# Pre-RMO or RMO algebra practice problem: infinite product

Find the product of the following infinite number of terms:

$\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \ldots = \prod_{m=2}^{\infty}\frac{m^{3}-1}{m^{3}+1}$

$m^{3}-1=(m-1)(m^{2}+m+1)$, and also, $m^{3}+1=(m+1)(m^{2}-m+1)=(m-1+2)((m-1)^{2}+(m-1)+1)$

Hence, we get $P_{m}=\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \ldots \times \frac{m^{3}-1}{m^{3}+1}$, which in turn, equals

$(\frac{1}{3} \times \frac{7}{3}) \times (\frac{2}{4} \times \frac{13}{7}) \times (\frac{3}{5} \times \frac{21}{13})\times \ldots (\frac{m-1}{m+1} \times \frac{m^{2}+m+1}{m^{2}-m+1})$, that is, in turn equal to

$\frac{2}{3} \times \frac{m^{2}+m+1}{m(m+1)}$, that is, in turn equal to

$\frac{}{} \times (1+ \frac{1}{m(m+1)})$, so that when $m \rightarrow \infty$, and then $P_{m} \rightarrow 2/3$.

personal comment: I did not find this solution within my imagination !!! 🙂 🙂 🙂

The credit for the solution goes to “Popular Problems and Puzzles in Mathematics” by Asok Kumar Mallik, IISc Press, Foundation Books. Thanks Prof. Mallik !!

Cheers,

Nalin Pithwa

# Algebra Training for RMO and Pre-RMO: let’s continue: Fibonacci Problem and solution

Fibonacci Problem:

Leonardo of Pisa (famous as Fibonacci) (1173) wrote a book “Liber Abaci” (1202), wherein he introduced Hindu-Arabic numerals in Europe. In 1225, Frederick II declared him as the greatest mathematician in Europe when he posed the following problem to defeat his opponents.

Question:

Determine the rational numbers x, y and z to satisfy the following equations:

$x^{2}+5=y^{2}$ and $x^{2}-5=z^{2}$.

Solution:

Definition: Euler defined a congruent number to be a rational number that is the area of a right-angled triangle, which has rational sides. With p, q, and r as a Pythagorean triplet such that $r^{2}=p^{2}+q^{2}$, then $\frac{pq}{2}$ is a congruent number.

It can be shown that square of a rational number cannot be a congruent number. In other words, there is no right-angled triangle with rational sides, which has an area as 1, or 4, or $\frac{1}{4}$, and so on.

Characteristics of a congruent number: A positive rational number n is a congruent number, if and only if there exists a rational number u such that $u^{2}-n$ and $u^{2}+n$ are the squares of rational numbers. (Thus, the puzzle will be solved if we can show that 5 is a congruent number and we can determine the rational number $u(=x)$). First, let us prove the characterisitic mentioned above.

Necessity: Suppose n is a congruent number. Then, for some rational number p, q, and r, we have $r^{2}=p^{2}+q^{2}$ and $\frac{pq}{2}=n$. In that case,

$\frac{p+q}{2}, \frac{p-q}{2}$ and n are rational numbers and we have

$(\frac{p+q}{2})^{2}=\frac{p^{2}+q^{2}}{4}+\frac{pq}{2}=(\frac{r}{2})^{2}+n$ and similarly,

$(\frac{p-q}{2})^{2}=(\frac{r}{2})^{2}-n$.

Setting $u=\frac{r}{2}$, we get $u^{2}-n$ and $u^{2}+n$ are squares of rational numbers.

Sufficiency:

Suppose n and u are rational numbers such that $\sqrt{u^{2}-n}$ and $\sqrt{u^{2}+n}$ are rational, when

$p=\sqrt{u^{2}+n}+\sqrt{u^{2}-n}$ and $q=\sqrt{u^{2}+n}-\sqrt{u^{2}-n}$

and 2n are rational numbers satisfying $p^{2}+q^{2}=(2n)^{2}$ is a rational square and also $\frac{pq}{2}=n$, a rational number which is a congruent number.

So, we see that the Pythagorean triplets can lead our search for a congruent number. Sometimes a Pythagorean triplet can lead to more than one congruent number as can be seen with $(9,40,41)$. This set obviously gives 180 as a congruent number. But, as $180=5 \times 36=5 \times 6^{2}$, we can also consider a rational Pythagorean triplet $(\frac{9}{6}, \frac{40}{6}, \frac{41}{6})$, which gives a congruent number 5 (we were searching for this congruent number in this puzzle!). We also determine the corresponding $u=\frac{r}{2}=\frac{41}{12}$.

