I came across this post about ABC conjecture. If you haven’t read it, you should it’s good writing.

Now somewhere at the start of the 2nd paragraph he states that if a and b are relatively prime(i.e, they **both** are not divisible by a common number n>1), then the sum c = a + b is also relatively prime to both a and b.

Now, my first reaction was obviously and skip it over, but then i asked myself why, and after some 30-40 mins of mental wrangling* proved to myself it is the case.

Here’s how:

Initially, just for surety, i asked is it true? So, i tried out some basic prime numbers, started with 2 and 3 went on to 2,3,5,8,13,21 etc.. and stopped and asked why?

I started of with wondering trying to figure out how the fact that we added those two numbers said something about their factors(i.e: a multiplicative operation). Then after some minutes of brain mashing, it hit me.

Ok, what exactly is relatively prime? it is not having a common clean divisor. now what does a clean divisor mean? It means the ability to express a number as a repeated addition of another (absolute??)prime number. Once i got that, it was straight forward algebra to work out the proof.

Relatively prime means, there is no number > 1 that can form all(in this case, both) the given numbers by repetitive addition.

So if you add relatively prime numbers, the resulting sum cannot but help being relatively prime to the added numbers. Because if it did it would mean that the sum has prime factors that is in common with the individual numbers prime factors, which is impossible since the individual numbers are relatively prime and prime numbers are non-factorizable by definition.

Ok, so far titling this post as category theory is outrageous, and far-fetching. I’ll explain my reasoning why, i was reminded of category theory, but am afraid it still might be far-fetched. Anyways, once i saw the relationship between multiplication and addition, my first recall from memory was groups. Numbers + operators(in this case addition and multiplication on ~~whole~~ natural numbers).

I’ll go on and try to explain the abc-conjecture now. or rather re-explain it. you really should read the original article(linked at the top).

To get to the conjecture, we need to define one more set/concept/number called Radical of a,b,c.

The way you find the Radical is you find all the prime factors of the given numbers(a,b,c with c=a+b, in this case) and then eliminate duplicates and finally multiply the rest.

To use the same operations, i used above, it is like the splitting the numbers into their prime factors.(i.e: smallest possible numbers, that can make up the given number by repetitive addition, and cannot themselves be made up by repetitive addition of other numbers).

Expressed that way, without the multiplication operator, a number like 24 can be expressed two ways.

i.e: 24 = 2 + 2+ 2+2+2+2+2+2 + 2 + 2 +2 +2 or

24 = 3 + 3 + 3 + 3 +3 +3 + 3 +3

So both 2 and 3 are prime factors of 24.

So, when you are given a set of numbers, you find all of it’s prime factors and then remove the duplicates(in the case of our a,b and c) there wouldn’t be.(relatively prime remember?) and then multiply them together.

Or rather repeatedly add the first factor as many times as the second and then the result as many times as the third and so on..

The abc-conjecture states that for any number h>1 there’s a finite set {a,b,c} that satisfies the relationship R raised to the power of h < c.

PS: Man it’s a pain to replace multiplication by repetitive addition. I tried to use only addition operation throughout that post and gave up towards the end when i came across exponentiation.

PPS: Also, i know i originally promised to write more about applications of category theory, but got distracted with this paper getting popular :-). I will try to get back the original quest next week. Cya all.

*– Rather rewarding experience in terms of the pleasure center stimulation it has resulted in.