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Twinkle twinkle little star is a short musical piece I composed and recorded last evening.

This piece consists of 56 measures. It is 1 minute 52 seconds long. The music is composed of four tracks. This is my second attempt at recording music with multiple tracks. The last such attempt was more than two years ago when I composed and recorded 'A few notes'.

The links to the audio files, sheet music, etc. are provided below. The files reside in my website. In case, my website is down, the YouTube link provided below should still work.

The four tracks in this piece are:

  1. Grand piano
  2. Slow strings
  3. Xenon pad
  4. Music box

This arrangement is based on the popular melody of the nursery rhyme called Twinkle, twinkle, little star. The melody is played with the treble notes of the piano. I wrote the bass notes for the piano and the strings, and the high notes for the pad and the music box to fill the music with emotions of love and happiness. I recorded this after about two hours of practice.

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In my last blog post on solving the Impossible Puzzle, I promised that I would prove the following.

For any positive integer N, there exists only one pair of positive integers a and b such that 1 < a < b and ab = N if and only if N is a product of two different primes, a cube of a prime, or a fourth power of a prime.
In the last post, I gave an intuitive demonstration of why this should be true. However, intuition may be misleading and should not be relied upon. We need to prove this by pure logic.

From the fundamental theorem of arithmetic, we know that any integer greater than 1 can be expressed as a unique product of primes. A corollary of this theorem is:

Any positive integer N > 1 can be written uniquely in a canonical form N = p1k1p2k2 … pnkn where, each ki is a positive integer and each pi is a prime for positive integers i, 1 ≤ i ≤ n.
From this, let me prove the following lemma.

Lemma: The number of possible factors of N is (k1 + 1)(k2 + 1)…(kn + 1)

Proof: Any factor of N can be expressed as p1r1p2r2 … pnrn, where 0 ≤ ri ≤ ki for 1 ≤ i ≤ n.

In the expression for the factor of N, we can see that each pi can be raised by any power between 0 and ki (including 0 and ki). So, there are ki + 1 different ways to choose the exponent for each pi in the above expression.

Hence, the total number of possible factors is: (k1 + 1)(k2 + 1)…(kn + 1). ∎

Now, let me prove the theorem I mentioned in the beginning of this post.

Theorem: For any positive integer N, there exists only one pair of positive integers a and b such that 1 < a < b and ab = N if and only if N is a product of two different primes, a cube of a prime, or a fourth power of a prime.

Proof: The sufficiency of the condition is easy to show.

If N is a product of two different primes, p1 and p2, where p1 < p2, the set of factors of N is {1, p1, p2, N}. Thus, it has only two factors a = p1 and b = p2 such that 1 < a < b and ab = N.

If N is a cube of a prime p, N = p3 and the set of factors of N is {1, p, p2, p3}. a = p and b = p2 are the only factors such that 1 < a < b and ab = N.

If N is a fourth power of a prime p, N = p4 and the set of factors of N is {1, p, p2, p3, p4}. a = p and b = p3 are the only factors such that 1 < a < b and ab = N.

Now, let us prove the necessity of the condition.

If a positive integer N has only one two factors a and b such that 1 < a < b, it implies that there are two possibilities of the total set of factors of N. If N is not a perfect square, the total set of factors of N is {1, a, b, N} where 1 × N = a × b. If N is a perfect square, such that √N = r, the total set of factors of N is {1, a, r, b, N} where 1 × N = a × b = r × r.

So, in any case, the total number of factors N has, is either 4 or 5.

Let us write N as p1k1p2k2…pnkn for primes p1, p2, &hellip, pn and positive integers k1, k2, …, kn.

For n ≥ 3, the total number of factors N has, is (k1 + 1)(k2 + 1)…(kn + 1) ≥ (k1 + 1)(k2 + 1)(k3 + 1) ≥ 6. But we have shown that the total number of factors of N is either 4 or 5. So, n must be less than 3. Hence, we can write N as p1k1p2k2 and thus the total number of factors it has can be written as (k1 + 1)(k2 + 1). Without loss of generality, let us assume k1 ≤ k2.

All possible factorings of 4 and 5 are as follows.

  • 4 = 1 × 4 = (0 + 1)(3 + 1)
  • 4 = 2 × 2 = (1 + 1)(1 + 1)
  • 5 = 1 × 5 = (0 + 1)(4 + 1)
From the above enumeration of factorings we can see that one of the following must be true.
  • k1 = 0 and k2 = 3
  • k1 = k2 = 1
  • k1 = 0 and k2 = 4
If k1 = 0 and k2 = 3, N = p23.

If k1 = k2 = 1, N = p1p2.

If k1 = 0 and k2 = 4, N = p24.

So, we have shown that N is a product of two primes, a cube of a prime, or a fourth power of a prime.∎

For a few more interesting results on factors, see http://susam.in/downloads/mathematics/theorems/factors.pdf.