Ehhh, I won’t be impressed until someone - anyone! - solves the 3n + 1 problem. 
There are a lot of proofs of the Pythagorean theorem, and in spite of the theorem being over 2,000 years old, they keep coming (President Garfield came up with one) - but coming up with a new one (if these students have) is still remarkable and notable. It’s possible that they went down a path that others didn’t because it seemed obvious that there was nothing new in that area of basic trig - in which case, all credit to them for giving it a try, and finding something.
I think the issue here is the terrible reporting of whatever it is they did. Nothing against the kids, but I’ve seen numerous articles and none of them make any sense.
Quality of STEM reporting in the mass-market press is consistently terrible.
Reporting is fine, but the way it should work is, you give a few lectures, presentations, whatever (not sure by what process “STEM reporters” show up at some random conference, though), and as for a formal proof (which is surely worth writing down if people are so interested and you did something non-trivial or important), you can submit it to an AMS publication (or whatever) and/or just post it online. Therefore it should be easy to answer exactly exactly what they did.
I will tangentially add that Euclidean geometry is a well-established playground for automated theorem provers and you can formally verify a proof and tinker with the axioms to see what is independent of what.
Aside: One pastime of mine is, whenever I encounter a new proof of the Pythagorean Theorem, to search through it to find where the author sneaks in the Fifth Postulate or something equivalent to it. It’s usually not explicit, but it’s always lurking in there somewhere, because the Pythagorean Theorem is not valid in curved spaces, even though some curved spaces are just as self-consistent as Euclidean space.
They basically did what you think they should have done.
On March 18 they gave a presentation at AMS - 2023 Spring Sectional Meeting in a Special Session on Undergraduate Mathematics.
I found that early on but it’s just the abstract, which still doesn’t answer my question. Apparently the paper itself is not freely/publicly available.
I wasn’t replying to you, but there is only one question in your OP.
And that question is indeed answered by the abstract.
… in our lecture we present a new proof of Pythagoras’s Theorem which is based on a fundamental result in trigonometry—the Law of Sines—and we show that the proof is independent of the Pythagorean trig identity \sin^2x + \cos^2x = 1.
I guess I am just saying the tautological fact that if/when they post a formal proof or at least a copy of the slides, then we will know “exactly what they did”, and if not (why wouldn’t they?) then… not. Certainly a proof of the Pythagorean identity should not assume the Pythagorean identity, but I’m not sure what “independent” is supposed to mean. The Law of Sines (and Pythagorean Theorem) also work in spherical or hyperbolic geometries if formulated correctly, by the way.
But they have submitted it to the AMS where it was presumably peer reviewed and found not only to be correct but mathematically interesting. They wouldn’t have been given a speaking slot at that conference without at least that.
So I’m going to assume their claims are at a minimum mathematically valid. There is a chance, maybe a good chance, that it will turn out that their work is not original, or at least original enough. There’s a lot of mathematical stuff that has been been published and even a diligent search could have failed to turn up something somewhere that presented essentially the same argument.
I forgot to add… the proceedings of AMS conferences are typically published so it’s not like they’re trying to hide something. The paper will be available eventually. Some things take time.
No, it hasn’t gone through peer review.
They gave a presentation during a special session on undergraduate mathematics at a local chapter meeting and the director of the meeting said the results had not yet been peer reviewed.
The school is apparently helping them prepare for review but it hasn’t happened yet.
Somebody from the AMS looked at it before awarding them a speaking slot at that conference.
I’m 100% certain it is a valid, trigonometric proof of the Pythagorean theorem.
I’m significantly less certain that it is the first such proof.
The abstract just says what they didn’t do, that the proof is independent of the Pythagorean identity. It doesn’t say how they did it. I’m not saying they didn’t really do it, but using the abstract to answer my question begs the question. Ultimately I am curious about what reasoning they developed that everyone else has missed.
I does say how they did it. They used the Law of Sines. That’s not something they didn’t do. It’s something they did. They derived the Pythagorean Theorem from the Law of Sines.
You seem to want details or the whole proof, which is fine, but that’s not what you asked for.
They have new proof of the Pythagorean Theorem based on a fundamental result of trigonometry. That’s not what they didn’t do. It’s what they did do.
Saying that they did it using the Law of Sines is like saying that you make a soufflé with eggs. I would be interested in the whole proof, or at least an informal outline of the strategy.
That’s what I said.
Ok, I think I figured it out (or at least something pretty close to what they presented).
You can see some of their slides in various news articles and in particular you can see a diagram that looks like this…
Start with a right triangle (upper left) with legs of lengths a and b, hypotenuse of length c, and non-right angles \alpha opposite a and \beta opposite b.
Then reflect that triangle across the b leg and create a larger (also right) triangle with sides length d and e by drawing a line perpendicular to the hypotenuse of the first triangle at the vertex of \beta and extending the hypotenuse of the reflected triangle to the point where these two meet.
In the diagram this is the largest right triangle with legs length c and d and hypotenuse length e.
This large triangle contains the original triangle, its reflection, and another triangle that we can partition into an infinite sequence of triangles drawing perpendicular lines as show in the diagram. The important fact to note here is that each triangle in this infinite sequence is similar to the original triangle (we get this from angle angle similarity applied over and over).
Because of this similarity we can compute the lengths of the sides of the infinite sequence of triangles in terms of a, b, and c and in doing so we can express lengths d and e as infinite sums.
I’m leaving out some algebra here but it turns out that both these series are geometric so we can express their sums in closed form.
d = \frac{2ac}{b} + \frac{2a^3c}{b^3} + \frac{2a^5c}{b^5} + ... = \frac{2abc}{b^2-a^2}
e = c + \frac{2a^2c}{b^2} + \frac{2a^4c}{b^4} + \frac{2a^6c}{b^6} + ... = \frac{a^2c+b^2c}{b^2-a^2}
These are both extremely nice, and the universe is not done smiling on us because the large triangle is a right triangle and thus:
\sin(2\alpha) = \frac{d}{e} = \frac{2ab}{a^2+b^2},
which is exceedingly nice.
Now we’re at the part where the law of sines comes in.
Consider the triangle formed by the original triangle and its reflection. From the law of sines we have:
\frac{\sin(2\alpha)}{2a} = \frac{\sin\beta}{c}.
Since \sin\beta = \frac{b}{c} we can express \sin(2\alpha) in a different way,
\sin(2\alpha) = \frac{2ab}{c^2}.
Now we have two expressions for \sin(2\alpha) so we can set them equal to each other.
\frac{2ab}{a^2+b^2} = \frac{2ab}{c^2}
And since the numerators are equal to each other, the denominators must be as well which gives, a^2+b^2=c^2, as desired.
You’re a hero.
Is the solution novel?