# A good way to start mathematical studies …

I would strongly suggest to read the book “Men of Mathematics” by E. T. Bell.

It helps if you start at a young age. It doesn’t matter if you start later because time is relative!! 🙂

Well, I would recommend you start tinkering with mathematics by playing with nuggets of number theory, and later delving into number theory. An accessible way for anyone is “A Friendly Introduction to Number Theory” by Joseph H. Silverman. It includes some programming exercises also, which is sheer fun.

One of the other ways I motivate myself is to find out biographical or autobiographical sketches of mathematicians, including number theorists, of course. In this, the internet is an extremely useful information tool for anyone willing to learn…

Below is a list of some famous number theorists, and then there is a list of perhaps, not so famous number theorists — go ahead, use the internet and find out more about number theory, history of number theory, the tools and techniques of number theory, the personalities of number theorists, etc. Become a self-learner, self-propeller…if you develop a sharp focus, you can perhaps even learn from MIT OpenCourseWare, Department of Mathematics.

Famous Number Theorists (just my opinion);

1) Pythagoras
2) Euclid
3) Diophantus
4) Eratosthenes
5) P. L. Tchebycheff (also written as Chebychev or Chebyshev).
6) Leonhard Euler
7) Christian Goldbach
8) Lejeune Dirichlet
9) Pierre de Fermat
10) Carl Friedrich Gauss
11) R. D. Carmichael
12) Edward Waring
13) John Wilson
14) Joseph Louis Lagrange
15) Legendre
16) J. J. Sylvester
11) Leonoardo of Pisa aka Fibonacci.
15) Srinivasa Ramanujan
16) Godfrey H. Hardy
17) Leonard E. Dickson
18) Paul Erdos
19) Sir Andrew Wiles
20) George Polya
21) Sophie Germain
24) Niels Henrik Abel
25) Richard Dedekind
26) David Hilbert
27) Carl Jacobi
28) Leopold Kronecker
29) Marin Mersenne
30) Hermann Minkowski
31) Bernhard Riemann

Perhaps, not-so-famous number theorists (just my opinion):
1) Joseph Bertrand
2) Regiomontanus
3) K. Bogart
4) Richard Brualdi
5) V. Chvatal
6) J. Conway
7) R. P. Dilworth
8) Martin Gardner
9) R. Graham
10) M. Hall
12) F. Harary
13) P. Hilton
14) A. J. Hoffman
15) V. Klee
16) D. Kleiman
17) Donald Knuth
18) E. Lawler
19) A. Ralston
20) F. Roberts
21) Gian Carlo-Rota
22) Bruce Berndt
23) Richard Stanley
24) Alan Tucker
25) Enrico Bombieri

Happy discoveries lie on this journey…
-Nalin Pithwa.

# Any integer can be written as the sum of the cubes of 5 integers, not necessarily distinct

Question: Prove that any integer can be written as the sum of the cubes of five integers, not necessarily.

Solution:

We use the identity $6k = (k+1)^{3} + (k-1)^{3}- k^{3} - k^{3}$ for $k=\frac{n^{3}-n}{6}=\frac{n(n-1)(n+1)}{6}$, which is an integer for all n. We obtain

$n^{3}-n = (\frac{n^{3}-n}{6}+1)^{3} + (\frac{n^{3}-n}{6}-1)^{3} - (\frac{n^{3}-n}{6})^{3} - (\frac{n^{3}-n}{6})$.

Hence, n is equal to the sum

$(-n)^{3} + (\frac{n^{3}-n}{6})^{3} + (\frac{n^{3}-n}{6})^{3} + (\frac{n-n^{3}}{6}-1)^{3}+ (\frac{n-n^{3}}{6}+1)^{3}$.

More later,
Nalin Pithwa.

# A random collection of number theory problems for RMO and CMI training

1) Find all prime numbers that divide 50!

2) If p and $p^{2}+8$ are both prime numbers, prove that $p^{3}+4$ is also prime.

3) (a) If p is a prime, and $p \not|b$, prove that in the AP a, $a+b$, $a+2b$, $a+3b$, $\ldots$, every pth term is divisible by p.

