finding ALL pythagorean triples (solutions to a^2+b^2=c^2)

How to generate ALL the Pythagorean Triples, namely a^2+b^2=c^2 where a, b, c, are whole numbers? Here we will find all the solutions to a^2+b^2=c^2. This is a very classic number theory question and it is very suitable for high school algebra and geometry students. A very similar video is "Solutions to x^y=y^x" check it out here: 👉 • Solutions to x^y=y^x
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[@lambdamax]

"We're all adults now"

[@blackpenredpen]

Yes!

[@OonHan]

nah

[@Davidamp]

Oon Han lmaoo

[@orcishh]

Im 15 lol

[@rohunse5555]

However, I disagree! (Like if you get the reference)

[@hdthor]

This can be done easier (solved in 1 line) by using midpoints: rewrite it as (m+x)^2 - (m-x)^2 = a^2, which is a^2 = 4mx. QED. So any time 4 * midpoint * (distance from the midpoint) multiply to a square, it forms a Pythagorean triple. E.g. 16 = 4*4*1, so sqrt(16),(4-1),(4+1) is a Pythagorean triple: 4,3,5. Another: 15^2 = 225 = 4*(15/2)*(15/2), and we can move factors of 15 around to: 4*(3*5*3/2)*(5/2), so 15,(45/2 - 2.5),(45/2 + 2.5) is a triple: 15,20,25. Just take any square a^2, and write it as 4*(a/2)*(a/2) and then move any factors of a from x to m. This generator makes it easy to exhaustively do all triples for a particular a^2 before moving on to the next square.

[@pappapez]

Thank you! This is wonderful, and you are a wonderful teacher!

[@blackpenredpen]

Thanks!

[@antimatter2376]

I always love your videos. Always so clear and doesn't just tell you the answer. Really like the format.

[@blackpenredpen]

I am glad to hear!!

[@yrcmurthy8323]

Really agree it

[@valeriobertoncello1809]

this.

[@Mia-tz9le]

I got lost after the third sentence

[@عبدالواحدالمجراوي]

هذا رائع جدا استفدت من هذا الفيديو

[@camerongray7767]

“We are all adults now” WOT! I’m 16! Lol

[@blackpenredpen]

You're a math adult tho!

[@camerongray7767]

blackpenredpen I learn from the best

[@pmj9925]

@@camerongray7767 now you are adult

[@camerongray7767]

Oh shit I am 19

[@frzv12]

Now you are 21

[@threepointonefour607]

My method of instantly finishing Pythagorean triples (very limited) Let a be some odd number greater than 1. Then b and c are the 2 consecutive numbers that add to a^2. In other words, b=(a^2-1)/2 and c=(a^2+1)/2 For example: Let a=3 Then b = (3^2-1)/2 = 4 and c = (3^2+1)/2 = 5 It's the 3,4,5 triple. It also works for 5,12,13, since 12 and 13 are the consecutive numbers that add to 5^2 = 25. 7,24,25 and 9,40,41 and so on also work similarly. Short proof: Given: n^2 = m + (m+1) n^2 = 2m+1 n^2 + m^2 = m^2 + 2m + 1 n^2 + m^2 = (m+1)^2 It is limited but pretty fast.

[@dmytro_shum]

Simple, but bounded method, because it has 1 parameter only. It has 2 parameters in the video - so it covers all the existing triples (at least, I hope that all =D).

[@BigDBrian]

Yup, I've known about this for a long time now, it's great, though the more generalised version is better (but less simple), of course.

[@Kantharr]

I found one that's similar to that which can cover all pythaogrean triplets given a value A and a value x that shares the same parity as A. For example, using the formula b=(a^2-x^2)/(2x) and c=(a^2+x^2)/(2x) you can use a=15 and x=1 to get the pythagorean triplet 15,112,113. You can also use x=3 to get 15,36,39. The equation b=(a^2-x^2)/(2x) also works with even numbers. For example, with a=8 and x=2 you get the pythagorean triplet 8, 15, 17. With x=4 you also get the pythagorean triplet 8, 6, 10.

