The substitution isn't really necessary in this problem. The given equation can be rewritten as (x + 1)(x - 1) = 2y^2. Because x is odd, 4 divides (x + 1)(x - 1), and it follows that 2 divides y. Given that y is prime, y = 2, etc.
that is also basically what i did . factor then set x= 2n+1 then Y^2=2n(n+1) then 2n=n+1 then n=1 ...now need to watch video to learn the more complicated approach and what the heck (from the comments) Pells rule is
Also becouse x and y are prime we can simple use the fact that x² - 1 = 2y² (x+1)(x-1) = 2y • y And becouse x + 1 > x-1 we have x + 1 = 2y and x - 1 = y => x + 1 = 2(x - 1) => x = 3 and we have done
There is actually an infinite amount of integer solutions given by the general method of solving Pell Equations. I wonder if there are an infinite amount of infinite prime solutions as well.
Just to get you started on the general case, where x must be odd and y must be even: { (1,0), (3,2), (17,12), (99,70), (577,408) ... } Obviously, the negative values are also solutions.
Basically, all the solutions are x and y, where x+y*sqrt(2) = (3+2*sqrt(2))^n * (-1)^k, for some n and k integers. This is from algebraic number theory, where x^2 - 2y^2 is the norm of the algebraic integer x+y*sqrt(2), and you want the norm 1 numbers, i.e. they are invertibles. All invertibles are of the form (1+sqrt(2))^n * (-1)^k, but this gives you elements of norm either 1 or -1. So you take n to be even to force a norm of 1, and that's how you get (3+2*sqrt(2))^n * (-1)^k
Yes if the constraint X,y prime wasn't there. With 2y^2 = (x+1)(y-1) and y any integer the solution is more complicated with the factors on the RHS = 2a * b where ab=y^2.
To summarise, with the restraint that x and y are prime, then we only need deal with natural numbers. Writing x² - 1 = 2y², it is immediately obvious that x is odd and x > y. So (x-1)(x+1) = 2y² but since y is prime, the only factorisations of 2y² are (1 * 2y²), (2,* y,²) and (y * 2y), with the first factor being smaller. That gives three cases: Case 1: (x-1) = 1 and (x+1) = 2y² giving x=2 which is not a solution for integer y. Case 2: (x-1) = 2 and (x+1) = y² giving x=3 and y=2. Case 3: (x-1) = y and (x+1) = 2y. Taking the difference we get 2 = y and hence x=3 as before. So that's the only solution for prime x and y.
My approach: We have x^2 - 1 = 2y^2, so x is odd (because if x is even, then x^2 is even, so x^2 - 1 is odd, a contradiction) If we factor the left hand side we have (x + 1)(x - 1) = 2y^2 Both (x + 1) and (x - 1) are even, so (x + 1)(x - 1) = 2y^2 is divisible by 4. So y^2 is even, which implies that y is even. As y is prime, y = 2. And from this we get x = 3. Thank you for the video !! :D
And if 2y^2 mod 4 = 0, this means it's *at least* divisible by 4. Since we know y is even, this means 2y^2 mod 8 = 0, because the greatest multiple of 2 that divides 2y^2 is 8. Another interesting fact is that the square of an odd integer is always 1 more than some multiple of 8. Said mathematically, if x is congruent to 1 modulo 2, then x^2 must also be congruent to 1 mod 8, and therefore also congruent to 1 mod 4. If we look at the square of an odd number as a binary numeral, its LSBs are "...001" because 8-1 is 3 bits wide, meaning "n & 7" ("&" is bitwise AND) is a 3b numeral
@@angelmendez-rivera351 I didn't know he only wanted integer solutions i wrote that before he posted the video when he just posted the blurb. So removing that restriction YES i IS INDEED A SOLUTION!
I found out that if we consider that x and y can be any integers, than if some (x;y) is a solution, than (4y^2+1;2xy) will also be a solution. This actually shows that there are infinite integer solutions, pretty interesting
This is interesting. Can you share how that works or how you discovered it? It misses some of the solutions like (99,70) but still a good recurrence formula!