3) (b) From part a, conclude that if b is an odd integer, then every other term in the indicated progression is even.

4) Let $p_{n}$ denote the nth prime. For $n>5$, show that $p_{n}.

Hint: Use induction and Bertrand's conjecture.

5) Prove that for every $n \geq 2$, there exists a prime p with $p \leq n < 2p$.

More later,
Regards,
Nalin Pithwa

# Find the last two digits of 9^{9^{9}}

Here is a cute example of the power of theory of congruences. Monster numbers can be tamed !!

Question :

Find the last two digits of $9^{9^{9}}$.

Solution:

A famous mathematician, George Polya said that a good problem solving technique is to solve an analagous less difficult problem.

So, for example, if the problem posed was “find the last two digits of 2479”. How do we go about it? Find the remainder upon division by 100. Now, how does it relate to congruences ? Modulo 100 numbers !

So, the problem reduces to — find out $9^{9^{9}} \equiv 9 \pmod {10}$.

Now, what is the stumbling block…the exponent $9^{9}$ makes the whole problem very ugly. But,

$9^{9} \equiv 9 \pmod {10}$, which means $9^{9}-9=10k$, that is, $9^{9} = 9 + 10k$,

also, use the fact $9^{9} \equiv 89 \pmod {100}$

Hence, $9^{9^{9}}=9^{9+10k} = 9^{9}.9^{10k}$

$9^{9^{9}} \pmod {100} = 9^{9}.9^{10k} \pmod {100} \equiv 89. 9^{10k} \pmod {100}$

So, now we need to compute $9^{10k} \pmod {100} = (9^{10})^{k} \pmod {100} = (89.9)^{k} \pmod {100} = 89^{k}. 9^{k} \pmod {100}$

Hence, $9^{9^{9}} \pmod {100} \equiv 89^{k+1}.9^{k} \pmod {100} \equiv (89.9)^{k}.89 \pmod {100} \equiv (1)^{k}.89 \pmod {100} \equiv 89 \pmod {100}$.

-Nalin Pithwa.

# A beautiful example of use of theory of congruences in engineering

The theory of congruences created by Gauss long ago is used in error control coding or error correction. The theory of congruences is frequently used to append an extra check digit to identification numbers, in order to recognize transmission errors or forgeries. Personal identification numbers of some kind appear in passports, credit cards, bank accounts, and a variety of other settings.

Some banks use (perhaps) an eight-digit identification number $a_{1}a_{2}\ldots a_{8}$ together with a final check digit $a_{0}$. The check digit is usually obtained by multiplying the digits $a_{i}$ for $1 \leq i \leq 8$ by certain “weights” and calculating the sum of the weighted products modulo 10. For instance, the check digit might be chosen to satisfy:

$a_{0} \equiv 7a_{1} + 3a_{2} + 9a_{3} + 7a_{4} + 3a_{5} + 9a_{6} + 7a_{7} + 3a_{8} {\pmod 10}$

The identification number 815042169 would be printed on the cheque.

This weighting scheme for assigning cheque digits detects any single digit error in the identification number. For suppose that the digit $a_{i}$ is replaced by a different $a_{i}$. By the manner in which the check digit is calculated, the difference between the correct $a_{9}$ and the new $a_9^{'}$ is

$a_{9} - a_9^{'} \equiv k(a_{i} - a_{i}^{'}) \pmod {10}$

where k is 7, 3, or 9 depending on the position of $a_{i}^{'}$. Because $k(a_{i}-a_{i}^{'}) \not\equiv 0 \pmod {10}$, it follows that $a_{9} \neq a_{9}^{'}$ and the error is apparent. Thus, if the valid number 81504216 were incorrectly entered as 81504316 into a computer programmed to calculate check digits, an 8 would come up rather than the expected 9.

The modulo 10 approach is not entirely effective, for it does not always detect the common error of transposing distinct adjacent entries a and b within the string of digits. To illustrate, the identification numbers 81504216 and 81504261 have the same check digit 9 when our example weights are used. (The problem occurs when $|a-b|=5$). More sophisticated methods are available, with larger moduli and different weights, that would prevent this possible error.

-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.