Prime

[cinelli]

This is a mathematical challenge. Find prime number p which satisfies the following equation:

arctan(1/2) + arctan(1/5) + arctan(1/9) + arctan(1/p) = pi/4

Cinelli

[9873210]

Are you sure you asked the right question? It doesn't seem very interesting as is.

Using the single valued definition of arctan this will have a single real answer that is easily evaluated numerically. It will either be integer or not, and prime or not.

It turns out to be approximately 73.000... or about 73 which is both integer and prime so if there is an answer that's it.

It's more interesting if you wanted proof that there is an answer. I.e. that 73.000... is exactly 73.

We can do this using the identity tan(x+y) = (tan(x)+tan(y))/(1-tan(x)tan(y))

let
θ_n = arctan(1/n)
so
tan(θ_n) = 1/n

Then
tan(θ_2 + θ_5) = (1/2+1/5)/(1-1/2*1/5) = 7/9
tan(θ_9 + θ_73) = (1/9+1/73)/(1-1/9*1/73) = 82/656
and
tan((θ_2 + θ_5) + (θ_9 + θ_73))=(7/9+82/656)/(1-7/9*82/656)=(7*656+82*9)/(9*656-7*82)=5330/5330=1

so
tan(θ_2 + θ_5+θ_9 + θ_73) = 1
θ_2 + θ_5+θ_9 + θ_73 = arctan(1) = π/4

[cinelli]

This seems to me an odd approach - find the answer with a calculator then show that it satisfies the equation. Why not solve the given equation for p using the tan(a+b) identity several times, or if you prefer, the tan(a+b+c+d) identity just once?

Cinelli

[9873210]

This seems to me an odd approach - find the answer with a calculator then show that it satisfies the equation. Why not solve the given equation for p using the tan(a+b) identity several times, or if you prefer, the tan(a+b+c+d) identity just once?

Cinelli

It's how you (or at least I) solve problems when I don't know the answer beforehand. Consider a few different approaches and start with the one that looks most promising. Sometimes it works, sometimes you learn something that leads to a different approach and sometimes you hit a dead end and have to regroup. The polished solution that makes it into a textbook is usually a much later version of the solution that has been tweaked many times.

In this case using a calculator to get a closed numerical solution (ignoring how the calculator computes arctan) is obvious, easy and solves the problem as actually worded. In almost any field (excluding pure math) the numeric solution is complete.

[MaraMan]

Interesting In our Time program on Radio 4 this morning about the great mathematician Paul Erdos

MM

[mc2fool]

Hmmm......

Me: Find prime number p which satisfies the following equation arctan(1/2) + arctan(1/5) + arctan(1/9) + arctan(1/p) = π/4
ChatGPT: The number p = 12,391 is a prime number that satisfies the equation of arctan(1/2) + arctan(1/5) + arctan(1/9) + arctan(1/p) = π/4.

Me: please show workings
ChatGPT: Sure, I can show the calculations. The fraction of π/4 is equal to 2*arctan(1/2) + arctan(1/3) -2*arctan(1/7). By rearranging, arctan(1/p) = (π/4 - 2*arctan(1/2) - arctan(1/3) + 2*arctan(1/7)). Solving for p gives p=12391.

....hmmmm.