[@Kitnitisgame]

When I was about 16-17 years old, or 4-5 years ago, I discovered the way to generate 'rational' Pythagorean triples. This was what I did... I observed that... 3²+4²=5² >> 3²=(4+5)(1) = (4+5)(5-4) 8²+15²=17² >> 8²=(15+17)(2) = (15+17)(17-15) From the pattern, we can see that... a²+b²=c² >> a²=(b+c)(c-b)=(c+b)(c-b) --(1) Also, we can see that c-b is the difference of the sides b and c; therefore I let... c-b=d, i.e. c=b+d --(2) Substituting (2) into (1); a²=((b+d)+b)((b+d)-b) a²=(2b+d)d Now I now get the pattern, which is: a²=d(2b+d) --(*) Let's generate an example.. Let a=26, with the difference between the other sides of 7 (d=7). Entering the values a and d into (*); 26²=7(2b+7) Solve the equation for b and we get b=627/14. Substituting b into (2), we get c=725/14. Now we get a set of rational triples: (a, b, c) = (26, 627/14, 725/14). However, if you want the triples to be all whole numbers, I will multiply all sides by the denominator of non-whole numbers (14) to all sides by proportion: (26×14, (627/14)×14, (725/14)×14) Now we successfully generate a new set of Pythagorean triples in all whole numbers, which is (a, b, c) = (364, 627, 725). That means 364²+627²=725².

[@mannyheffley9551]

Nice one! What I once figured out or probably saw somewhere (because I don't find myself worthy enough to generate ingenious mathematical results) Some Pythagorean triplets are m^2-1,2m,m^2+1.

[@dipanwitanandi4274]

I also discovered this similar tactic too . Also I also did it while I was 16 years old .

[@juniorstudent5950]

Bro When I was in 8th standard our teacher taught use how to generate pythagoras triplets

[@oathbringer7614]

bro i was ez doin it when i was 2 years old

[@cartatowegs5080]

@ཀཱ I was doing this during impregnation

[@-a5624]

"wouldn't it be nice if we had a generator..." me: *already opening pycharm*

[@supercool1312]

- A i did it in regular python for all triples up to 100 code: For a in range (1,100): for b in range(1,100): for c in range(1.100): if (a**2+b**2==c**2): if c>b and c>a: print: (a, b, c) that prints a list of every pythagorean triple up to 84, 35, 91

[@tpthpt5973]

@@supercool1312 I think we do not need to check condition "if c>b and c>a".

[@nwoyeezekiel320]

@@supercool1312 you are blessed :)

[@PankajKumar-os3tm]

@@supercool1312 Your code is correct but that will give one pythagorean triplet 2 times for eg it will give 3,4,5 as well as 4,3,5 both This code can help with it for a in range(1,100): for b in range(1,100): for c in range(1,100): if ((a**2)+(b**2)==(c**2)): if a

[@rdius4822]

​@@supercool1312​ time complexity goes brrrrrrrrrr

[@trueriver1950]

I used this result a lot when I taught relativity to physics students. Why? In special relativity two quantities are often used, beta and gamma. beta is v/c, or the fraction the speed of light at which the other reference frame is moving. gamma can be defined by beta^2 + 1/gamma^2 = 1 (though students are usually taught a formula which is harder to remember) It turns out that Pythagorean triples turn up in an unusual way (because of the 1/gamma^2) but it can still be used, and actually IS often used by professors setting homework. I know because I was one A clock on a spaceship travelling at 3 4/5 c is observed to run at 3/5 of its proper rate. Notice the 3,4,5 Triangle? A 1kg weight is observed to have a mass of 1.25 kg when observed from a spaceship zooming past at 0.6 c (Those numbers as fractions are 5/4 and 3/5) Using Pythagoras answer the following: A stationary twin ages 25yrs: how much does her travelling twin age travelling at 0.28c? (Hint think 0.28 as a fraction=7/25: 7,24,25 She ages 24 years. When I was teaching Relativity for a UK University I used to encourage students to memorize the following triangles which are popular with Relativity exam setters 3,4,5 5,12,13 7,24,25 and I taught the general rule that if any fraction in a special relativity paper contains consecutive integers, add them and take the square root to find the smallest number in the Pythagorean Triple. That number forms a fraction with the bigger of the previous numbers to give you the ratio you need. In exams (but not in real life) they usually come out as ratios of integers under 20. I once set a not-for-grading quiz q where a cosmic ray was moving at (112/113).c An incoming particle had an observed mass of 113 keV, what would its rest mass have been? Almost all the class got the answer in under a second Even if you don't know the physics, knowing Pythagoras applies you might instantly say an integer number of keV (go on post a guess... it's an integer below 20) Of course I advised students to show conventional working in assessed work, because in Physics most of the marks come from showing the understanding of the physics and displaying knowledge of the usual equations, not showing off a crafty shortcut ;) Even so knowing the result you are working towards can give you an edge, and serves to double check the result when you get it. And to come back to this video, I deduced the generator formula roughly the way demonstrated here when I first started setting questions myself...

