# Is there a Fermat in you?? A solution possible

Reference: Popular Problems and Puzzles in Mathematics, Asok Kumar Mallik, Foundation Books, IISc Press.

https://www.amazon.in/Popular-Problems-Puzzles-Mathematics-Mallik/dp/938299386X/ref=sr_1_1?s=books&ie=UTF8&qid=1519414577&sr=1-1&keywords=Popular+problems+and+puzzles+in+mathematics

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:

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

Solution:

We show that the equation $y^{3}=x^{2}+2$ has only one solution for positive integers x and y, viz., $x=5$ and $y=3$. First, we write $y^{3}=x^{2}+2=(x+i\sqrt{2})(x-i\sqrt{2})$. It can be shown that unique prime factorization occurs in the complex number system $a+ ib\sqrt{2}$. It can be also argued that since the product of the two complex numbers of that form is a cube, each factor must be a cube. So, we write:

$x+i\sqrt{2}=(a+ib\sqrt{2})^{3}=a^{3}-6ab^{2}+i(3a^{2}b-2b^{3})\sqrt{2}$.

Equating imaginary parts of both sides, we get

$3a^{2}b-2b^{3}=b(3a^{2}-2b^{2})=1$.

Thus, $b=\pm 1$ and $3a^{2}-2b^{2}=\pm 1$, both having the same sign. But, with $b=-1$, one gets $a^{2}=1/3$, which is not possible. So, we take $b=1$ when $a=\pm 1$. Finally, with $a=1$, $x=-5$ and with $a=-1, x=5$. The sign of x is irrelevant as only $x^{2}$ is involved in the original equation. With the solution $x=5$, we get $y=3$ as the only set of solution.

Cheers,

Nalin Pithwa.

# 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.

# 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.

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.

# Another special number(s): Wilson primes and playful programming!

Problem:

A prime number p is called a Wilson prime if $(p-1)! \equiv -1 \pmod {p^{2}}$. Using a computer and some programming language like C, C++, or Python find the three smallest Wilson primes.

Cheers,

Nalin Pithwa.

# 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: