Pythagoras weird proof?

[hard anvil]

I always had this kinda silly idea: is it possible to prove Pythagoras theorem from a purely "analytical" view without (or almost without) any involvation of geometry?

My idea is that, let f(a, b)=c denote the function to calculate hypotenuse from cathetes (i.e. sqrt(a^2+b^2)). It then must follow:
symmetry f(a, b)=f(b, a)
scaling similarity f(n*a, n*b)=n*f(a, b)=n*c
dimensional analysis, that given inputs in R the result maps to R and not R^2 or R^3 or smt else weird
Maybe satisfy one specific case of f(3, 4)=5

[elfin cipher]

Yes there exists a sort of dimensional proof of the theorem. I found one in a physics book written in French.

[hard anvil]

do you maybe have the name of the book?

[elfin cipher]

I will try to find it again. The page https://math.stackexchange.com/questions/43398/prove-pythagoras-theorem-through-dimensional-analysis also deals with the subject.

[hard anvil]

tysm!

[elfin cipher]

Le livre qu'il vous faut pour comprendre la mécanique by Brahim Lamine.

The proof in this book is the same as in https://scienceetonnante.com/blog/2014/02/03/la-plus-belle-demonstration-du-theoreme-de-pythagore

The last proof of the theorem I heard about is https://youtube.com/watch?v=p6j2nZKwf20

[hard anvil]

thank you!
the images are outdated but I do understand what they meant. However I believe they did use that the 2 triangles cut by the height to hypotenuse are similar so that they both have angle theta, which involves a bit geometry. But it's nevertheless a fascinating proof!

[hard anvil]

I've seen this before haha

[hard anvil]

wait I actually think it can be improved to throw away the angle theta part entirely, as it must be symmetrical (a and b can be exchanged and equation still hold)

so assume that one of the smaller triangles cut by the height does not contain angle theta and f(omega) produces something else, then switching a and b will produce a different c and thus make the equation not necessarily hold

[elfin cipher]

Meaning the height does not fall from a right angle?

[hard anvil]

the last part says:
C^2*f(theta)=A^2*f(theta)+B^2*f(theta)
they assumed that both triangles contain an angle theta, which is really obtained from that the two triangles are similar

My argment is that if f(theta) and let's say the function with some other angle f(omega) are not equal, then the symmetry in the expression is no longer present

so it must be that f(theta)=f(omega)

because we know that the two cathetes are equivalent, switching b with a and a with b should produce identical results

and the expression should be symmetrical

so if we treat f(theta) and f(omega) in C^2*f(theta)=A^2*f(theta)+B^2*f(omega) purely as coefficients that have nothing to do with angles or anything with geometry at all, we can still find that f(theta)=f(omega), as A^2*f(theta)+B^2*f(omega)=B^2*f(theta)+A^2*f(omega)