[@Homelander6341]

Wow

[@vrabiealexandru2755]

we need more people like you

[@williamperezhernandez7331]

Beta = 3/5 and 4/5 are frequently used because gamma results in easy fractions, no radicals. Beta=3/5 gives gamma=5/4, beta=4/5 gives gamma = 5/3.

[@That_One_Guy...]

My teacher used the pytaghorean identity (sin^2 = 1-cos^2 or cos^2 = 1 - sin^2) in order to make analogous similarity to a formula in relativity (i forget which formula)

[@pyros6139]

I remember noticing that when I was taking a class that touched on relativity.

[@notspaso6644]

A Number Theory video by blackpenredpen, i feel like a kid on Christmas day! #YAY

[@diegofigueroa8307]

"We're all adults now" yo I literally turn 18 yesterday and this is the first video from him that I see since then

[@flowerwithamachinegun2692]

No, I won't allow you to write the m^2 first!!!

[@tomasbeltran04050]

What?

[@ffggddss]

Very nice! This formula can also be derived using complex numbers, z, and the fact that |z₁| |z₂| = |z₁z₂| So you can go: |z| |z| = |z|² = |x + yi|² = x² + y² = |z·z| = |z²| = |(x + yi)²| = |x²-y² + 2xyi| Squaring both "ends," (x²+y²)² = (x²-y²)² + (2xy)² And then, just identify a = x²-y² b = 2xy c = x²+y² and you have a general PT. ⧠ Graphically, on a complex plot (Argand diagram), you can draw z = x + yi for positive integers x>y; and then its square, Z = z² = x²-y² + 2xyi = a + bi will make an integer right triangle with real & imaginary components as its legs, and the "radius" as its hypotenuse. Fred

[@CousinoMacul]

OMG, I was just reading the Wikipedia article on Euclid's formula for Pythagorean triples earlier today.

[@sugarfrosted2005]

Psh, I can generate all the solutions for all higher exponent variants.

[@scathiebaby]

You must be Andrew Wiles.

[@mitchellgerrard4664]

Girlfriend picked a nice new Mic. great video my dude keep up the great work.

[@blackpenredpen]

Thank you for your nice comment!

[@shre6619]

I miss the old mike

[@nandeesh2ninad]

The pokeball mic is the best 😆😆

[@thesinglemathnerd]

I got the same formulas a different way. Suppose you have a complex number a+bi, where a and b are integers that are chosen freely. When you plot a number in the complex plane you can draw a line from that point to the real axis perpendicular and another line straight to the number 0 from the number a+bi, thus forming a right triangle. By the Pythagorean theorem, this means that the line from a+bi to 0 is of course of length √(a^2+b^2), which I may refer to as r. Given the original condition that a and b must be integers, then r^2 must always be an integer because r^2=a^2+b^2. This is important for later. If we were to square the number a+bi we get a^2+2abi-b^2. For simplicity I will write it as a^2-b^2+2abi to keep real and imaginary parts separate. In our original number a+bi, a was of course the real part and b was of course the imaginary part. To get the length of the line from this point (a+bi)^2 to 0, we have to take (a^2-b^2)^2 (real part)^2 + (2ab)^2 (imaginary part)^2 and square root it. This yields that the length of this line is √((a^4-2a^2b^2+b^4)+(4a^2b^2)). Combining like terms reveals that the length is also equal to √(a^4+2a^2b^2+b^4). The inside factors to (a^2+b^2)^2. The ^2 and the √ cancel each other revealing that the length of the line from (a+bi)^2 to 0 is a^2+b^2 which, from before, is r^2, which, remember, is an integer. This is where it all comes together. So after all this (showing that squaring a+bi and r yields that a, b, and r are integers and that complex numbers can form a right triangle when plotted in the complex plane), we have a Pythagorean triple (since I have used a, b, and c I will use x, y, and z) in the form of x=a^2+b^2, y=2ab, and z=a^2+b^2. If I explained anything to vaguely or incorrectly please let me know.

