# A Special Number

Problem:

Show that for each positive integer n equal to twice a triangular number, the corresponding expression $\sqrt{n+\sqrt{n+\sqrt{n+ \sqrt{n+\ldots}}}}$ represents an integer.

Solution:

Let n be such an integer, then there exists a positive integer m such that $n=(m-1)m=m^{2}-m$. We then have $n+m=m^{2}$ so that we have successively

$\sqrt{n+m}=m$; $\sqrt{n + \sqrt{n+m}}=m$; $\sqrt{n+\sqrt{n+\sqrt{n+m}}}=m$ and so on. It follows that

$\sqrt{n+\sqrt{n+\sqrt{n+ \sqrt{n+\ldots}}}}=m$, as required.

Comment: you have to be a bit aware of properties of triangular numbers.

Reference:

1001 Problems in Classical Number Theory by Jean-Marie De Koninck and Armel Mercier, AMS (American Mathematical Society), Indian Edition:

https://www.amazon.in/1001-Problems-Classical-Number-Theory/dp/0821868888/ref=sr_1_1?s=books&ie=UTF8&qid=1508634309&sr=1-1&keywords=1001+problems+in+classical+number+theory

Cheers,

Nalin Pithwa.

# Another cute proof: square root of 2 is irrational.

Reference: Elementary Number Theory, David M. Burton, Sixth Edition, Tata McGraw-Hill.

(We are all aware of the proof we learn in high school that $\sqrt{2}$ is irrational. (due Pythagoras)). But, there is an interesting variation of that proof.

Let $\sqrt{2}=\frac{a}{b}$ with $gcd(a,b)=1$, there must exist integers r and s such that $ar+bs=1$. As a result, $\sqrt{2}=\sqrt{2}(ar+bs)=(\sqrt{2}a)r+(\sqrt{2}b)s=2br+2bs$. This representation leads us to conclude that $\sqrt{2}$ is an integer, an obvious impossibility. QED.

# Algebra : max and min: RMO/INMO problem solving practice

Question 1:

If $x^{2}+y^{2}=c^{2}$, find the least value of $\frac{1}{x^{2}} + \frac{1}{y^{2}}$.

Solution 1:

Let $z^{'}=\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{y^{2}+x^{2}}{x^{2}y^{2}} = \frac{c^{2}}{x^{2}y^{2}}$

$z^{'}$ will be minimum when $\frac{x^{2}y^{2}}{c^{2}}$ will be minimum.

Now, $let z=\frac{x^{2}y^{2}}{c^{2}}=\frac{1}{c^{2}}(x^{2})(y^{2})$….call this equation I.

Hence, z will be maximum when $x^{2}y^{2}$ is maximum but $(x^{2})(y^{2})$ is the product of two factors whose sum is $x^{2}+y^{2}=c^{2}$.

Hence, $x^{2}y^{2}$ will be maximum when both these factors are equal, that is, when

$\frac{x^{2}}{1}=\frac{y^{2}}{1}=\frac{x^{2}+y^{2}}{1}=\frac{c^{2}}{1}$. From equation I, maximum value of $z=\frac{c^{2}}{4}$. Hence, the least value of $\frac{1}{x^{2}} + \frac{1}{y^{2}}=\frac{4}{c^{2}}$.

Some basics related to maximum and minimum:

Basic 1:

Let a and b be two positive quantities, S their sum and P their product; then, from the identity:

$4ab=(a+b)^{2}-(a-b)^{2}$, we have

$4P=S^{2}-(a-b)^{2}$ and $S^{2}=4P+(a-b)^{2}$.

Hence, if S is given, P is greatest when $a=b$; and if P is given, S is least when $a=b$. That is, if the sum of two positive quantities is given, their product is greatest when they are equal; and, if the product of two positive quantities is given, their sum is least when they are equal.

Basic 2:

To find the greatest value of a product the sum of whose factors is constant.

Solution 2:

Let there be n factors $a,b,c,\ldots, k$, and suppose that their sum is constant and equal to s.

