A Vietnamese Olympiad Problem: 1968: RMO Training for Algebra


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.


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].


Nalin Pithwa.

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

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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:


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.

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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:


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.


Nalin Pithwa.

A Special Number


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.


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.


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

Amazon India link:



Nalin Pithwa.