[@soufian2733]

9:15 so shouldn’t you say a = k * mn b = k * (m^2 - n^2) c = k * (m^2 + n^2) with k a whole number?

[@edvink8766]

You can but not necessary since it is trivial that if (a,b,c) is a pythagorean triple then (ka,kb,kc) is as well.

[@helloitsme7553]

With k=1 you already have all answers since if you take m=kb and n=kc then you get the same results

[@prabhakarmishra9042]

we don't need to write k... because m and n are variable..

[@prabhakarmishra9042]

@@helloitsme7553 you have explained in better way

[@paddeytpaddeyoutube4695]

Question: when you have (2n-1)²+(2n²-2n)²=(2n²-2n+1)² you can generate infinite a²+b²=c² with whole numbers right?

[@angel-ig]

Right. This is a clear demostration of why there are infinite pythagorean triples :)

[@paddeytpaddeyoutube4695]

@@angel-ig okay😁👍🏻

[@sahibakaur2930]

Amazing! I have seen the way to get this same result using a different number theory approach where we realize that exactly one of the legs is even and the hypotenuse and the other leg is odd, then some algebraic and number theory manipulations, and that way was a lot longer than this! Once again, thank you! I love seeing multiple ways to solve the same problem :)

[@whebon7266]

You can expand the generator by adding a third variable k. a = 2kmn b= k(m^2-n^2) c = k(m^2+n^2) But this still doesn't include all the possible triples. ({3, 0, 3} is excluded for example) And allowing k to be a rational number introduces new problems...

[@frabol02]

a, b and c should be all different

[@helloitsme7553]

K should be whole since you want a b and c to be whole as well

[@HeyItzMeDawg]

It's a nice triples generator but to be complete you should include multiples of a, b, and c: a = k(2mn) b = k(m^2 - n^2) c = k(m^2 + n^2) For instance, if you double the traditional 3,4,5 solution you get 6^2 + 8^2 = 10^2, which is also solution but cannot be generated with m,n as whole numbers. 3,4,5 is generated by m,n as 1,2, but neither 2,2 nor 1,4 generate 6,8,12 - the former generates the trivial 0,8,8 and the latter generates 8,15,17. This issues because the "easiest solution" for resolving the c/a and b/a fractions is, as you explain at 9:00, not the only solution; in fact it works for all multiples of k on the top and k on the bottom as well.

[@ChadTanker]

My Pythagorean Triple: a= Some Komplex Number b= Synatx Error c= Somehow Infinity

[@ujjwalmk]

It took me 12 minutes to realise that we wrote Pythagoras theorem in terms of Pythagoras theorem but of different variables

[@widscience2373]

widodo's conjecture (method and equations) to find ALL permutation of pythagorean triples, see the video : kzbin.info/www/bejne/bKXbfKGtpc10gsk

[@kripalbabaria1235]

I will steal your knowledge 🏃🏻‍♂️🏃🏻‍♂️🏃🏻‍♂️🏃🏻‍♂️

[@Rogue_Art]

Here are my implementations of the *algorithm* in the video *in Python, Java, and Javascript* just to help out those in the comments. *Python:* for m in range(1, 10): for n in range(1, 10): a = 2 * m * n b = m**2 - n**2 c = m**2 + n**2 if b

[@nandeesh2ninad]

Could only understand the python one...🤔🤔🤔🤔🤔🤔🤔

[@davidbrisbane7206]

Cool

[@PraneshPyaraShrestha]

i am already opening my Jupyter notebook 😁

[@SuperDeadparrot]

3Blue1Brown also does an excellent video on this topic.

[@lostwizard]

Your note about the different fraction solutions fits with the fact that you can multiply any pythagorean triple by a constant and still have a pythagorean triple. Taking the general solution to the fractions would introduce another unknown, say x, which would multiply all three values (unless I'm smoking something unhelpfu, it would have to be the same in both equations to maintain consistency in the system). Assuming I didn't make any transcription errors: m = 67890, n = 12345 gives the triple 1676204100, 4456653075, 4761451125. And our friend 3,4,5 comes from m = 2, n = 1.