@@SyberMath yeah, there are other solutions as well. This is how I found it: By looking (mod 4) we see that x is odd and y is even, so we substitute: x=2k+1 y=2m 4k^2+4k=8m^2 k^2+k=2m^2 k(k+1)=2m^2 k and k+1 are two consecutive integers, so they have no common devisors and when their product is a doubled square, than one them is also a doubled square and the other one is a square number. Let’s try the case k=2p^2 x=4p^2+1 (2p^2)(2p^2+1)=4p^4+2p^2=2m^2 2p^4+p^2=m^2 Divide both sides by p^2 1=(m/p)^2-2p^2 This has the same form as the original equality So if x1=m/p and y1=p were solutions x1=(y/2)/y1, so y=2x1y1 and x=4y1^2+1 are also solutions In the case when k+1=2p^2 we get different solutions, but I didn’t find any pattern in them and don’t know how to find all of them
It is possible to reasoning starting by factoring the equation like this: (x+y)(x-y)=y^2+1. Thus (x+y)=(x-y)mod y^2 and consequently 2y=0 mod y^2. This means that 2y divides y^2, particularly 2 divides y^2. Since y is prime, it follows y=2.
in (x,y) ordered pair the following are solutions :(3;2), (-3;2), (3;-2), (-3; -2), because the restriction are that the number must be integers, not only prime numbers.
2y²=x²-1=(x-1)(x+1) (1) 2 divides 2y² then 2 divides (x-1)(x+1). But that implies that 2 divides both x-1 and x+1 (since their parity is the same). Then 4 divides 2y² then 2 divides y² then 2 divides y (for a similar reason). Since we supposed that y is a prime number, that forces y=2 (2 is the only even prime member). So we have: x²-8=1 and therefore x=3.
I saw the thumbnail and thought that we were meant to search for 1 unsigned integer solution. I rewrote the equation multiple times hoping to find such solution but only got these: x^2 - 2(y^2) - 1 = 0 (xx - 1) / 2 = yy x mod 2 = 1, therefore, xx mod 8 = 1 xx - 2yy = 1 (xx - 1)^(1/2) = 2^(1/2) * y ((xx - 1) / 2)^(1/2) = y (xx - 1) / yy = 2 ("lb" is binary log) lb((xx - 1) / yy) = 1 lb(xx - 1) = 2 * lb(y) + 1 lb(xx - 1) = (lb(y) + 1/2) * 2 lb(xx - 1) / 2 = lb(y) + 1/2 I gave up so I used the brute force approach to find a trivial solution: x = 1, y = 0. But I wasn't satisfied so I found {3, 2} and stopped there. Then I saw the video and you said "primes" and I was like "bruh"
A simpler way to solve is as follows: x² - 2y² = 1 then (x² - 1)/2 = y² Thus (x² - 1)/2 is a perfect squared integer, (x² - 1) is even hence x is an odd number. With these conditions then x = 3 as (3² - 1)/2 = 4 = 2² a perfect squared integer. Thus (x ,y)=(3,4)
Fursly you can factor x*x-1 to get: (x-1)*(x+1) = 2*y*y If x = 2, then LHS is odd, but RHS is even. If X > 2 - than X must be odd. Because of that, LHS whill be prodoct of 2 even numbers, and it means that LHS is divisible by 4. But RHS can be divisible by 4 only if Y is 2. Then RHS is 8, and X is 3. And this is all.
I thought the prime number limitation was unnecessary, but I was wrong. Eg 17^2-2×12^2=1 . Solutions of this type depend on finding triangular numbers "T" which are also square. In this case, T=36. y= 2×sqrt(T), 2y^2+1=289 36=(m)(m+1)/2 where m=8 and x=2m+1=17. The given solution corresponds to T=1, m=1 Square, triangular numbers are rare. Here are the first few. 1, 36, 1225, 41616, 1413721, 48024900.
(x-1)(x+1) = 2 y*y has a feasible solution for (x,y) in integral domain, for odd x only . Writing x= 2n+1 one gets n(n+1) = y*y . This essentially means y*y lies in between two square numbers n^2 and (n+1)^2. Hereby no feasible solution is there to x^2 -2*y^2 = 1 for integral (x,y)
The unity fundamental of Z[√2] is 1+√2 but (1+√2)*(1-√2)=-1. That's the Norm of 1+√2 is (-1) then "the unity fundamental" with Norm 1 is (1+√2)^2=3+2*√2. Then the unities in Z[√2] with Norm 1 are the form +-(3+2*√2)^n with n\in Z. But if x^2-2*y^2=1 then (x+y*√2)*(x-y*√2)=1. That is x+y*√2 is a unity of Z[√2] with Norm 1. Then if x_n=(1/2)*((3+2*√2)^n+(3-2*√2)^n) and y_n=(1/(2*√2))*((3+2*√2)^n-(3-2*√2)^n). Then (+-x_n,+-y_n) are solutions where n\in Z.
