# Number theory: let’s learn it the Nash way !

Reference: A Beautiful Mind by Sylvia Nasar.

Comment: This is approach is quite similar to what Prof. Joseph Silverman explains in his text, “A Friendly Introduction to Number Theory.”

Peter Sarnak, a brash thirty-five-year-old number theorist whose primary interest is the Riemann Hypothesis, joined the Princeton faculty in the fall of 1990. He had just given a seminar. The tall, thin, white-haired man who had been sitting in the back asked for a copy of Sarnak’s paper after the crowd had dispersed.

Sarnak, who had been a student of Paul Cohen’s at Stanford, knew Nash by reputation as well as by sight, naturally. Having been told many times Nash was completely mad, he wanted to be kind. He promised to send Nash the paper. A few days later, at tea-time, Nash approached him again. He had a few questions, he said, avoiding looking Sarnak in the face. At first, Sarnak just listened politely. But within a few minutes, Sarnak found himself having to concentrate quite hard. Later, as he turned the conversation over in his mind, he felt rather astonished. Nash had spotted a real problem in one of Sarnak’s arguments. What’s more, he also suggested a way around it. “The way he views things is very different from other people,” Sarnak said later. ‘He comes up with instant insights I don’t know I would ever get to. Very, very outstanding insights. Very unusual insights.”

They talked from time to time. After each conversation, Nash would disappear for a few days and then return with a sheaf of computer printouts. Nash was obviously very, very good with the computer. He would think up some miniature problem, usually very ingeniously, and then play with it. If something worked on a small scale, in his head, Sarnak realized, Nash would go to the computer to try to find out if it was “also true the next few hundred thousand times.”

{What really bowled Sarnak over, though, was that Nash seemed perfectly rational, a far cry from the supposedly demented man he had heard other mathematicians describe. Sarnak was more than a little outraged. Here was this giant and he had been all but forgotten by the mathematics profession. And the justification for the neglect was obviously no longer valid, if it had ever been.}

Cheers,

Nalin Pithwa

PS: For RMO and INMO (of Homi Bhabha Science Foundation/TIFR), it helps a lot to use the following: (it can be used with the above mentioned text of Joseph Silverman also): TI nSpire CAS CX graphing calculator.

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

# Find a flaw in this proof: RMO and PRMO tutorial

What ails the following proof that all the elements of a finite set are equal?

The following is the “proof”;

All elements of a set with no elements are equal, so make the induction assumption that any set with n elements has all its elements equal. In a set with n elements, the first and the last n are equal by induction assumption. They overlap at n, so all are equal, completing the induction.

End of “proof:

Regards,

Nalin Pithwa

# Number theory : A set of friendly examples

Even and odd numbers

Two whole numbers are added together. If their sum is odd, which statements below are
always true? Which are always false? Which are sometimes true and sometimes false?
1 Their quotient is not a whole number.
2 Their product is even.
3 Their difference is even.
4 Their product is more than their sum.
5 If 1 is added to one of the numbers and the product is found, it will be even.
The Collatz conjecture
Choose any whole number to start with.

If it is odd, multiply it by 3 and then add 1.
If it is even, divide it by 2. Then repeat this process on the number just obtained. Keep repeating the procedure.

For example, if you start with 58, the resulting chain of numbers is
58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, …

The Collatz conjecture, made by Lothar Collatz in 1937, claims that, if you repeat this
process over and over, starting with any whole number greater than zero, eventually you will finish up with the sequence … 4, 2, 1, 4, 2, 1.

A conjecture is a statement that is thought to be true but has not been proved mathematically to be true for all cases. Although the Collatz conjecture has been shown to work − often very quickly − for many whole numbers, there are some quite small numbers that take a very long time to come down to … 4, 2, 1, 4, 2, 1.Apply this process to all the whole numbers greater than zero and less than or equal to 30.
For each one, find:
• how many steps it takes to reach 1 the frst time
• the largest number in the sequence. (For the sequence above, 58 takes 19 steps and reaches a maximum of 88.)
Look for shortcuts and work with a partner if you like.

Long division
Here is a way to check how good your long division skills are. If you are able to follow it
through and get to the end without making a mistake, you can consider yourself a qualiꏨed long division champion.
• Start with any two-digit number (for example, 58). Write it three times so that a six-digit  number is formed (585 858).

• Divide this number by 21. There should not be any remainder. If there is, try and find out where you made your mistake and fix it.
• Now divide this new four- or possibly five-digit number by 37. Once again, there should be no remainder.
• Finally, divide this number − which should by now have only three or four digits − by 13.
You will know if you got it right by looking at the number you are left with.
Explain why this exercise works.
(Doing any of this exercise on a calculator is still interesting but is de뀠nitely wimping out!)

