Is there a “Fermat” in you? ? :-)

Fermat’s problem:

Pierre de Fermat (1601-65), the great French mathematical genius, is described as the “Prince of Amateurs” as he was not a professional mathematician and never published any work during his lifetime. As a civil servant, he served in the judiciary and spent his leisure with mathematics as his hobby. He corresponded with the best of mathematicians of his time. He teased the contemporary English mathematicians, Wallis and Digby, with the following problem, who had to admit defeat.

Problem:

26 is a number that is sandwiched between a perfect square (25) and a perfect cube (27). Prove that there is no other such number.

Please do try and let me know your solutions. Even partial solutions are welcome.

Cheers,

Nalin Pithwa.

A note on Diophantine problems

In solving Diophantine problems, one must guard against falling into certain formal traps. The identity{}

(a+b)^{2}=a^{2}+2ab+b^{2}

might lead one to think that, because all numbers of the form a^{2}+2ab+b^{}{2} are squares, any number of say, the form a^{2}+3ab+b^{2} could never be a perfect square because it is not an algebraic square. But, numbers are more versatile than that. Try a=7, b=3 in the expression a^{2}+3ab+b^{2}.

It was once proposed as a theorem that “the product of four consecutive terms of an AP of integers plus the fourth power of the common difference is always a perfect square but never a perfect fourth power.” If a is the first term of the four and b the common difference, we have a(a+b)(a+2b)(a+3b)+b^{4} which works out to be the same thing as (a^{2}+3ab+b^{2})^{2}.

The truth of the first assertion of the theorem is now evident, but we have already given a counter example showing the second assertion to be incorrect.

Ref: Excursions in Number Theory, C. Stanley Ogilvy and John T. Anderson,

http://www.amazon.in/Excursions-Number-Theory-Dover-Mathematics/dp/0486257789/ref=sr_1_4?s=books&ie=UTF8&qid=1491402900&sr=1-4&keywords=excursion+in+mathematics

Nalin Pithwa.

A geometry and algebra problem — RMO training

Several Olympiad problems deal with functions defined on certain sets of points. These problems are interesting in that they combine both geometrical and algebraic ideas.

Problem.

Let n>2 be an integer and f:P \rightarrow \Re a function defined on the set of points in the plane, with the property that for any regular n-gon A_{1}A_{2} \ldots A_{n}

f(A_{1})+f(A_{2})+ \ldots + f(A_{n})=0.

Prove that f is the zero function.

Proof:

Core Concept: 

In Euclidean geometry, the only motions permissible are rigid motions — translations, rotations, and reflections.

Solution:

Let A be an arbitrary point. Consider a regular n-gon AA_{1}A_{2}\ldots A_{n-1}. Let k be an integer, 0 \leq k \leq n-1. A rotation with center A of angle \frac{2\pi k }{n} sends the polygon AA_{1}A_{2}\ldots A_{n-1} to A_{k0}A_{k1} \ldots A_{k,n-1}, where A_{k0}=A and A_{ki} is the image of A_{i} for all I=1, 2, \ldots, n-1

From the condition of the statement, we have

\sum_{k=0}^{n-1} \sum_{i=0}^{n-1}f(A_{ki})=0.

Observe that in the sum the number f(A) appears n times, therefore,

nf(A)+ \sum_{k=0}^{n-1} \sum_{i=1}^{n-1}f(A_{kl})=0

On the other hand, we have

\sum_{k=0}^{n-1} \sum_{i=1}^{n-1}f(A_{ki})=\sum_{i=1}^{n-1} \sum_{k=0}^{n-1}f(A_{ki})=0

since the polygons A_{0i}A_{1i} \ldots A_{n-1,i} are all regular n-gons. From the two equalities above we deduce f(A)=0, hence, f is the zero function.

More later,

Nalin Pithwa

 

 

Brain teasers based on Bezout’s identity — RMO training

Problem 1.

In a special football game, a team scores 7 points for a touch-down and 3 points for a field goal. Determine the largest mathematically unreachable number of points scored by a team in an (infinitely) long game.

Problem 2.

There is an ample of supply of milk in a milk tank. Mr. Fat is given a 5 litre (unmarked) container and a 9-litre (unmarked) container. How can he measure out 2 litres of milk?

The above two are classic brain teasers based on Bezout’s identity.

More later,

Nalin Pithwa