[@tylershepard4269]

Love the video! This is super cool. I do wonder, what are the broader implications of having this formula, i.e., what problems can it be used to solve?

[@hybmnzz2658]

Not sure about practical applications, but one thing is that it solves Fermats last theorem for n=2^k. It can be shown that x^4 + y^4 = z^2 has no integer solutions due to (x^2,y^2,z) being a pythagorean triple leading to a contradiction. This implies x^4+y^4=z^4 has no solution, and thus x^4k+y^4k=z^4k has no solution. Historically, this is the first stepping stone to proving FLT and is even easier than n=3 due to how well we understand pythagorean triples. Then we "only" need to prove FLT for n=odd prime. An incorrect proof of the fact that FLT holds for odd primes (due to Lamé) led to the concept of regular primes, unique factorization domains, and more complicated number theory. So maybe practically not so useful, but motivates algebraic number theory.

[@suhaimimazed1136]

I find it rather hard to solve this Q, can you solve it please? "A right angle triangle has a base of 11 cm. The other two sides are integers. Find the Perimeter of the triangle."

[@reetasingh1679]

You can apply the method given in the video. Take the base to be equal to m²-n² (don't take it as 2mn because 11 isn't divisible by 2, so that way, m or n could be fractions instead of whole numbers, which we don't want) (m-n)(m+n) = 11. This gives m-n = 1, m+n = 11, since 11 is a prime so it only has 2 factors. Solving for m and n gives m=6, n=5. Plug these values into 2mn and m²+n² to get the other 2 sides. So you get the Pythagorean triple (11,60,61)

[@suhaimimazed1136]

Reeta Singh Thanks for the help!

[@jeremybuchanan4759]

An alternative approach for 'a' (short leg) of odd length is to just square the number ... b (long leg) = (a^2-1)/2 c (hypotenuse) = (a^2+1)/2 in this case a=11 ... a^2 =121 ... 121/2 = 60.5 ... the integers either side are b and c :)

[@trueriver1950]

Jeremy Buchanan It follows that there are infinitely many right triangles where the hypotenuse is exactly one more than one leg. The two that BPRP had on the board are just the first two in an infinite series, the next being 7,24,25 which is the smallest where the hypotenuse is already a square. (You might already know the smallest right triangle where any side is a square)

[@harmitchhabra989]

@@reetasingh1679 Thanks for the approach man!!

[@plaustrarius]

I really like this video! i've always been introduced this proof using geometry, trig functions, parametric equations and so on, this is the first explanation i've seen that's almost entirely algebraic!

[@contemporarilyancient]

We have been taught a similar method in school but we just set m = 1 for all the forms of Pythaogorean triplets m² + 1, m² - 1 and 2m. This makes calculations a lot easier and works for all even values of m. I know it is very limited but works all the time. Example Let m = 6 2m = 3 m² + 1 = 3² + 1 = 10 m² - 1 = 3² - 1 = 8 2m = 2(3) = 6 The Pythaogorean triplet is 6,8,10

[@dekeltal]

Unfortunately, just picking integer m and n values will not produce ALL Pythagorean Triples, since some of them require irrational m and n. For example, (30,40,50) will require m = SQRT(45) and n = SQRT(5).

[@wowo29]

Nice video, have you ever seen the video "all possible Pythagorean triples, visualized" by 3Blue1Brown? also very interesting

[@michaelbergwell]

Great video...I haven't seen this approach. It seems the same result is reached if you square any complex number (and consider the lateral lines as pointed out in the 3B1B video). For example, pick any two whole numbers, say 3 and 8. If we square 3+8i, we get -55+48i. Now take the absolute value of the real part and the imaginary part to get the a and b of a pythagorean triple (48, 55, 73). And there you go, math is beautiful...though most people will never know it.

[@vnknovn]

you were so close to the general solution and, instead, you decided to TAKE a subset of it :( you just have to multiplý each a, b and c on your solutions by k for every whole number k and there you have it!

[@vnknovn]

My bad, I just realized that I was missing solutions: it also works for all rational number k such that all the products are whole numbers. This tells me why you didn't show this, because the result is not so easy to compute as a for loop

[@rudimetzger-wang9414]

can you do the same for pythagorean quadruples? (find it interesting because we use a lot of 3D Space and need absolute value of vectors 😁)

[@friedkeenan]

Does this get rid of all the ones that are just multiples of other Pythagorean triples? Would those come in if we didn't just make top=top and bottom=bottom?

