, 13, 17, 19.

So that $\phi(20) = 8$. If we multiply these by any number x which is relatively prime to 20, the new numbers are congruent to the original numbers in some other order. For example, if x is 3, the new numbers are congruent respectively to

$3, 9, 1, 7, 13, 19, 11, 17 {\pmod 20}$;

and the argument proves that $3^{8} \equiv 6561$.

Reference:

1. The Higher Arithmetic, H. Davenport, Eighth Edition.
2. Elementary Number Theory, Burton, Sixth Edition.
3. A Friendly Introduction to Number Theory, J. Silverman

Shared for those readers who enjoy expository articles.

Nalin Pithwa.

AMS Menger Awards 2018

(shared from the AMS website for motivational purposes)

The AMS presented the Karl Menger Memorial Awards at the 2018 Intel International Science and Engineering Fair (Intel ISEF), May 13-18, 2018 in Pittsburgh, PA. The First Place Award of US$2000 was given to Ryusei Sakai, Sota Kojima, and Yuta Yokohama, Shiga Prefectural Hikone Higashi High School, Japan, for “Extension of Soddy’s Hexlet: Number of Spheres Generated by Nested Hexlets.” [Photo: bottom row (left to right): Dr. Keith Conrad (committee chair), Rachana Madhukara, Yuta Yokohama, Sota Kojima, Ryusei Sakai; top row (left to right): Chavdar Lalov, Gianfranco Cortes-Arroyo, Gopal Goel, Savelii Novikov, Boris Baranov. Not pictured: Muhammad Abdulla. Photo by the Society for Science & the Public.] The Menger Awards Committee also presented the following awards: • Second Award of$1,000: Gopal Krishna Goel (Krishna Homeschool, OR), “Discrete Derivatives of Random Matrix Models and the Gaussian Free Field” and Rachana Madhukara, Canyon Crest Academy, CA, “Asymptotics of Character Sums”
• Third Award of \$500: Chavdar Tsvetanov Lalov, Geo Milev High School of Mathematics, Bulgaria, “Generating Functions of the Free Generators of Some Submagmas of the Free Omega Magma and Planar Trees”; Gianfranco Cortes-Arroyo, West Port High School, FL, Generalized Persistence Parameters for Analyzing Stratified Pseudomanifolds”; Muhammad Ugur Oglu Abdulla, West Shore Junior/Senior High School, FL, “A Fine Classification of Second Minimal Odd Orbits”; Boris Borisovich Baranov and Savelii Novikov, School 564, St. Petersburg, Russian Federation, “On Two Letter Identities in Lie Rings”
• Certificate of Honorable Mention: Dmitrii Mikhailovskii, School 564, St. Petersburg, Russian Federation, “New Explicit Solution to the N-Queens Problem and the Millennium Problem”; Chi-Lung Chiang and Kai Wang, The Affiliated Senior High School of National Taiwan Normal University, Chinese Taipei, “’Equal Powers Turn Out’ – Conics, Quadrics, and Beyond”; Kayson Taka Hansen, Twin Falls High School, ID, “From Lucas Sequences to Lucas Groups”; Gustavo Xavier Santiago-Reyes and Omar Alejandro Santiago-Reyes, Escuela Secundaria Especializada en Ciencias, Matematicas y Tecnología, Puerto Rico, “Mathematics of Gene Regulation: Control Theory for Ternary Monomial Dynamical Systems”; Karthik Yegnesh, Methacton High School, PA, “Braid Groups on Triangulated Surfaces and Singular Homology”

A booklet on Karl Menger was also given to each winner. This is the 28th year of the presentation of the Karl Menger Memorial Awards. The Society’s participation in the Intel ISEF is supported in part by income from the Karl Menger Fund, which was established by the family of the late Karl Menger. For more information about this program or to make contributions to this fund, contact the AMS Development Office.

Cheers to the winners,

Nalin Pithwa.