Alternatively, if you dont want to check the cases mod 4, restricting b

I did the same and ended up with (a-4)(b-4)=8 or b = 8/(a-4) + 4

@@Skandalos Nice! That's equivalent to mine, but the decomposition into (a-4) and (b-4) makes all the possible cases super easy! I'm just usually bad at spotting polynomial decompositions

@@alucs6362 I find those factorizations always neat problems by themselves.

"Since a must be a natural number you can analyse the cases mod 4 ..."??? What? You got a = (4b-8)/(b-4) You can a = (4b-16+8)/(b-4) = 4+8/(b-4) The only possible values for b are given by b-4 b 1 5 2 6 4 8 8 12

Was looking for this, you need a multiple k also in N and a=2pqk; b= k(p^2-q^2); c= k(p^2+q^2)

P and q need not be integers here

Heron of Alexandria begs to differ ... but also consider the relationship between the area of a circle and its perimeter ... it's an integral/derivative thing, more generally an n D boundary sweeps out an n+1 D volume. but yeah area = perimeter is click-bait lol

Indeed a meter will never be a square meter

this means the magnitude of area and perimeter are same. he did not ue dimesions anywhere in the equations

Just consider the area of the rectangles extruded 1 unit outwards from the sides and the dimensions work out

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

Exactly. And the problem remains valid even for physicists who are unnecessarily concerned about preserving the units.

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

You should find some "better" things to bother you.

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

If p or q is rational but not integral, the side lengths of the triangle might not be integral, although of course you could multiply by a factor to fix that. If one of them is irrational, it might not be normalizable.

Yes, and by thinking this way, the equation becomes valid even for those physicists who complained about preserving the units.

It is a necessary-but-not-sufficient. I..e. primitive triples require p and q to be coprime, but many coprime p and q do not yield a primitive triple. It is however the case that any primitive triple is generated by some coprime p and q. There is a proof of this on the Wiki page for Pythagorean triples.

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

@@UltimaGaina So, all triangles with an incircle of radius 2 have the property that their area equals the perimeter. I didn't think of it that way, but of course you're right. Still, this is just another way of introducing a length scale into the problem.

Yes, I agree. How the devil did he obtain that parameterisation. Where is the discourse?

Don't the leg lengths of 5 and 12 also fit those equations?

@RexxSchneider Yes I forgot to include those 2.

ps: nevermind, I found the bug, after adding an additional factor k to Euclid's formula - so it also generates all non-primitive triples - the final formula becomes n (m - n) = 2k, which obviously has more solutions.

In geometry? IN ... GEOMETRY? Hahahaha.

@@samueldeandrade8535 What?

The possible values for n² (mod 10) are 0, 1, 4, 5, 6, 9 When n²=0 or n²=5 (mod 10), n is divisible by 5. So let's see when x²+y² can be a square z² (mod 10), with x and y NOT divisible by 5: 1 4 6 9 1 2× 4 5✓ 8× 6 7× 0✓ 2× 9 0✓ 3× 5✓ 8× In all possible cases, z is divisible by 5.

@ingiford175 - You have garbled something. A Pythagorean triple is still a Pythagorean triple if both of the smaller numbers are odd. It's the INPUTS going into Euclid's formula for FINDING Pythagorean triples that must be one even natural number and one odd natural number. When both of them are odd, all three of the natural numbers that are outputted by the formula/method will be even, and thus NOT a "primitive" Pythagorean triple, even though they will still satisfy the condition that the sum of the two smaller ones squared is equal to the largest one squared. But that's a condition for the inputs to the formula to work, i.e. to result in a primitive Pythagorean triple, not a condition that a triple must meet to be considered a primitive Pythagorean triple.