[@faith3174]

Keenan Horrigan Technically, yes. With this formula you can't get the triple 6,8,10 because it's a multiple of the 3,4,5 triple. You'll skip over multiples of the triples who are coprime. If you want to include the multiples of the triples multiply the equation a² + b² = c² by a constant (say, k) to get: ka² + kb² = kc²

[@zanti4132]

35cut: You can get some of the multiples of Pythagorean triples, but not all of them. You CAN get the 6,8,10 triple with the values m=3 and n=1. However, there are no integral values for m and n that produce the 9,12,15 triple.

[@kumargs2887]

@@zanti4132 m=4 n= 3

[@SamuelAndradeGTutos]

Are all pythagorean triples of kind (2mn,m²-n²,m²+n²) for some m and n integers?

[@angel-ig]

Yeah

[@jimmylu7954]

blackpenredpenbluepen #yay

[@jyl123]

and i was waiting for this

[@tamircohen1512]

Jin Yang Li same!

[@blackpenredpen]

: )

[@tamircohen1512]

blackpenredpen thanks to your videos I already know most of the calculus course that I will be studying for 2 years starting when I return to school in September and more :)

[@matthewstevens340]

I remember doing an intuitive proof ages ago based on 'fitting' an odd length line (generated from each odd square) around a corresponding square to get the next square up. But I'm loving this!

[@RussellSubedi]

Alternatively, you can find m and n using a, b and c. m = √((c + a) / 2) n = √((c - a) / 2) Turns out you don't really need b, but if you insist, m = √((√(a^2 + b^2) + a) / 2) n = √((√(a^2 + b^2) - a) / 2) Here you don't need c. You can have both, but not all three.

[@malcomthonger]

"We're all adults now" heck yeah I recently turned 18 and my gift is hard math👍

[@blackpenredpen]

😆

[@ПетросТепоян]

Make video about solving integrate (x^-x) from 0 to inf. I think it will be very interesting

[@hellomoto131313]

I like to generate Pythagorean tripples by taking any odd number, squaring it, then think of the 2 consecutive numbers that add up to your squared odd number. Examples: 3^2 = 9 = 4 + 5. Thus, 3, 4, 5. 5^2 = 25 = 12 + 13. Thus, 5, 12, 13. 7^2 = 49 = 24 + 25. Thus, 7, 24, 25. 9^2 = 81 = 40 + 41. Thus, 9, 40, 41. ... (odd number)^2 = (integer Z) + (integer Z+1). Thus, (odd number), (integer Z), (integer Z+1).

[@PeChips]

Wow cool

[@yrcmurthy8323]

#YAY THANKS ! We can even have The triplet (2m,m²-1, m²+1) ? Example : If one of the triplet is 6, Let's say 2m = 6 Therefore m = 3 m²-1 = 8 m²+1 = 10 Therefore (6,8,10) is a triplet.

[@MegaJamil12]

So, m should be bigger then n is it?

[@binaryblade2]

and of course you can always scale the term by and integer, so that means theres a transform from 3d space to the pythagorean triple space.

[@factsheet4930]

My question now is this, using this method you get an algorithm that's O(n) complexity, although you cannot get all pythagorean triples this way. The way I found that gets you all of them is O(n^2). So my question now is this, is it possible to solve in O(n) such that it produces all triples?

[@louisvictor3473]

~~You most certainly can get any of the possible triples this way,~~ (edit: not 100% sure about that part after a bit of coffee) but there is no algorithm or method to "get them all" as there are infinitely many of them. You need a better defined question, cause O(infinity^2) doesn't make sense (that would be to get "all") and if you don't define how each "n" maps to a triple, it is still insufficient. There is no natural "nth" triple.

[@factsheet4930]

@@louisvictor3473 The question was that given some n, can you print out all Pythagorean triplets up to n? I did actually find an O(nlogn) algorithm and I think that you can't do any better than that?

[@ant0625]

I developed an algorithm to find pythagorian triplets where if a^2 + b ^2 = c^2 . and c - b = 3. Pick a value of a>6 and then c = ((a^2 + 9)/6)^2 and b = ((a^2 - 9)/6)^2

[@iboroudoh6505]

Terrific! Yet complex, but absolutely terrific.