Good morning! If they are prime numbers it is easy. x must to be odd. If you use mod4, cleary you see that y is even as we do not have squares gongruent to 3 mod4 for odd numbers, only congurents to 1. And then y can be only 2. By the way, where to study pell's equation? If you replace x^2 for a and y^2 for b we have that a=9+ 2K and b= 4 + k ; k integer. Is it possiple that a and b are both perfect squares for any k0?
0:35 "in this particular equation, we're looking for prime solutions" - that came out of nowhere. How did you establish that we would only look for prime solutions?
You're welcome! Good question. We can solve this by using the following x^2 - 5y^2 = 5 ... (x, y) = (5, 2) x^2 - 5y^2 = 1 ... (x, y) = (9, 4) x + y*sqrt(5) = (9 + 4sqrt(5))^n*(5+2sqrt(5)) ... n is an integer Also check this out: www.wolframalpha.com/input?i2d=true&i=Power%5Bx%2C2%5D-5Power%5By%2C2%5D%3D5+solve+for+integers
x^2 = 412·y - 3 x^2 (mod 412) = -3 (mod 412). Thus you only need to consider x = 0, 1, ..., 206 (mod 412) and check the solutions to the above equation of modular arithmetic. For x = 0, ..., 14 (mod 412), there is trivially no answer, because 14^2 < 206. Equations where one side is linear and the other is quadratic can always be reduced to solving elementary quadratic equations in modular arithmetic.
x^2 - 2*(y^2) = 1 x^2 - 1 = 2*(y^2) (x - 1)(x + 1) = 2*(y^2) Since y is prime, there are two ways to factor this. 2*y and y, or 2 and y^2. First version. 2*y > y, and x + 1 > x - 1, so 2*y = x+1 and y = x-1. 2*y - y = (x + 1) - (x - 1) y = 2 Putting it back into x^2 - 2*(y^2) = 1, we get x = 3. Second version. y^2 > 2 for any prime y. Since x+1 > x-1, we get y^2 = x+1 and 2 = x-1. Since 2 = x-1, we get 3 = x. y^2 = 4, so y = 2. Same solution.
can you please answer me what is the solution for the other case when x^2-2y^2=-1 i found that 5 and 7 are solutions but i cant prove that it is the only solution
Brother can you please take this equation for a video Its a iit jee question and I can't figure it out. Q;- integrate x^2/{1-x^4√(1+x4)} dx Please make a vid on it brother ❤️
To be serious, the only factors of a prime number are itself and 1, by definition. All even numbers have 2 as a factor, by definition. So an even prime number must have itself equal to 2. Q.E.D.
There are infinite number of integer solutions. For every solution (x;y), (4y^2+1;2xy) is also a solution, you can try this with (3;2) for example and find out that (17;12) is also a solution
The substitution isn't really necessary in this problem. The given equation can be rewritten as (x + 1)(x - 1) = 2y^2. Because x is odd, 4 divides (x + 1)(x - 1), and it follows that 2 divides y. Given that y is prime, y = 2, etc.
A nice approach!
that is also basically what i did . factor then set x= 2n+1 then Y^2=2n(n+1) then 2n=n+1 then n=1 ...now need to watch video to learn the more complicated approach and what the heck (from the comments) Pells rule is
Also becouse x and y are prime we can simple use the fact that
x² - 1 = 2y²
(x+1)(x-1) = 2y • y
And becouse x + 1 > x-1
we have x + 1 = 2y and x - 1 = y
=> x + 1 = 2(x - 1) => x = 3 and we have done
We can directly say 3 and 2 satisfies the equation sir
DO U DAILY WORK ON THIS STUFF .....THIS IS INSANE MAN.....THNKS FOR EVERYTHING
Absolutely and thanks!
There is actually an infinite amount of integer solutions given by the general method of solving Pell Equations. I wonder if there are an infinite amount of infinite prime solutions as well.
Nope! Only one prime pair since y must be even and the only even prime is 2.
@@SyberMath I agree prime and even is the key here.
there arent infinite solution there are like 3
@@SyberMath correct
Just to get you started on the general case, where x must be odd and y must be even:
{ (1,0), (3,2), (17,12), (99,70), (577,408) ... }
Obviously, the negative values are also solutions.
Basically, all the solutions are x and y, where x+y*sqrt(2) = (3+2*sqrt(2))^n * (-1)^k, for some n and k integers.
