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

"We're all adults now"

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

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.

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

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.

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...

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

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 :)

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!

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!

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

Wonderful job. I was confused at the beginning, but it made sense once I watched a whole thing

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.

When you got to the exemple --> mindblow. Awesome class, thank you very much.

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

What an elegant proof! If m, n are N, 0

Brilliantly presented!

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

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

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.

this is one of the most beautiful things i have seen in maths

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.

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...

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

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

Awesome approach! My idea was just to take any complex number with integer parts (a + bi) and square it (a^2 - b^2 + 2abi), give you a right-triangle on the complex plane with side lengths of a^2 - b^2, 2ab and a^2 + b^2, which are all positive integers, provided a > b > 0

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

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

I must admit I didn't see where the math was going until towards the end but once I saw it my mind was blown! What I find particularly fascinating is that this shows that for an integer value of c, c must be the sum of two square numbers, so a^2+b^2=(m^2+n^2)^2. So many squares!

Excellent!! Love this.

Terrific! Yet complex, but absolutely terrific.

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

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

You can further set m=a and n=1/2 so that you get a simpler solution set of Pythagorean triples in one variable: (a, a^2 + 1/4, a^2 - 1/4)

I am not sure if anyone has mentioned this (I checked most of the comments) but it should be m>n, if we want this to make sense geometrically. In the other case, b can be negative, which cannot be the length of an edge... No matter, keep up the good work!

I love your enthusiasm buddy

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.

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?

Playing around with M and N there’s a fun relationship where if M = N+1 then you get that C = A+1 as well (B is the smallest number of the trio). For instance, with M=2 and N=1 you get the 3,4,5 triplet (A = 2MN = 4, C = M^2 + N^2 =5). Likewise M = 3 and N = 2 gives the 5,12,13 triple (A=12 C=13). And so on. Also coincidentally in each case B is the odd number 2N + 1, and thus for every odd number Q there is a Pythagorean triple of the form Q, P, P+1 where P in generated as above.

Such a great vid, love the videos you post.

Thanks a brilliant lesson!

love it. Just like all the other ones.

holy crap... this video made me want to get back into math a little more. I'm working on project euler, what got me here, and i'll make this elegant solution into a fine program ^^ thanks!

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.

Take any complex number in a form a+ib where a>b. Square it and you get a new complex number c+id. You find a pythagorean triple: c, d, sqrt(c^2 + d^2) For example: (2+i)^2 = 4 + 4i -1 = 3+4i ; sqrt(3^2 + 4^2)=5 For a

Really awesome video nice work

This was great, thanks

and i was waiting for this

Thanks for the video, helped me solve a project euler challenge.

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

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

Thank you!

Very cool. Thanks!

This video is what I was looking for 😃

Awesome sir. Beautiful and wonderful!!!!!!!!!!!!!!!!!!

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

Great video!

That's superb!

You can be more general than taking the "easy solution". You can fix the value for a and then use m and n to generate the values for b and c. You have intoduced m and n as known constants, and with them two equations that constrain abc. The original equation is also a constraint. Three unknowns abc and three constraints fully determine your system.

Truly awesome!

This is awesome!

This was great fun, very interesting ☆

Loving the new microphone!

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

3:17 good awareness!

wow, you are a great RU-vid teacher 😊😊😊

YAY from swiss ! Your channel is awesome

it's so complicated! nice GENERATOR ( i mean if you want to make a program to follow these rules it will take forever)

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.

Just glorious!

Recién conozco tu canal, gracias chino

#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.

Thank you sir..

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?

this is AMAZING !

Not sure if this was noted or not, but m and n must not be equal and m must be greater than n for this to work for triangles.

Nicely done. Now we can code it and make a table

Nice to notice that this generator is one of the most important theorems in chromogeometry

Woah that's super cool!

well explained

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

This kind of questions are also fun: give the collection of possible values of a, b and c, such that b is one greater than a and a^2+b^2=c^2

This is quite commonly taught in high school to convert and eqn into parametric form. Though well explained.

Do more number theory videos!

I found a generator where you can get a Pythagorean triple by letting a be any number besides a perfect power of two. That is, a can be any odd number, or any multiple of an odd number, so it will have the form h(2k+1), where k and h are integers. Then b will be 2h(k^2 + k), and c will be b+h. It's not quite as clean, but you can find a triple for almost every a. I was thinking we would be able to combine that with the 2mn approach, where m and n are powers of 2, to get a pythagorean triple for every integer a, but it doesn't seem to work for a = 2, 4, 8, or 16, because m^2-n^2 (b) ends up being smaller than a, (though it does for all larger powers of 2). Are there no pythagorean triples for a=2^1 through 2^4? If so, why would that be? I guess there are if you relax and let b be smaller than a, but that doesn't seem fully satisfying.

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 😁)

This is so satisfying.

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?

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

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)

love it!

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

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).

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.

#YAY I have a similar generator. It works with all whole numbers greater than 2. If the number is even (n) than use the formula (n/2)^2=x. You then take x and add 1 to produce the second triplet and subtract 1 to produce the third triplet and (n) being the first triplet. It very similar for odd numbers but slightly different. If the number is odd (n') then use the formula (n^2)/2=x. You then take x and add .5 to produce the second triplet and then subtract .5 to produce then third triplet. Unfortunately it does not produce a unique triplet for every number. For instance if you use 3 you get the familiar 3^2+4^2=5^2 and the same is true when you use 4. Hope this makes sense and it's a little easier to remember for me.

Superb maths buddy! #mathssupport

This would also prove that b=0 is a right triangle when m=n. What are exact angles of the triangle when n=m/2 and m->Inf? Could that be called "golden angles"?

Great video

You explain it in a really playful way! #yay

The generator can be generalized as follows: a = m^2 - n^2; b = 2m(m*cos(C) + n); c = m^2 + n^2 +2mn*cos(C). If cos(C) is rational, this will generate scalene triangles having angle 'C' and three rational sides. For example, if C is 60 degrees, cos(C) = 1/2. The equations above generate triangles which are obtuse. If you want acute triangles, use an acute angle for C and change the formula for 'a' to: a = n^2 - m^2 + 2b*cos(C).