[@ajiwibowo8736]

Yay, bprp upload video again 😂 Greeting from Indonesia sir! ❤

[@jacksainthill8974]

Pythagoras Square also happens to be a location name in Samos Town, Greece. Unfortunately, so far as I can tell the space is roughly rectilinear, rather than in the form of a right angled triangle. Source: www.dreamstime.com/editorial-stock-photo-pythagoras-square-samos-town-greece-view-greek-island-image60861603

[@yifanhuang2721]

What is the restrictions of m&n in this case? If m&n both equal to 1, the formula doesn't work anymore. Should both m&n be greater than 1?

[@angel-ig]

For non-zero positive sides of the triangle, do m>n choices

[@angel-ig]

Also, m and n should be naturals

[@victoirevim9698]

Black pen Red pen Blue pen YAAAAAAY!

[@a.a.347]

actually, you are the ONE who teached me loving mathematics, thank you so much!

[@blackpenredpen]

Glad to hear! Thank you!!

[@OptimusPhillip]

3Blue1Brown did a video with a similar conclusion, but his method involved squaring complex numbers.

[@6612770]

I chose n=2 and got an (8,0,8) triangle!! Lol

[@angel-ig]

For non-zero positive sides of the triangle, do m>n choices

[@LamNguyen-jp5vh]

I stopped at your question: "How can we find the triples". My idea is to let b equal to any odd number such as 3,5,7,9. Then square b, we will get another odd number. Since it is an odd number, it can be the sum of two consecutive number. The larger number will be c and the smaller number will be a. Let's get an example: b=7 --> b^2=49=24+25 --> 24^2+7^2=25^2 b=9 --> b^2=81=40+41 --> 40^2+9^2=41^2 Thank you for reading this comment and yayyyyyy

[@sabihhasan3262]

Exactly what i was looking for, so beautifully explained. Thank you! :)

[@Chalisque]

@3blue1brown's video: kzbin.info/www/bejne/h3u8nqyeo8aUm80 is worth a look if you like Pythagorean triples.

[@saharshbehal8766]

Sir ... at 3:49 when you wrote a/(c-b) = (c+b)/a ...we can directly take that when (c-b),a,(c+b) are in geometric progression .....a,b,c are gonna be Pythagorean tripples ...

[@dasegu94]

By this formula are we reaching all possible triplets? I'm trying to solve Challenge 54 from CodeAbbey. The input is S, where S=a+b+c, and S is above 10e7 values. There are missing triplets? (Primitive as Scalar)

[@mtaur4113]

If you assume that gcd(a,b)=1, you also have gcd(a,c)=1. If not, call them capital A,B,C and divide by gcd(A,B) to get lower case Pythagorean equation with gcd(a,b)=1. From there you can prove that a=2mn and c=m^2 + n^2 is the only possibility later on. We get that gcd(m,n)=1 here, and if a,b,c positive, m>n. Then we see that all other cases are scaling by k.

[@lazaremoanang3116]

For example (3,4,5), (5,12,13), (6,8,10), (10,24,26), (26,168,170). I can write it all my life.

[@0011peace]

Mu;itpleples of the base trip always workso really don't count. 3^2 + 4^2 = 5^2 is th same ratio as 6^2 + 8^2 = 10^2

[@defishkoi7536]

Pick any random complex number a+bi. e.g: 3+4i square it to get another complex number c+di, then c,d are 2 in 3 Pythagorean Triple (3+4i)²= -7+24i 7²+24²=625=25²

[@babupriya8814]

Sir, I am from India. I am very interested in mathematics. But I wasn't studied mathematics. I surely said, This Euclid PPT formula is not well. I prove It . But I didn't know the way, and 2n , (n*2 - 1), (n*2 + 1) This Plaeto' s formula is also not well. Sir answer me. Thank you.

[@neeraj8278]

Well haven't watched the video but here's what I do, take the example of 3^2+4^2=5^,just keep multiplying by 2 or 3 or any whole number both sides and you will keep generating Pythagoran triplets....

[@christian.dev8808]

m > n always while n > 0 and m > 1. They work like indexes on an infinite set of triples!