Totient numbers

1 A totient number is the number of fractions between 0 and 1 (not including 0 or 1) for
a given denominator that cannot be reduced to a simpler equivalent fraction. The totient
number of 2 is 1, since we have $\frac{1}{2}$; of 3 it is 2, since we have $\frac{1}{3}$ and $\frac{2}{3}$; and of 4 it is also 2, since we have $\frac{1}{4}$
and $\frac{3}{4}$ ($\frac{2}{4}$ can be reduced to $\frac{1}{2}$). The totient number of 5 is 4, since we have $\frac{1}{5}$, $\frac{2}{5}$, $\frac{3}{5}$, $\frac{4}{5}$; and of 6 it is 2, since we have $\frac{1}{5}$ and
$\frac{5}{6}$. Find the totient numbers forall denominators up to 12.

2 For any denominator n, there are n fractions between 0 and 1 (including 0 but not 1). Of
these fractions, some will be counted towards the totient number of n, but others will
cancel down and count towards the totient number of one of the factors of n. Using this
information and the totient numbers from the previous question, calculate the totient
numbers for 15, 18, 20 and 24.

3 The totient number is related to the prime factors of the original number, since these will
determine which fractions can be cancelled. Using this information, calculate the totient numbers of 72, 81, 98 and 100.

$\bf{Last \hspace{0.1in}Digits \hspace{0.1in} of \hspace{0.1in}powers}$
$\bf{Square \hspace{0.1in}Numbers}$

Without using a calculator, can you say which of this set of numbers could not be square numbers?

8116801, 251301659, 3186842, 20720704.

You can just by checking the last digit (units digit) of each number.

Do a bit of experimentation with a calculator and find the four digits that square numbers end in. (This eliminates the third number in this set).

Now check out the pairs of digits that your odd square numbers end with. What digits are possible in the tens position of an odd square number? (This number eliminates the second number in the set).

Complete these sentences with what you have discovered:

* In a square number, the last digit (units digit) can only be _____, _______, _______, _______, _______ or _______.
* The second last digit (tens place) of an odd square number is always _______.

$\bf{Cube \hspace {0.1in} Numbers}$

Cube numbers behave rather differently.

A bit of experimentation will show that cube numbers can end in ANY digit (units place). This digit depends on the last digit (units place) of the original number being cubed.

Complete this table:
$\left| \begin{array}{cc} \mbox {if a number ends in} & \mbox{its cube will end in}\\ 0 & \\ 1 & \\ 2 & \\ 3 & \\ 4 & \\ 5 & \\ 6 & \\ 7 & \\ 8 & \\ 9 & \end{array}\right|$

$\bf{Fourth \hspace{0.1in}Powers}$

Fourth powers are in fact just square numbers that have been squared. For example, $7^{4}=7^{2} \times 7^{2}= 49^{2}=2401$.

Since $4^{2}=16$ and $9^{2}=81$, the last digit of a fourth power can only be 0, 1, 5 or 6.

$\bf{Fifth \hspace{0.1in}powers}$

Fifth powers have a magic of their own. Do a bit of experimentation to find out what it is. In p, particular, I suggest you create tables of second powers, third powers, fourth powers and fifth powers of all numbers from zero to 20. Check, compare…actually, it is fun to “compare rate of growth of powers with increasing integers”…this idea involves rudimentary ideas of calculus…

$\bf{Obstinate \hspace{0.1in} numbers}$

An odd number can usually be written as a sum of a prime number and another number, which is a power of two. This is true for all odd numbers greater than one but less than 100.

For example, if we choose 23, we can say that it is equal to $23=19 + 2^{2}$. There is one more way to represent 23: it is $7 + 2^{4}$. So, there are two ways to represent 23 as a sum of a prime number and a power of two. But, $21+2^{1}$ and $15+2^{3}$ do not work as both 21 and 15 are not prime numbers.

Some odd numbers like this can be expressed in many ways.

Try to find various representations as sum(s) of prime number and a power(s) of two of the following integers: 45, 29, 59, 95.

If you are adventurous or courageous, try to find such representations of all odd numbers lying from 1 to 100. You need a lot of patience and stamina and grit…but you will develop an “intuitive feel or tactile feel for numbers”…that’s the way math begins…

There are some odd numbers which cannot be expressed as a sum of a prime number and a power of two. Such numbers are called $\bf{obstinate \hspace{0.1in} numbers}$.

