( In the middle of preparations for my 4th semester Endsems right now ! So a lighter blog today )
Cute Little Equation
I was burnt out after doing Analog Electronics for 4 hours in one straight sitting, so after touching grass and going on youtube, I found this :-
which was a philosophical way of thinking about Maxwell's equation, but that's not what this short blog is about, it about this :-
I mean if you like math and if you see something like this, you'll surely give a try - at least mentally. And when you think you're really close but you're brain power isn't that of a genius, you start doing it on a paper. Solving this led me to few other interesting results, nothing of particular value but really fun to play with.
First Intuition
So clearly :-
$$13(x+y) = xy$$
Now how to ensure only positive integers ? I thought about reducing it just 1D graph, I thought should I look it only from one variable perspective, which is sought of the key. I played around by plugging in numbers. What this equation is saying is :-
so you just plug in values and try to catch any pattern ;
Getting Closer
I plugged y = 20, and observed x :-
$$13\cdot(x+20) = x\cdot 20$$
$$13x + 13 \times 20 = 20x$$
$$13 \times 20 = ( 20 - 13 )x$$
$$x = \frac{13 \times 20}{20-13}$$
Which is wrong, but now if you reverse engineer ( wherever you see 20, put back the y )
$$x = \frac{13 y}{y-13}$$
This is really great ! Since for x to be some positive integer,
$$\frac{y}{y-13}$$
the above term has to be a positive integer integer. This is the key term we need to look at. Now that means that :-
$$y = 13k$$
which would make the fraction :-
$$\frac{13k}{13k-13}$$
$$\frac{k}{k-1}$$
$$\implies k = 2$$
Since no other values of k, can make the fraction an integer.
$$\implies y = 13k = 13\times 2 = 26$$
$$\implies x = 13\left( \frac{y}{y-13} \right) = 13 \left( \frac{26}{13}\right) = 26$$
I feel Stupid
So the answer for our question
$$\frac{1}{13} = \frac{1}{x} + \frac{1}{y}$$
is just (26,26) , that is :-
$$\frac{1}{13} = \frac{1}{26} + \frac{1}{26}$$
And this is kind of embarrassing since throughout high school while dealing with resistors of equal resistance in parallel, the equivalent resistance which we get from :-
$$\frac{1}{R_{eff}} = \frac{1}{R} + \frac{1}{R}$$
turns out to be :-
$$\frac{1}{R_{eff}} = \frac{2}{R}$$
Which is exactly what happened here.
$$\frac{1}{13} = \frac{1}{26} + \frac{1}{26} = \frac{2}{26}$$
What I am trying to point out is :-
$$\frac{1}{\text{any number}} = \frac{1}{\text{twice the number}} + \frac{1}{\text{twice the number}}$$
Generalizing this Idea
Now I think I can pose a better question :-
Given
$$A = \begin{bmatrix} a_{1}& \\ a_{2}& \\ \vdots & \\ a_{n} & \end{bmatrix}$$
$$\frac{1}{73939133} = \sum_{i=1}^{n} \frac{1}{a_{i}}$$
Find the vector A such that all components of A are positive integers. It can be proven that such a solution always exists UNIQUELY.
Fun fact, the number I chose might seem arbitrary but it has some really interesting property not related to the problem though. ( Check it out here ( Flammable Maths YT Channel ) )
Hint
we know
$$\frac{1}{\text{any number}} = \frac{1}{\text{twice the number}} + \frac{1}{\text{twice the number}}$$
that is :-
$$\frac{1}{x} = \frac{1}{2x} + \frac{1}{2x}$$
Now if we extend it
$$\frac{1}{x} = \frac{1}{3x} + \frac{1}{3x} + \frac{1}{3x}$$
And that's how you generate it !
I got Analog Endsem Examination in one week, and I spend one and a half hours looking at some random math equation and writing a blog about it. Now it's time to have the next 4 hour sitting sigh.
Thank you,
__CPP_Try_Hard__ ;