WOW WOW WOW WOW!!! ❤❤❤❤ Number Theory, Modulo Arithmetic - HOUDINI OF MATHEMAGIC!!! While the Pythogarean triplets gave away 1 solution by just a look at the problem, but then proving this was the ONLY Integer solution was sheer magic!!! Absolutely LOVED IT ❤❤❤
He took it from Babylons , much older lads who solved it. It's just our ignorant history indoctrinations from school who told us it's "Pythagoras". Hellenistic culture didn't discover anything pretty much , mostly copy and occasionally "re-discover". (edit: sorry , I just assumed it was from Nineveh, where most stuff stolen from Iraq is from, esp. during Alexander's pilfering from it, which made Hellenistic "philosophy and science" explode/accelerate occurred).
Cute. By the way, maybe this is a nitpick or maybe I am wrong but since you mention the a>b clauses, shouldn't you mention the n > 0 when doing mod 3 at around @7:20 mark? if n = 0, 3^0 mod 3 = 1 != 0, so n has to be greater than 0. Of course, later you solve it for being precisely that but I still think you should have added it since the mod isn't lossless... or am I just been too long away from doing maff? :)
Slight problem... -1 = 1 mod 2. The first time you use this, it's fine, because what you did is also valid mod 4. From there... 3^2n = 9^n, so we have 9^k + 4^y = 5^z. Next, if y > 1, then 4^y is divisible by 8. So, taking mod 8 we get 1^k + 0 = 5^z. This only holds if z is even, and we're good to continue... unless y = 1. So, you still need to handle the case of 4 = 5^z - 9^k.
there is a very simple fact in math. the number of equations to solve MUST match ( or be more than ) the number of variables. Here 1 vs 3 - not possible to solve. Another- you cannot say that using mod and then doing math on mod results will work. Here is good example 5 + 5 = 10 - now take mod 3. you get 2 + 2 = 1 - not quite true is it? So, anymone want to tell me I am wrong?
I just plugged in 2 for everything.
Thank you for your great videos, but 1 is congruent to -1 mod 2 so it's true for any integers z and y.
I also noticed that error. Showing that x and z must be even can be done by reducing mod 4 and mod 3, respectively.
WOW WOW WOW WOW!!! ❤❤❤❤
Number Theory, Modulo Arithmetic - HOUDINI OF MATHEMAGIC!!!
While the Pythogarean triplets gave away 1 solution by just a look at the problem, but then proving this was the ONLY Integer solution was sheer magic!!!
Absolutely LOVED IT ❤❤❤
Glad to hear that! 💗🤩😊
Very good, and thoroughly explained 👍😁.
I wonder if Pythagoras could have proven that?
Good question!
He took it from Babylons , much older lads who solved it. It's just our ignorant history indoctrinations from school who told us it's "Pythagoras". Hellenistic culture didn't discover anything pretty much , mostly copy and occasionally "re-discover". (edit: sorry , I just assumed it was from Nineveh, where most stuff stolen from Iraq is from, esp. during Alexander's pilfering from it, which made Hellenistic "philosophy and science" explode/accelerate occurred).
Cute. By the way, maybe this is a nitpick or maybe I am wrong but since you mention the a>b clauses, shouldn't you mention the n > 0 when doing mod 3 at around @7:20 mark? if n = 0, 3^0 mod 3 = 1 != 0, so n has to be greater than 0. Of course, later you solve it for being precisely that but I still think you should have added it since the mod isn't lossless... or am I just been too long away from doing maff? :)
Imagine though, if this happened to show the full generalised "pythagoras",. that would be wicked sick. ^^
Fermat would have liked this problem.
Is'nt minus 1 is congruent to plus 1 modulo 2?
it is
The most obvious solution is x = y = z = 2, from Pythagoras.
we can also know that x and z are even using mod 4 and mod 3.
Slight problem... -1 = 1 mod 2. The first time you use this, it's fine, because what you did is also valid mod 4. From there... 3^2n = 9^n, so we have 9^k + 4^y = 5^z. Next, if y > 1, then 4^y is divisible by 8. So, taking mod 8 we get 1^k + 0 = 5^z. This only holds if z is even, and we're good to continue... unless y = 1. So, you still need to handle the case of 4 = 5^z - 9^k.
You're right!
3^x + 4^y = 5^z
( x , y , z ) = { ( 0 , 1 , 1 ) , ( 2 , 2 , 2 ) }
😊👍👋
You are nice.
Nice
Thanks
I thought it couldn't be too easy
How 6:37 that's equal to one 😮
5^x=64 , (20)^2x/x+1=?
Is the last expression 1 + (20^2x)/x ?
2x/x+1
(20)
there is a very simple fact in math. the number of equations to solve MUST match ( or be more than ) the number of variables. Here 1 vs 3 - not possible to solve. Another- you cannot say that using mod and then doing math on mod results will work. Here is good example 5 + 5 = 10 - now take mod 3. you get 2 + 2 = 1 - not quite true is it? So, anymone want to tell me I am wrong?
Diophantine Equation
@@SyberMath😂😂😂diophantine equation
Bro is confidently wrong
The next time use more variables. Just kidding. Nice explanation 👌
😲😁😜😜Ahahaha. Thanks
Just gues it , 2
Be humble and stop your I solved !
x = y = z = 2