A Vietnamese Olympiad Problem: 1968: RMO Training for Algebra

Problem:

Let a and b satisfy a \geq b >0 and a+b=1.

  1. Prove that if m and n are positive integers with m<n, then a^{m}-a^{n} \geq b^{m}-b^{n}>0.
  2. For each positive integer n, consider a quadratic function: f_{n}(x)=x^{2}-b^{n}x-a^{n}.

Show that f(x) has two roots that are in between -1 and 1.

Solution:

Let k=n-m \in (0,n). Consider a^{m}b-ab^{m}=ab(a^{m-1}-b^{m-1}) with m-1 \geq 0. Since a \geq b >0, we have a^{m-1} \geq b^{m-1}. Hence, a^{m}b \geq ab^{m}. Call this relationship I.

On the other hand, notice that a \geq b, a^{2} \geq b^{2}, \ldots, a^{k-1} \geq b^{k-1}, which implies

1+a+a^{2}+\ldots+a^{k-1} \geq 1+b+b^{2}+ \ldots + b^{k-1}….call this relationship II.

From relationships I and II, it follows that

a^{m}b(1+a+a^{2}+\ldots+a^{k-1}) \geq ab^{m}(1+b+b^{2}+\ldots+b^{k-1})

which can be written as

a^{m}(1-a)(1+a+a^{2}+\ldots+a^{k-1}) \geq b^{m}(1-b)(1+b+b^{2}+\ldots+b^{k-1}), or equivalently, a^{m}(1-a^{k}) \geq b^{m}(1-b^{k}). That is, a^{m}-a^{n} \geq b^{m}-b^{n}.

It remains to prove that b^{m}-b^{n}>0. Indeed, b^{m}-b^{n}=b^{m}(1-b^{k})>0 as 0<b<1.

The equality occurs if and only if a=b=\frac{1}{2}.

2) Since discriminant \Delta=b^{2n}+4a^{n}>0, f_{n}(x) has two distinct real roots x_{1} \neq x_{2}. Also, note that if a, b \in (0,1), then the following holds:

f_{n}(1)=1-b^{n}-a^{n}=a+b-b^{n}-a^{n}=(a-a^{n})+(b-b^{n}) \geq 0,

f_{n}(-1)=1+b^{n}-a^{n}=(1-a^{n})+b^{n} \geq 0,

\frac{S}{2} = \frac{x_{1}+x_{2}}{2}=\frac{b^{n}}{2} \in (-1,1).

We conclude that x_{1}, x_{2} \in [-1, 1].

Cheers,

Nalin Pithwa.

Reference: Selected Problems of the Vietnamese Mathematical Olympiad (1962-2009), Le Hai Chau, Le Hai Khoi. 

Amazon India link:

https://www.amazon.in/Selected-Problems-Vietnamese-Olympiad-Mathematical/dp/9814289590/ref=sr_1_1?s=books&ie=UTF8&qid=1509907825&sr=1-1&keywords=selected+problems+of+the+vietnamese+mathematical+olympiad

Happy Numbers make Happy Programmers ! :-)

Here is one question which one of my students, Vedant Sahai asked me. It appeared in his computer subject exam of his recent ICSE X exam (Mumbai):

write a program to accept a number from the user, and check if the number is a happy number or not; and the program has to display a message accordingly:

A Happy Number is defined as follows: take a positive number and replace the number by the sum of the squares of its digits. Repeat the process until the number equals 1 (one). If the number ends with 1, then it is called a happy number.

For example: 31

Solution : 31 replaced by 3^{2}+1^{2}=10 and 10 replaced by 1^{2}+0^{2}=1.

So, are you really happy? 🙂 🙂 🙂

Cheers,

Nalin Pithwa.

Yet another special number !

The eminent British mathematician had once remarked: Every integer was a friend to Srinivasa Ramanujan.

Well, we are mere mortals, yet we can cultivate some “friendships with some numbers”. Let’s try:

Question:

Squaring 12 gives 144. By reversing the digits of 144, we notice that 441 is also a perfect square. Using C, C++, or python, write a program to find all those integers m, such that 1 \leq m \leq N, verifying this property.

PS: in order to write some simpler version of the algorithm, start playing with small, particular values of N.

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

Amazon India link:

https://www.amazon.in/1001-Problems-Classical-Number-Theory/dp/0821868888/ref=sr_1_1?s=books&ie=UTF8&qid=1509189427&sr=1-1&keywords=1001+problems+in+classical+number+theory

Cheers,

Nalin Pithwa.

 

Fundamental theorem of algebra: RMO training

It is quite well-known that any positive integer can be factored into a product of primes in a unique way, up to an order. (And, that 1 is neither prime nor composite) —- we all know this from our high school practice of “tree-method” of prime factorization, and related stuff like Sieve of Eratosthenes. But, it is so obvious, and so why it call it a theorem, that too “fundamental” and yet it seems it does not require a proof. It was none other than the prince of mathematicians of yore, Carl Friedrich Gauss, who had written a proof to it. It DOES require a proof — there are some counter example(s). Below is one, which I culled for my students:

Question:

Let E= \{a+b\sqrt{-5}: a, b \in Z\}

(a) Show that the sum and product of elements of E are in E.

(b) Define the norm of an element z \in E by ||z||=||a+b\sqrt{-5}||=a^{2}+5b^{2}. We say that an element p \in E is prime if it is impossible to write p=n_{1}n_{2} with n_{1}, n_{2} \in E, and ||n_{1}||>1, ||n_{2}||>1; we say that it is composite if it is not prime. Show that in E, 3 is a prime number and 29 is a composite number.

(c) Show that the factorization of 9 in E is not unique.

Cheers,

Nalin Pithwa.