You are correct: to guarantee a unique primitive, 0 < q < p, gcd(p, q) = 1, and p - q is not cong to 0 mod 2. However, you can plug in any positive real values for p and q and still get a right triangle, such as p = sqrt(3) and q = 1 which yields a 2, 2sqrt(3), 4 right triangle. So long as p > q (otherwise you might get a leg with length 0) this always works. And you can plug in any natural numbers and still get a Pyth triple that may or may not be primitive. Here Michael is looking for primitives and non-primitives so no need to worry about the parity and divisibility. Again, the 0 < q < p remains critical as we want to avoid degenerate triangles and impossible lengths. If p and q are both odd, you get the same triangle as if you chose (p + q)/2 and (p - q)/2 except it's doubled in length. Ex. (3, 1) is the same as ((3 + 1)/2, (3 - 1)/2) or (2, 1) except sides are twice the length. If p and q are both even OR have any prime factors in common (say the gcd = s --- if they're both even then 2 divides into s), then it's a little more hairy to describe but if you divide p and q by s, you'll either have your preferred even and an odd OR you'll be in the above case.

@@topherthe11th23 I said primitive. if both generators are odd, then both a and b are even.

@@ingiford175 I am talking about Penn's definition of a primitive Pythagorean triple, given at 1:13 to 1:26. What you are talking about is where Penn said much later, at 2:57, that the numbers p and q that he could use to generate Pythagorean triples must be have no common factors, and then he did indeed forget the part where p and q must be an odd and an even natural number. In my error I was aided by what YOU SAID in your original comment. It didn't say that he omitted a 2nd condition required for the numbers p and q. Your original comment said that's a condition on Pythagorean triples to qualify them as primitive, and that's not true. Your later comment said it's an additional condition on the "generators", not the resulting Pythagorean triple itself, and that IS true. When I read your first comment it seemed to me that you must be talking about the two short sides of the right triangle having a requirement that one be odd and one be even. Maybe you could edit your original comment.

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

@@UltimaGaina no, a radius can't be"2". It has to be 2 Units, and an area has to be 2 units squared. THey can not be equal, ever.

@@lynnrathbun that's what I said: set r to 2 units of your liking, and you'll have to solve the same equation as proposed in the clip. Otherwise, rest assured that the teacher and the vast majority of his followers (probably 100%) are fully aware of this very basic rule everybody learned in the first elementary school physics lessons. You bring nothing new on the table.

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

Haha. Just shut up.

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

@@UltimaGaina Area and perimeter fundamentally can't have the same units. One is the square of the other. For example, if you picked meters, then the perimeter would be in meters and the area would be in meters^2. Different units.

@@stevenschmidt You didn't get my point. One more time: asume that the inradius is r=2m. And since the area A = r * P/2 we have A (m^2)= 1/2 * P (m) × 2m Both sides are now expressed in m^2 and you must solve exactly the same equation without having any reason to complain about units.

@@UltimaGaina The whole point was to find a triangle where the Perimeter (which has units of "meters") is equal to the Area (which has units of meters^2). I understand that ignoring the units you can do that, but with units they are fundamentally different. Your equation doesn't show that. It does relate the perimeter to the area, but by multiplying by the inradius -- which is how the meters becomes meters^2. Yes, your equation has meters^2 on both sides, because A = r*P/2, so on the left A has units of m^2, and on the right we have r (with units of 'm') multiplied by P/2 (which has units of 'm'), making m^2 on the right. But like I said, this still doesn't show that P = A, because P is in units of meters while A is in units of m^2. I feel like I'm just repeating myself over and over now.

@@stevenschmidt Obviously, I didn't manage to make my point clear. One more try: Basically all triangles with an inradius equal to 2 units have the area NUMERICALLY equal to the perimeter. He could have simply stated the problem as follows: "Find the length (in meters) of the sides of a right triangle, if its inradius is equal to 2 (in meters)", and all sides are natural numbers (in meters)." You would be forced to solve exactly the same equation without having any reason to complain about the units integrity.

p and q is the standard parametrisation for Pythagorean triples. Take p and q arbitrary integers, p>q. Then a=2pq, b=p²-q², c=p²+q² is a Pythagorean triple. (Check that.) One can obtain all irreducible Pythagorean triples ("irreducible" means that a, b, c do not have a common factor) in this way, and for that one does not even need all possible pairs (p, q), but only those which do not have a common factor and where either p or q is even. (Actually, if these conditions on p and q aren't fulfilled, the Pythagorean triple obtained is not irreducible.)