The puzzle/problem is now solved with $x=\frac{41}{12}$, which gives $y=\frac{49}{12}$, and $z=\frac{31}{12}$.

One can further show that if we take three rational squares in AP, $u^{2}-n$, and $u^{2}$, and $u^{2}+n$, with their product defined as a rational square $v^{2}$ and n as a congruent number, then $x=u^{2}$, $y=v$ is a rational point on the elliptic curve $y^{2}=x^{3}-n^{2}x$.

Reference:

1) Popular Problems and Puzzles in Mathematics: Asok Kumar Mallik, IISc Press, Foundation Books, Amazon India link:

https://www.amazon.in/Popular-Problems-Puzzles-Mathematics-Mallik/dp/938299386X/ref=sr_1_1?s=books&ie=UTF8&qid=1525099935&sr=1-1&keywords=popular+problems+and+puzzles+in+mathematics

2) Use the internet, or just Wikipedia to explore more information on Fibonacci Numbers, Golden Section, Golden Angle, Golden Rectangle and Golden spiral. You will be overjoyed to find relationships amongst all the mentioned “stuff”.

Cheers,

Nalin Pithwa

# An easy inequality from Nordic mathematical contests !?

Reference: Nordic Mathematical Contest, 1987-2009, R. Todev.

Question:

Let a, b, and c be real numbers different from 0  and $a \geq b \geq c$. Prove that inequality

$\frac{a^{3}-c^{3}}{3} \geq abc(\frac{a-b}{c} + \frac{b-c}{a})$

holds. When does the equality hold?

Proof:

We know that a, b and c are real, distinct and also non-zero and also that $a \geq b \geq c$.

Hence, $c-b \leq 0 \leq a-b$, we have $(a-b)^{3}\geq (c-b)^{3}$, or

$a^{3}-3a^{a}b+3ab^{2}-b^{3} \geq c^{3}-3bc^{2}+3b^{2}c-b^{3}$

On simplifying this, we immediately have

$\frac{1}{3}{(a^{3}-c^{3})} \geq a^{2}b-ab^{2}+b^{2}c-bc^{2}=abc(\frac{a-b}{c}+\frac{b-c}{a})$.

A sufficient condition for equality is $a=c$. If $a>c$, then $(a-b)^{3}>(c-b)^{3}$. which makes the proved inequality a strict one. So, $a=c$ is a necessary condition for equality too.

-Nalin Pithwa.

# There are many “inequalities” ! :-( :-) !

Reference: R. Todev, Nordic Mathematical Contests, 1987-2009.

Question:

Let a, b, and c be positive real numbers. Prove that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \leq \frac{a^{2}}{b^{2}} + \frac{b^{2}}{c^{2}} + \frac{c^{2}}{a^{2}}$.

Solution:

The arithmetic-geometric inequality yields

$3=3\sqrt[3]{\frac{a^{2}}{b^{2}}.\frac{b^{2}}{c^{2}}.\frac{c^{2}}{a^{2}}}\leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}$,

or $\sqrt{3} \leq \sqrt{\frac{a^{2}}{b^{2}} + \frac{b^{2}}{c^{2}} + \frac{c^{2}}{a^{2}}}$…call this relation I.

On the other hand, the Cauchy-Schwarz inequality implies

$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \leq \sqrt{1^{2}+1^{2}+1^{2}}\sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}=\sqrt{3}\sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}$….call this relation II.

We arrive at the inequality we desire by combining relations I and II. Hence, the proof. QED.

Cheers,

Nalin Pithwa.

# Functions — “s’wat” Math is about !! :-)

Reference: Nordic Mathematical Contest 1987, R. Todev:

Question:

Let f be a function, defined for natural numbers, that is strictly increasing, such that values of the function are also natural numbers and which satisfies the conditions $f(2)=a>2$ and $f(mn)=f(m)f(n)$ for all natural numbers m and n. Define the smallest possible value of a.

Solution:

Since, $f(n)=n^{2}$ is a function satisfying the conditions of the problem, the smallest possible a is at most 4. Assume that $a=3$. It is easy to prove by induction that $f(n^{k})={f(n)}^{k}$ for all $k \geq 1$. So, taking into account that f is strictly increasing, we get

${f(3)}^{4}=f(3^{4})=f(81)>f(64)=f(2^{6})={f(2)}^{6}=3^{6}=27^{2}>25^{2}=5^{4}$

as well as ${f(3)}^{8}=f(3^{8})=f(6561).