Consider the product $abc\ldots k$, and suppose that a and b are any two unequal factors. If we replace the two unequal factors a and b by the two equal factors $\frac{a+b}{2}, \frac{a+b}{2}$, the product is increased, while the sum remains unaltered; hence, so long as the product contains two unequal factors it can be increased without altering the sum of the factors; therefore, the product is greatest when all the factors are equal. In this case, the value of each of the n factors is $\frac{s}{m}$, and the greatest value of the product is $(\frac{s}{n})^{n}$, or $(\frac{a+b+c+\ldots+k}{n})^{n}$.

Corollary to Basic 2:

If $a, b, c, \ldots k$ are unequal, $(\frac{a+b+c+\ldots+k}{n})^{2}>abc\ldots k$;

that is, $\frac{a+b+c+\ldots +k}{n} > (\frac{a+b+c+\ldots + k}{n})^{\frac{1}{n}}$.

By an extension of the meaning of the terms arithmetic mean and geometric mean, this result is usually stated as follows: the arithmetic mean of any number of positive quantities is greater than the geometric mean.

Basic 3:

To find the greatest value of $a^{m}b^{n}c^{p}\ldots$ when $a+b+c+\ldots$ is constant; m,n, p, ….being positive integers.

Solution to Basic 3:

Since m,n,p, …are constants, the expression $a^{m}b^{n}c^{p}\ldots$ will be greatest when $(\frac{a}{m})^{m}(\frac{b}{n})^{n}(\frac{c}{p})^{p}\ldots$ is greatest. But, this last expression is the product of $m+n+p+\ldots$ factors whose sum is $m(\frac{a}{m})+n(\frac{b}{n})+p(\frac{c}{p})+\ldots$, or $a+b+c+\ldots$, and therefor constant. Hence, $a^{m}b^{n}c^{p}\ldots$ will be greatest when the factors $\frac{a}{m}, \frac{b}{n}, \frac{c}{p}, ldots$ are all equal, that is, when

$\frac{a}{m} = \frac{b}{n} = \frac{c}{p} = \ldots = \frac{a+b+c+\ldots}{m+n+p+\ldots}$

Thus, the greatest value is $m^{m}n^{n}p^{p}\ldots (\frac{a+b+c+\ldots}{m+n+p+\ldots})^{m+n+p+\ldots}$.

Some examples using the above techniques:

Example 1:

Show that $(1^{r}+2^{r}+3^{r}+\ldots+n^{r})>n^{n}(n!)^{r}$ where r is any real number.

Solution 1:

Since $\frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n}>(1^{r}.2^{r}.3^{r}\ldots n^{r})^{\frac{1}{n}}$

Hence, $(\frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n})^{n}>1^{r}.2^{r}.3^{r} \ldots n^{r}$

that is, $>(n!)^{r}$, which is the desired result.

Example 2:

Find the greatest value of $(a+x)^{3}(a-x)^{4}$ for any real value of x numerically less than a.

Solution 2:

The given expression is greatest when $(\frac{a+x}{3})^{3}(\frac{a-x}{4})^{4}$ is greatest; but, the sum of the factors of this expression is $3(\frac{a+x}{3})+4(\frac{a-x}{4})$, that is, $2a$; hence, $(a+x)^{3}(a-x)^{4}$ is greatest when $\frac{a+x}{3}=\frac{a-x}{4}$, that is, $x=-\frac{a}{7}$. Thus, the greatest value is $\frac{6^{3}8^{4}}{7^{7}}a^{r}$.

Some remarks/observations:

The determination of maximum and minimum values may often be more simply effected by the solution of a quadratic equation than by the foregoing methods. For example:

Question:

Divide an odd integer into two integral parts whose product is a maximum.

Let an odd integer be represented as $2n+1$; the two parts by x and $2n+1-x$; and the product by y; then $(2n+1)x-x^{2}=y$; hence,

$2x=(2n+1)\pm \sqrt{(2n+1)^{2}-4y}$

but the quantity under the radical sign must be positive, and therefore y cannot be greater than $\frac{1}{4}(2n+1)^{2}$, or, $n^{2}+n+\frac{1}{4}$; and since y is integral its greatest value must be $n^{2}+n$; in which case $x=n+1$, or n; thus, the two parts are n and $n+1$.

Sometimes we may use the following method:

Find the minimum value of $\frac{(a+x)(b+x)}{c+x}$.