An example of an obstinate number is $\bf{251}$ as the working below shows:

$251-2^{1}=249=3 \times 83$

$251-2^{2}=247=13 \times 19$

$251-2^{3}=243=3 \times 81$

$251-2^{4}=235= 5 \times 47$

$251-2^{5}=219= 3 \times 73$

$251-2^{6}= 187= 11 \times 17$

$251-2^{7}=123=3 \times 41$

The next power of 2 is $2^{8}=256$, which is clearly greater than 251. Hence, 251 is an obstinate number.

In fact, 251 is the third obstinate number. The first two lie between 100 and 150. Find these two odd numbers keeping track of how you eliminated the other twenty three odd numbers between 100 and 150.

$\it{Remember \hspace{0.1in} to \hspace{0.1in} be \hspace{0.1in} systematic \hspace{0.1in}}$.

Making a list of the powers of two up to $2^{8}$ might be a good place to start with. Look for short cuts and patterns as you proceed further.

Have fun with numbers !!

Regards,
Nalin Pithwa.

# Patterns in the primes: Clay Math, James Maynard, 2015

(shared from Clay Math website and shared for my readers. Many thanks to James Maynard and Clay Math :-))

Patterns in primes

# Tutorial on Euclidean algorithm for GCD: pre RMO or PRMO 2019

For each of the following pairs of integers a and n, show that a is relatively prime to n, or that a and n are coprime to each other:

1. a=13, n=20
2. a=69, n=89
3. a=1891, n=3797
4. a=6003722857, n=77695236973 (the Euclidean algorithm only requires 3 steps for these integers!).

Cheers,

Nalin Pithwa.

# Some Number Theory Questions for RMO and INMO

1) Let $n \geq 2$ and k be any positive integers. Prove that $(n-1)^{2}\mid (n^{k}-1)$ if and only if $(n-1) \mid k$.

2) Prove that there are no positive integers a, b, $n >1$ such that $(a^{n}-b^{n}) \mid (a^{n}+b^{n})$.

3) If a and $b>2$ are any positive integers, prove that $2^{a}+1$ is not divisible by $2^{b}-1$.

4) The integers 1,3,6,10, $\ldots$, $n(n+1)/2$, …are called the triangular numbers because they are the numbers of dots needed to make successive triangular arrays of dots. For example, the number 10 can be perceived as the number of acrobats in a human triangle, 4 in a row at the bottom, 3 at the next level, then 2, then 1 at the top. The square numbers are $1, 4, 9, \ldots, n^{2}, \ldots$ The pentagonal numbers 1, 5, 12, 22, $\ldots$, $(3n^{2}-n)/2$, $\ldots$, can be seen in a geometric array in the following way: Start with n equally spaced dots $P_{1}, P_{2}, \ldots, P_{n}$ on a straight line in a plane, with distance 1 between consecutive dots. Using $P_{1}P_{2}$ as a base side, draw a regular pentagon in the plane. Similarly, draw $n-2$ additional regular pentagons on base sides $P_{1}P_{3}$, $P_{1}P_{4}$, $\ldots$, $P_{1}P_{n}$, all pentagons lying on the same side of the line $P_{1}P_{n}$. Mark dots at each vertex and at unit intervals along the sides of these pentagons. Prove that the total number of dots in the array is $(3n^{2}-n)/2$. In general, if regular k-gons are constructed on the sides $P_{1}P_{2}$, $P_{1}P_{3}$, …, $P_{1}P_{n}$, with dots marked again at unit intervals, prove that the total number of dots is $1+kn(n-1)/2 -(n-1)^{2}$. This is the nth k-gonal number.

5) Prove that if $m>n$, then $a^{2^{n}}+1$ is a divisor of $a^{2^{m}}-1$. Show that if a, m, n are positive with $m \neq n$, then

$( a^{2^{m}}+1, a^{2^{n}}+1) = 1$, if a is even; and is 2, if a is odd.

6) Show that if $(a,b)=1$ then $(a+b, a^{2}-ab+b^{2})=1$ or 3.

7) Show that if $(a,b)=1$ and p is an odd prime, then $( a+b, \frac{a^{p}+b^{p}}{a+b})=p$ or 1.

8) Suppose that $2^{n}+1=xy$, where x and y are integers greater than 1 and $n>0$. Show that $2^{a}\mid (x-1)$ if and only if $2^{a}\mid (y-1)$.

9) Prove that $(n!+1, (n+1)!+1)=1$.

10) Let a and b be positive integers such that $(1+ab) \mid (a^{2}+b^{2})$. Show that the integer $(a^{2}+b^{2})/(1+ab)$ must be a perfect square.

Note that in the above questions, in general, (a,b) means the gcd of a and b.

More later,
Nalin Pithwa.