Closing the gap: quest to understand prime numbers; Dr Vicky Neale

You and your research or you and your studies for competitive math exams

You and your research ( You and your studies) : By Richard Hamming, AT and T, Bell Labs mathematician;

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.

Method of undetermined coefficients for PreRMO, PRMO and IITJEE Foundation maths

Posted by Nalin Pithwa

Find out when the expression $x^{3}+px^{2}+qx+r$ is exactly divisible by $x^{2}+ax+b$

Solution 1:

Let $x^{3}+px^{2}+qx+r=(x^{2}+ax+b)(Ax+B)$ where A and B are to be determined in terms of p, q, r, a and b. We can assume so because we know from the fundamental theorem of algebra that the if the LHS has to be of degree three in x, the remaining factor in RHS has to be linear in x.

So, expanding out the RHS of above, we get:

$x^{3}+px^{2}+qx+r=Ax^{3}+aAx^{2}+bAx+Bx^{2}+Bax+bB$

$x^{3}+px^{3}+qx+r=Ax^{3}+(aA+B)x^{2}+x(bA+aB)+bB$

We are saying that the above is true for all values of x: hence, coefficients of like powers of x on LHS and RHS are same; we equate them and get a system of equations:

$A=1$

$p=aA+B$

$bA+aB=q$

$bB=r$

Hence, we get $p=a+\frac{r}{b}$ and $bp-ba=r$ or that $b(p-a)=r$

Also, $b+aB=q$ so that $q=b+\frac{ar}{b}$ which means $q-b=\frac{a}{b}r$

but $\frac{r}{b}=B=p-a$ and hence, $q-b=\frac{a}{b}(p-a)$

So, the required conditions are $b(p-a)=r$ and $q-b=\frac{a}{b}(p-a)$ .

2) Find the condition that $x^{2}+px+q$ may be a perfect square.

Solution 2:

Let $x^{2}+px+q=(Ax+B)^{2}$ where A and B are to be determined in terms of p and q; finally, we obtain the relationship required between p and q for the above requirement.

$x^{2}+px+q=A^{2}x^{2}+B^{2}+2ABx$ which is true for all real values of x;

Hence, $A^{2}=1$ so $A=1$ or $A=-1$

Also, $B^{2}=q$ and hence, $B=\sqrt{q}$ or $B=-\sqrt{q}$

Also, $2AB=p$ so that $2\sqrt{q}=p$ so $q=\frac{p^{2}}{4}$ , which is the required condition.

3) To prove that $x^{4}+px^{3}+qx^{2}+rx+s$ is a perfect square if $(q-\frac{p^{2}}{4})^{2}=4s$ and $r^{2}=p^{2}s$ .

Proof 3:

Let $x^{4}+px^{3}+qx^{2}+rx+s=(Ax^{2}+Bx+C)^{2}$

$x^{4}+px^{3}+qx^{2}+rx+s=A^{2}x^{4}+B^{2}x^{2}+C^{2}+2ABx^{3}+2BCx+2ACx^{2}$

$A^{2}=1$

$2AB=p$

$q=B^{2}+2AC$

$2BC=r$

$C^{2}=s$

$A=1$ or $A=-1$

$2AB=p \longrightarrow 2B=p \longrightarrow B=\frac{p}{2}$

$q=B^{2}+2AC=\frac{p^{2}}{4}+2\times \sqrt{s} \longrightarrow (q-\frac{p^{2}}{4})^{2}=4s$

$2 \times \frac{p}{2} \times \sqrt{s}=r \longrightarrow r^{2}=p^{2}s$

More later,

Nalin Pithwa.

PS: Note in the method of undetermined coefficients, we create an identity expression which is true for all real values of x.

Axiomatic Method : A little explanation

Posted by Nalin Pithwa

I) Take an English-into-English dictionary (any other language will also do). Start with any word and note down any word occurring in its definition, as given in the dictionary. Take this new word and note down any word appearing in it until a vicious circle results. Prove that a vicious circle is unavoidable no matter which word one starts with , (Caution: the vicious circle may not always involve the original word).

For example, in geometry the word “point” is undefined. For example, in set theory, when we write or say : $a \in A$ ; the element “a” ‘belongs to’ “set A” —- the word “belong to” is not defined.

So, in all branches of math or physics especially, there are such “atomic” or “undefined” terms that one starts with.

After such terms come the “axioms” — statements which are assumed to be true; that is, statements whose proof is not sought.

The following are the axioms based on which equations are solved in algebra:

If to equals we add equals, we get equals.
If from equals we take equals, the remainders are equal.
If equals are multiplied by equals, the products are equal.
If equals are divided by equals (not zero), the quotients are equal.

More later,

Nalin Pithwa.

Check your mathematical induction concepts

Posted by Nalin Pithwa

Discuss the following “proof” of the (false) theorem:

If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:

PROOF BY INDUCTION:

Step 1:

If $n=1$ , the result is evident.

Step 2: By the induction hypothesis the result is true when $n=k$ ; we must prove that it is correct when $n=k+1$ . Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}$ . If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \ldots, a_{k}$ ; by induction hypothesis these numbers are all equal:

$a_{1}=a_{2}= \ldots = a_{k}$ .

If we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \ldots, a_{k}, a_{k+1}$ ; by the induction hypothesis these numbers are all equal:

$a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}$ .

It follows easily that all $k+1$ numbers in S are equal.

*************************************************************************************

Comments, observations are welcome 🙂

Regards,

Nalin Pithwa

Miscellaneous Algebra: pRMO, IITJEE foundation maths 2019

Posted by Nalin Pithwa

For the following tutorial problems, it helps to know/remember/understand/apply the following identities (in addition to all other standard/famous identities you learn in high school maths):

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

By the way, I hope you also know how to derive the above.Let me mention two methods to derive the above :

Method I: Using polynomial division in three variable, divide the dividend $a^{3}+b^{3}+c^{3}-3abc$ by the divisor $a+b+c$ .