Solution:

Put $c+x=y$; then the expression $=\frac{(a-c+y)(b-c+y)}{y}=\frac{(a-c)(b-c)}{y}+y+a-c+b-c$

which in turn equals

$(\frac{\sqrt{(a-c)(b-c)}}{\sqrt{y}}-\sqrt{y})^{2}+a-c+b-c+2\sqrt{(a-c)(b-c)}$.

Hence, the expression is a a minimum when the square term is zero; that is when $y=\sqrt{(a-c)(b-c)}$.

Thus, the minimum value is $a-c+b-c+2\sqrt{(a-c)(b-c)}$, and the corresponding value of x is $\sqrt{(a-c)(b-c)}-c$.

Problems for Practice:

1. Find the greatest value of x in order that $7x^{2}+11$ may be greater than $x^{3}+17x$.
2. Find the minimum value of $x^{2}-12x+40$, and the maximum value of $24x-8-9x^{2}$.
3. Show that $(n!)^{2}>n^{n}$ and $2.4.6.\ldots 2n<(n+1)^{n}$.
4. Find the maximum value of $(7-x)^{4}(2+x)^{5}$ when x lies between 7 and -2.
5. Find the minimum value of $\frac{(5+x)(2+x)}{1+x}$.

More later,

Nalin Pithwa.

# Algebra question: RMO/INMO problem-solving practice

Question:

If $\alpha$, $\beta$, $\gamma$ be the roots of the cubic equation $ax^{3}+3bx^{2}+3cx+d=0$. Prove that the equation in y whose roots are $\frac{\beta\gamma-\alpha^{2}}{\beta+\gamma-2\alpha} + \frac{\gamma\alpha-\beta^{2}}{\gamma+\\alpha-2\beta} + \frac{\alpha\beta-\gamma^{2}}{\alpha+\beta-2\gamma}$ is obtained by the transformation $axy+b(x+y)+c=0$. Hence, form the equation with above roots.

Solution:

Given that $\alpha$, $\beta$, $\gamma$ are the roots of the equation:

$ax^{3}+3bx^{2}+3cx+d=0$…call this equation I.

By relationships between roots and co-efficients, (Viete’s relations), we get

$\alpha+\beta+\gamma=-\frac{3b}{a}$ and $\alpha\beta+\beta\gamma+\gamma\alpha=\frac{3c}{a}$, and $\alpha\beta\gamma=-\frac{d}{a}$

Now, $\gamma=\frac{\beta\gamma-\alpha^{2}}{\beta+\gamma-2\alpha}=\frac{\frac{\alpha\beta\gamma}{\alpha}-\alpha^{2}}{(\alpha+\beta+\gamma)-3\alpha}=\frac{-\frac{d}{a\alpha}-\alpha^{2}}{-\frac{3b}{a}-3\alpha}=\frac{d+a\alpha^{3}}{3\alpha(b+a\alpha)}$, that is,

$3xy(b+ax)=d+ax^{3}$, or $ax^{3}-3ayx^{2}-3byx+d=0$…call this equation II.

Subtracting Equation II from Equation I, we get

$3(b+ay)x^{2}+3(c+by)x=0$

$(b+ay)x+c+by=0$ since $x \neq 0$

$axy+b(x+y)+c=0$ which is the required transformation.

Now, $(ay+b)x=-(by+c)$, that is, $x=-\frac{by+c}{ay+b}$

Putting this value of x in Equation I, we get

$-a(\frac{by+c}{ay+b})^{3}+3b(\frac{by+c}{ay+b})^{2}-3c(\frac{by+c}{ay+b})+d=0$, that is,

$a(by+c)^{3}-3b(by+c)^{2}(ay+b)+3c(by+c)(ay+b)^{2}-d(ay+b)^{3}=0$, which is the required equation.

Cheers,

Nalin Pithwa.

# RMO 2017 Warm-up: Two counting conundrums

Problem 1:

There are n points in a circle, all joined with line segments. Assume that no three (or more) segments intersect in the same point. How many regions inside the circle are formed in this way?

Problem 2:

Do there exist 10,000 10-digit numbers divisible by 7, all of which can be obtained from one another by a re-ordering of their digits?

Solutions will be put up in a couple of days.

Nalin Pithwa.