This is from algebraic number theory, where x^2 - 2y^2 is the norm of the algebraic integer x+y*sqrt(2), and you want the norm 1 numbers, i.e. they are invertibles.
All invertibles are of the form (1+sqrt(2))^n * (-1)^k, but this gives you elements of norm either 1 or -1. So you take n to be even to force a norm of 1, and that's how you get (3+2*sqrt(2))^n * (-1)^k
Yes if the constraint X,y prime wasn't there. With 2y^2 = (x+1)(y-1) and y any integer the solution is more complicated with the factors on the RHS = 2a * b where ab=y^2.
as and y are prime numbers and y = 2k, the only prime number for y that is even is 2
To summarise, with the restraint that x and y are prime, then we only need deal with natural numbers.
Writing x² - 1 = 2y², it is immediately obvious that x is odd and x > y.
So (x-1)(x+1) = 2y² but since y is prime, the only factorisations of 2y² are (1 * 2y²), (2,* y,²) and (y * 2y), with the first factor being smaller. That gives three cases:
Case 1: (x-1) = 1 and (x+1) = 2y² giving x=2 which is not a solution for integer y.
Case 2: (x-1) = 2 and (x+1) = y² giving x=3 and y=2.
Case 3: (x-1) = y and (x+1) = 2y. Taking the difference we get 2 = y and hence x=3 as before. So that's the only solution for prime x and y.
My approach:
We have x^2 - 1 = 2y^2, so x is odd (because if x is even, then x^2 is even, so x^2 - 1 is odd, a contradiction)
If we factor the left hand side we have
(x + 1)(x - 1) = 2y^2
Both (x + 1) and (x - 1) are even, so (x + 1)(x - 1) = 2y^2 is divisible by 4.
So y^2 is even, which implies that y is even.
As y is prime, y = 2. And from this we get x = 3.
Thank you for the video !! :D
And if 2y^2 mod 4 = 0, this means it's *at least* divisible by 4. Since we know y is even, this means 2y^2 mod 8 = 0, because the greatest multiple of 2 that divides 2y^2 is 8.
Another interesting fact is that the square of an odd integer is always 1 more than some multiple of 8. Said mathematically, if x is congruent to 1 modulo 2, then x^2 must also be congruent to 1 mod 8, and therefore also congruent to 1 mod 4. If we look at the square of an odd number as a binary numeral, its LSBs are "...001" because 8-1 is 3 bits wide, meaning "n & 7" ("&" is bitwise AND) is a 3b numeral
Also we can use Pell equation yo solve the exercise
Yes!
@@SyberMath Did you see my spoiler comment about how i aka swuare root of -1 is an answer?
But who knows about Pells equation? I've heard but forgot about it.
@@leif1075 i is not an answer, because i is not an integer.
@@angelmendez-rivera351 I didn't know he only wanted integer solutions i wrote that before he posted the video when he just posted the blurb. So removing that restriction YES i IS INDEED A SOLUTION!
I found out that if we consider that x and y can be any integers, than if some (x;y) is a solution, than (4y^2+1;2xy) will also be a solution. This actually shows that there are infinite integer solutions, pretty interesting
This is interesting. Can you share how that works or how you discovered it? It misses some of the solutions like (99,70) but still a good recurrence formula!
I have no clue how you discovered that but it certainly works !
I was also looking for a more general solution (ie x,y not necessarily prime)
@@SyberMath yeah, there are other solutions as well. This is how I found it:
By looking (mod 4) we see that x is odd and y is even, so we substitute:
x=2k+1
y=2m
4k^2+4k=8m^2
k^2+k=2m^2
k(k+1)=2m^2
k and k+1 are two consecutive integers, so they have no common devisors and when their product is a doubled square, than one them is also a doubled square and the other one is a square number.
Let’s try the case k=2p^2
x=4p^2+1
(2p^2)(2p^2+1)=4p^4+2p^2=2m^2
2p^4+p^2=m^2 Divide both sides by p^2
1=(m/p)^2-2p^2
This has the same form as the original equality
So if x1=m/p and y1=p were solutions
x1=(y/2)/y1, so y=2x1y1 and
x=4y1^2+1 are also solutions
In the case when k+1=2p^2
we get different solutions, but I didn’t find any pattern in them and don’t know how to find all of them
best video! love it :)
very cool
i hope you get way more popular way faster
Thank you! 😊
It is possible to reasoning starting by factoring the equation like this: (x+y)(x-y)=y^2+1. Thus (x+y)=(x-y)mod y^2 and consequently 2y=0 mod y^2. This means that 2y divides y^2, particularly 2 divides y^2. Since y is prime, it follows y=2.
in (x,y) ordered pair the following are solutions :(3;2), (-3;2), (3;-2), (-3; -2), because the restriction are that the number must be integers, not only prime numbers.