[@tamirerez2547]

I know this generator for years, but I don't understand one thing: After we generated the numbers 5, 12 and 13, Where did the m and n disappeared? I mean in the 5, 12, 13 triangle, where is the "memory" of 2 and 3? The two numbers that generated these 3 pithagorians numbers. In other words: when I see 5,12,13 triangle, how can I find or calculate (fast) the two numbers that genareted this triangle?

[@PhilBoswell]

Check out this beautiful visualisation by 3Blue1Brown: kzbin.info/www/bejne/h3u8nqyeo8aUm80

[@thinkitwell8734]

Well instead of using two numbers, we can use a single number to get a pythagorean triple generator. For m € N-{1} , 2m, m^2-1 and m^2+1 are pythagorean triplets

[@cutsandstonebetablocksumo9891]

a = 2x+1 b = (a^2-1)/2 c = (a^2+1)/2 always generates whole solutions with whole number x

[@nataliaokay2076]

I'm not an adult I'm 14 😌 WHAT I AM DOING HERE WHY I FIND IT INTERESTING IM LITERALLY A CHILD I SHOULD HAVE MY CHILDHOOD NOT MATHS

[@OonHan]

don't forget m > n xD

[@butter5014]

The Pythagorean triple I produced could be reduced to the base Pythagorean triple of 17, 144, 145.

[@michalnemecek3575]

If you plug in m = 1/sqrt(2) and n = i/sqrt(2) you get a 1 i 0 triangle! (source: I was bored, so I set a = i, b = 1 and c = 0 and just calculated m and n)

[@sanpetchpanyapattanaporn7275]

m and n cannot be anything. if m=1 and n=2 ,the formula(a^2 + b^2 = c^2) is true but the meaning as a length if a side of triangle is wrong. then it's not a Pythagorean Triples. Pythagorean Triples must be an integer.

[@enejidjsi5939]

9:00 i think taking the simplest solution is a bitt redudnant cuz u can just divide by a^2 on the pythagorean equation and get 1 + b^2/a^2 = c^2/a^2 and then place the val.ues of each and then get the same result

[@angelmayca2198]

there is another formula: a=K, b=(K¨2-1)/2 and c=(k¨2+1)/2 of course K belongs to Z

[@bbartt80]

Hmm I don't get it: we take 2 integers to calculate a, b and c based on derived formulas. Why not to set a and b (also 2 integers) and calculate c instead? :)

[@seanchen4487]

what about 9, 12, 15? I only found irrational solutions for m and n, sqrt(12) for m and sqrt(3) for n

[@molgera3]

Captions: "pacific rim theorem" lmao

[@blackpenredpen]

😆 I just corrected the auto caption. Thanks.

[@miaash3870]

Brilliantly presented!

[@MisterPenguin42]

3:17 good awareness!

[@BingusLover-yx7vf]

Very boring and made it more difficult than it should have been. Never showed how to calculate the other 2 values if you know one

[@tv..6531]

KPop+KMath: 피타고라스의 수(Pythagoras Number)를 구하는 색다른 방법을 제시하고, 피타고라스 수를 구해봅니다. x^2 + y^2 = z^2 ∴ x = a + 2n, y = b + 2n, z = a + b + 2n (단, ab = 2n^2) kzbin.info/www/bejne/p4bXnZ-vgtWIgpI

[@maksatilmyradov9021]

And m > n is sufficient in order all possible numbers of a, b and c to agree with the equation of inequality of triangle.

[@adamkovacs4265]

It was wonderful and terrifying to watch because how simppe is this actually.

[@arshiaabedini2686]

This can be a powerful tool, but not for these things😂in relativity or related ones might be practical

[@franciszekkrawczyk2556]

my favourite pythagorean triple is 30i-1,15+18i,30+10i

[@gilberttheisen9270]

A compléter avec ;""l'énigme de FERMAT "" sur ""YOU TUBE "" .

[@rajveermandal9538]

Background sound

[@gameryl3900]

Lmao at 10:23 I chose m=3 and n=2!!! I was so shocked when @blackpenredpen chose it as well 😂

[@michaferenc6112]

I would also add multiplication by some k in these values because if we multiply all elements of pythagorean triple by some natural number we also get pythagorean triple

[@_invencible_]

yeah, the k should have appeared when he equaled the numerators and denominators to each other, even though they don't have to be exactly the same but rather multiples of each other.

[@DarkRizzard69]

I found another method a=3n b=4n c=5n Works also for decimal