So, we arrive at $5. But, this is not possible, since $f(3)$ is an integer. So, $a=4$.

Cheers,

Nalin Pithwa.

# Francois Viete, France, Vietnam and Algebra: RMO training for algebra

Reference 1: https://en.wikipedia.org/wiki/Vieta%27s_formulas

Reference 2: Selected Problems of the Vietnamese Mathematical Olympiad (1962-2009), Le Hai Chau, and Le Hai Khoi, published by World Scientific;

Question:

Without solving the cubic equation, $x^{3}-x+1=0$, compute the sum of the eighth powers of all roots of the equation.

Approach: we want to be able to express the sum of the eighth powers of the three roots in terms of the three Viete’s relations here.

If $x_{1}$, $x_{2}$, $x_{3}$ are roots of the given cubic equation then, by Viete’s relations between roots and coefficients, we can say the following:

$x_{1}+x_{2}+x_{3}=0$

$x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}=-1$

$x_{1}x_{2}x_{3}=-1$.

Furthermore, from $x_{i}^{3}-x_{i}+1=0$, it follows that

$x_{i}^{3}=x_{i}-1$

$x_{i}^{5}=x_{i}^{3}.x_{i}^{2}=(x_{i}-1)x_{i}^{2}=x_{i}^{3}-x_{i}^{2}=-x_{i}^{2}+x_{i}-1$

Then, $x_{1}^{8}+x_{2}^{8}+x_{3}^{8}=2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}) - 3(x_{1}+x_{2}+x_{3})+6$

But, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{2}-2(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1})=2$ and so $x_{1}^{8}+x_{2}^{8}+x_{3}^{8}=4-0+6=10$.

More later,

Nalin Pithwa.

# An inequality for harmonic numbers — RMO Inequalities — Basics

The Harmonic Numbers $H_{j}$ for $j=1,2,3, \ldots$ are defined by

$H_{j}=1+\frac{1}{2}+\frac{1}{3} +\ldots + \frac{1}{j}$

For instance, $H_{4}=1+\frac{1}{2}+\frac{1}{3} + \frac{1}{4} = \frac{25}{12}$

Use mathematical induction to show that $H_{2^{n}} \geq 1 + \frac{n}{2}$ whenever n is a non-negative integer.

Solution:

To carry out the proof, let P(n) be the proposition that $H_{2^{n}}=1+\frac{n}{2}$

Basis step:

P(0) is true because $H_{2^{0}}=H_{1}=1 \geq 1 + \frac{0}{2}$

Inductive Step:

The inductive hypothesis is the statement that $P(k)$ is true, that is, $H_{2^{k}} \geq 1 + \frac{k}{2}$

where k is a non-negative integer. We must show that if $P(k)$ is true, then $P(k+1)$, which states that

$H_{2^{k+1}} \geq 1 + \frac{k+1}{2}$, is also true. So, assuming the inductive hypothesis, it follows that

$H_{2^{k+1}} = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{2^{k}} + \frac {1}{2^{k}+1} + \ldots + \frac{1}{2^{k+1}}$…this step follows from the definition of harmonic number

$=H_{2^{k}} + \frac{1}{2^{k}+1} + \ldots + \frac{1}{2^{k+1}}$….this step again follows by the definition of $2^{k}$th harmonic number

$\geq (1+\frac{k}{2}) + \frac{1}{2^{k}+1} + \ldots + \frac{1}{2^{k+1}}$…this step follows by the inductive hypothesis

$\geq (1+ \frac{k}{2}) + 2^{k}. \frac{2}{2^{k+1}}$…because there are $2^{k}$ terms each greater than or equal to $\frac{1}{2^{k+1}}$

$\geq (1+\frac{k}{2})+ \frac{1}{2}$….canceling a common factor of $2^{k}$ in second term

$= 1 + \frac{k+1}{2}$

This establishes the inductive step of the proof.

We have completed the basis step and the inductive step. Thus, by mathematical induction $P(n)$ is true for all non-negative integers. That is, the inequality $H_{2^{n}} \geq 1 + \frac{n}{2}$ for the harmonic numbers is valid for all non-negative integers n. QED.

Remark:

The inequality established here shows that the harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} + \ldots$ is a divergent series. This is an important example in the study of infinite series.

Note:

Google, or through some other literature, find out why harmonic numbers are termed so. You will understand something more beautiful, more deeper !!

Nalin Pithwa.