Method II: Assume that $P(X)$ is a polynomial with roots a, b and c. So, we know by the fundamental theorem of algebra that $P(X)=(X-a)(X-b)(X-c)$ . Now, we also know that a, b and c satisfy P(X). Now, proceed further and complete the proof.

Let us now work on the tutorial problems below:

1) If $2s=a+b+c$ , prove that $\frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = \frac{abc}{s(s-a)(s-b)(s-c)}$

2) If $x^{2}+a^{2}=2(xy+yz+zu-y^{2}-z^{2})$ , prove that $x=y=z=u$ .

Prove the following identities:

3) $b(x^{3}+a^{3})+ax(x^{2}-a^{2})+a^{3}(x+a)=(a+b)(x+a)(x^{2}-ax+a^{2})$

4) $(ax+by)^{2}+(ay-bx)^{2}+c^{2}x^{2}+c^{2}y^{2}=(x^{2}+y^{2})(a^{2}+b^{2}+c^{2})$

5) $(x+y)^{3}+ 3(x+y)^{2}z+3(x+y)z^{2}+z^{3}=(x+z)^{3}+3(x+z)^{2}y+3(x+z)y^{2}+y^{3}$

6) $(a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)$

7) $(a+b+c)^{2}-a(b+c-a)-b(a+c-b)-c(a+b-c)=2(a^{2}+b^{2}+c^{2})$

8) $(x-y)^{3}+(x+y)^{3}+3(x-y)^{2}(x+y)+3(x+y)^{2}(x-y)=8x^{3}$

9) $x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)+(y-z)(z-x)(z-y)=0$

10) $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)(a+b+c)$

11) Prove that $(b-c)^{3}+(c-a)^{3}+(a-b)^{3}=3(b-c)(c-a)(a-b)$

12) If3 $2s=a+b+c$ , prove that $(s-a)^{2}+(s-b)^{2}+(s-c)^{2}+s^{2}=a^{2}+b^{2}+c^{2}$

13) If $2s=a+b+c$ , prove that $(s-a)^{3}+(s-b)^{3}+(s-c)^{3}+3abc=s^{3}$

14) If $2s=a+b+c$ , prove that $16s(s-a)(s-b)(s-c)=2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-a^{4}-b^{4}-c^{4}$

15) If $2s=a+b+c$ , then prove that $2(s-a)(s-b)(s-c)+a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)=abc$

16) If $a+b+c=0$ , then prove that $(2a-b)^{3}+(2b-c)^{3}+(2c-a)^{3}=3(2a-b)(2b-c)(2c-a)$

17) If $a+b+c=0$ , then prove that $\frac{a^{2}}{2a^{2}+bc} + \frac{b^{2}}{2b^{2}+ca} + \frac{c^{2}}{2c^{2}+ab} =1$

18) Prove that $(x+y+z)^{3}+(x+y-z)^{3}+(x-y+z)^{3}+(x-y-z)^{3}=4x(x^{2}+3y^{2}+3z^{2})$

19) If $a+b+c=0$ prove that $(s+3a)^{3}-(s-3b)^{3}-(s-3c)^{3}-3(s-3a)(s-3b)(s-3c)=0$

20) If $X=b+c-2a$ , $Y=c+a-2b$ , $Z=a+b-2c$ , find the value of $X^{2}+Y^{2}+Z^{2}-3XYZ$

21) Prove that $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2(c-b)(c-a)+2(b-a)(b-c)+2(a-b)(a-c)$

22) Prove that $a^{2}(b^{3}-c^{3})+b^{2}(c^{3}-a^{3})+c^{2}(a^{3}-b^{3})=(a-b)(b-c)(c-a)(ab+bc+ca)=a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3} = -[a^{2}b^{2}(a-b)+b^{2}c^{2}(b-c)+c^{2}a^{2}(c-a)]$

23) if $(a+b)^{2}+(b+c)^{2}+(c+a)^{2}=4(ab+bc+cd)$ , prove that $a=b=c=d$ .

24) If $x=a+d$ , $y=b+d$ , $z=c+d$ , prove that $x^{2}+y^{2}+z^{2}-yz-zx-xy=a^{2}+b^{2}+c^{2}-bc-ca-ab$

25) If $a+b+c=3$ , prove that $\frac{1}{b^{2}+c^{2}-a^{2}}+ \frac{1}{c^{2}+a^{2}-b^{2}} + \frac{1}{a^{2}+b^{2}-c^{2}}=0$

26) If $a+b+c=0$ , simplify: $\frac{b+c}{bc}(b^{2}+c^{2}-a^{2}) + \frac{c+a}{ca} (c^{2}+a^{2}-b^{2})+ \frac{a+b}{ab}(a^{2}+b^{2}-c^{2})$

27) Prove that the equation $(x-a)^{2}+(y-b)^{2}+(a^{2}+b^{2}-1)(x^{2}+y^{2}-1)=0$ is equivalent to the equation $(ax+by-1)^{2}+(bx-ay)^{2}=0$ , hence show that the only possible values of x and y are: $\frac{a}{a^{2}+b^{2}}$ , $\frac{b}{a^{2}+b^{2}}$

28) If $2(x^{2}+a^{2}-ax)(y^{2}+b^{2}-by)=x^{2}y^{2}+a^{2}b^{2}$ , prove that $(x-a)^{2}(y-b)^{2}+(bx-ay)^{2}=0$ and therefore that $a=x$ and $y=b$ are the only possible solutions.