@@user-gd9vc3wq2h Ahhhh! Right! Thank yo for being so kind. Now I can follow like I wished. Thanks man :) A 3,4,5 is a triple :).....I see!

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

There are no units this is math

Even if there were no units, the original commenter is right because without mentioning numerical value, it wouldn't make sense to have an area (space) equal to a perimeter (length). Not bashing on the video though

@@MushookieMan Your comment is ridiculous. There is no way to conceive of a triangle as having a length instead of an area, and no way to conceive of a perimeter as having an area instead of a length. The result that is proven in this video is that the number of units of length of the perimeter is equal to the number of units of area of the triangle, where one unit of area of the triangle is the SQUARE of the unit of length.

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

@@UltimaGaina A is something in square units. R is in linear units. P is in linear units. When you multiply the linear units of r and the linear units of P on the right, you will get something in SQUARE units. So far so good. Square units on both sides. But then how do we get from there to A = P x 1? If r equals 2 units (meaning, r doesn't equal 2, but, rather, r equals 2 units) then when we divide r by 2 we get 1 UNIT. So what you SHOULD have said is A square units is equal to P linear units times 1 linear unit, NOT the same as P linear units times 1.

Technical right. I assumed the "magnitude of the area" and the "magnitude of the perimeter" was implied.

Calm down dude. It's just a slip of the tongue. He OBVIOUSLY meant *the numerical value of the area* and the *numerical value* of the perimeter. You don't need to write an angry treatise for such a trivial thing.

@@RafaelSCalsaverini My "treatise" wasn't even close to angry. I'd say it was just a statement of the facts. Given that I'm responding to a PhD and faculty-member (who should therefore know better), I'd say it's a DEFIANT statement of the facts, because a person must advocate REALLY HARD for their opinion when it's against someone with actual credentials (because I'm anticipating a lot of push-back and pre-responding to it so I don't have to come back). But "angry"? No. "Firm"? Yes.

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units

@@UltimaGaina You end up with A square units = r units times P/2 units. So far so good, you have square units on both sides. But you can't get from there to A square units = P units. It's P units times 1 unit, so that on the right there would also still be square units. Professor Penn should have said "Find a right triangle that has an area "A". "A" divided by the unit of area is no longer an area, but is just a natural number, "x". The triangle has a perimeter "P". "P" divided by the unit of length is no longer a length, but is just a natural number "y". Find the triangles where the natural numbers "x" and "y" are the same.

Whats your point? Nobody is allowed to do math because they like it? That everything MUST be applied math? Without these, according to you useless, exercises we would not have most of the math that gets applied. Like the entire branch of Number Theory was in the realm of pure math right up until the late 20th century when computers became common and things like digital cryptography came about and use a lot number theory.

Yeah the numerical values of length and area were the things equated in the video. I get your point though 👍🏼

Respectfully, while maths does have practical applications, maths isn't centered around practicality. I mean, those Olympiad number theory and geometry problems serve more as brain ticklers and food for thought than a practical application. So I don't think "What practical application can this have?" is the right question to ask. That said, I once again recognise the importance of mentioning that the numerical values are the things to be compared, which I believe was the point you were trying to make.

Application: having fun. Tax dollars go to mathematicians so we can have fun discovering, and every once in a while something useful pops up. Not this time though, thanks for the funding. I can't believe that 3^2+4^2=5^2 and 3,4,5 are consecutive. Every day we are graced by the coincidences that small numbers grant us. We should be thankful for universal truths like this.

Just asume that the inradius is r=2 using the units you prefer. And since the area A = r * P/2 we have A = P × 1 without messing up the units