# Math concept(s) : simple yet subtle

Most math concepts are intuitive, simple, yet subtle. A similar opinion is expressed by Prof. Michael Spivak in his magnum opus, Differential Geometry (preface). It also reminds me — a famous quote of the ever-quotable Albert Einstein: “everything should be as simple as possible, and not simpler.”

I have an illustrative example of this opinion(s) here:

Consider the principle of mathematical induction:

Most students use the first version of it quite mechanically. But, is it really so? You can think about the following simple intuitive argument which when formalized becomes the principle of mathematical induction:

Theorem: First Principle of Finite Induction:

Let S be a set of positive integers with the following properties:

1. The integer 1 belongs to S.
2. Whenever the integer k is in S, the next integer $k+1$ must also be in S.

Then, S is the set of all positive integers.

The proof of condition 1 is called basis step for the induction. The proof of 2 is called the induction step. The assumptions made in carrying out the induction step are known as induction hypotheses. The induction situation has been likened to an infinite row of dominoes all standing on edge and arranged in such a way that when one falls it knocks down the next in line. If either no domino is pushed over (that is, there is no basis for the induction), or if the spacing is too large (that is, the induction step fails), then the complete line will not fall.

So, also remember that the validity of the induction step does not necessarily depend on the truth of the statement that one is endeavouring to prove.

More later,

Nalin Pithwa.

# Famous harmonic series — solutions

Solutions to previous blog questions on harmonic series are presented below:

Basic Reference: Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books; Amazon India link:

https://www.amazon.in/Popular-Problems-Puzzles-Mathematics-Mallik/dp/938299386X/ref=sr_1_2?ie=UTF8&qid=1505622919&sr=8-2&keywords=popular+problems+and+puzzles+in+mathematics

Detailed Reference:

Mallik, A. K. 2007: “Curious consequences of simple sequences,” Resonance, (January), 23-37.

Personal opinion only: Resonance is one of the best Indian magazines/journals for elementary/higher math and physics. It behooves you to subscribe to it. It will help in RMO, INMO and Madhava Mathematics Competition of India.

http://www.ias.ac.in/Journals/Resonance_–_Journal_of_Science_Education/

Solutions:

1. The thirteenth century French polymath Nicolas-Oresme proved that the harmonic series :$1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots$ does not converge. Prove this result.

Solution 1:

Nicolas Oreme had provided a simple proof as it involves mere grouping of terms, noticing patterns and making comparisons:

$\lim_{n \rightarrow \infty}H_{n}=H_{\infty}=1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \ldots$

$> 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \ldots$

$> 1+ \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \ldots$

Therefore, $H_{\infty}$ diverges as we go on adding one half indefinitely. Here is another way to prove this:

Consider $\lim_{n \rightarrow \infty}H_{n}=H_{\infty}=1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots$

By multiplying and dividing both sides by 2 and then by regrouping the terms, we get:

$H_{\infty}=\frac{2}{2} + \frac{2}{4} + \frac{2}{6} + \frac{2}{8} + \ldots$

$H_{\infty}= \frac{1+1}{2} + \frac{1+1}{4} + \frac{1+1}{6} + \frac{1+1}{8} + \ldots$

$H_{\infty}= (\frac{1}{2} + \frac{1}{2}) + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{6} + \frac{1}{6}) + (\frac{1}{8} + \frac{1}{8}) + \ldots$

$H_{\infty}<1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \ldots$

leading to a contradiction that $H_{\infty} The contradiction arose because only finite numbers remain unaltered when multiplied and divided by 2. So, $H_{\infty}$ is not a finite number, that is, it diverges.

2. Prove that $\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots$ does not converge.

Solution 2:

$H_{E}=\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots=\frac{1}{2}(1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} +\ldots)=\frac{1}{2}H_{n}$.

Since $H_{\infty}$ diverges, so does $H_{E}$.

3. Prove that $1+ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \ldots$ does not converge.

Solution 3:

$H_{O}=\frac{1}{1}+ \frac{1}{3} +\frac{1}{5}+ \frac{1}{7}+ \ldots$ diverges as each term in this series is greater than the corresponding term of $H_{E}$, which we have just sent to diverge.

Cheers,

Nalin Pithwa.