(1,0) is also a solution. In fact, this équation have an infinity of solutions.
2y²=x²-1=(x-1)(x+1) (1)
2 divides 2y² then 2 divides (x-1)(x+1).
But that implies that 2 divides both x-1 and x+1 (since their parity is the same).
Then 4 divides 2y² then 2 divides y² then 2 divides y (for a similar reason).
Since we supposed that y is a prime number, that forces y=2 (2 is the only even prime member).
So we have: x²-8=1 and therefore x=3.
I saw the thumbnail and thought that we were meant to search for 1 unsigned integer solution. I rewrote the equation multiple times hoping to find such solution but only got these:
x^2 - 2(y^2) - 1 = 0
(xx - 1) / 2 = yy
x mod 2 = 1, therefore, xx mod 8 = 1
xx - 2yy = 1
(xx - 1)^(1/2) = 2^(1/2) * y
((xx - 1) / 2)^(1/2) = y
(xx - 1) / yy = 2
("lb" is binary log)
lb((xx - 1) / yy) = 1
lb(xx - 1) = 2 * lb(y) + 1
lb(xx - 1) = (lb(y) + 1/2) * 2
lb(xx - 1) / 2 = lb(y) + 1/2
I gave up so I used the brute force approach to find a trivial solution: x = 1, y = 0. But I wasn't satisfied so I found {3, 2} and stopped there. Then I saw the video and you said "primes" and I was like "bruh"
What happens when x,y are not prime?
Loved it
Glad to hear that!
A simpler way to solve is as follows:
x² - 2y² = 1 then (x² - 1)/2 = y²
Thus (x² - 1)/2 is a perfect squared integer, (x² - 1) is even hence x is an odd number. With these conditions then x = 3 as (3² - 1)/2 = 4 = 2² a perfect squared integer. Thus (x ,y)=(3,4)
Fursly you can factor x*x-1 to get:
(x-1)*(x+1) = 2*y*y
If x = 2, then LHS is odd, but RHS is even.
If X > 2 - than X must be odd. Because of that, LHS whill be prodoct of 2 even numbers, and it means that LHS is divisible by 4.
But RHS can be divisible by 4 only if Y is 2. Then RHS is 8, and X is 3.
And this is all.
Primes can’t be negative?
Just amazing ❤️❤️🤩
💖 Thank you! 💖
I thought the prime number limitation was unnecessary, but I was wrong. Eg 17^2-2×12^2=1
. Solutions of this type depend on finding triangular numbers "T" which are also square.
In this case, T=36.
y= 2×sqrt(T), 2y^2+1=289
36=(m)(m+1)/2 where m=8 and x=2m+1=17.
The given solution corresponds to T=1, m=1
Square, triangular numbers are rare. Here are the first few. 1, 36, 1225, 41616, 1413721, 48024900.
(x-1)(x+1) = 2 y*y has a feasible solution for (x,y) in integral domain, for odd x only .
Writing x= 2n+1 one gets
n(n+1) = y*y .
This essentially means y*y lies in between two square numbers n^2 and (n+1)^2.
Hereby no feasible solution is there to x^2 -2*y^2 = 1 for integral (x,y)
Nope, you lost the 2 on the way. It is 2n(n+1)=y*y
The unity fundamental of Z[√2] is 1+√2 but (1+√2)*(1-√2)=-1. That's the Norm of 1+√2 is (-1) then "the unity fundamental" with Norm 1 is (1+√2)^2=3+2*√2.
Then the unities in Z[√2] with Norm 1 are the form +-(3+2*√2)^n with n\in Z.
But if x^2-2*y^2=1 then (x+y*√2)*(x-y*√2)=1. That is x+y*√2 is a unity of Z[√2] with Norm 1.
Then if x_n=(1/2)*((3+2*√2)^n+(3-2*√2)^n) and
y_n=(1/(2*√2))*((3+2*√2)^n-(3-2*√2)^n).
Then (+-x_n,+-y_n) are solutions where n\in Z.
If you allow x,y to be composite, how do you generate the solution set?
Good question! That will be the topic of another video.
Es mucho más didactico y atractivo el hallazgo de Fermat en el trabajo de Nicómaco al respecto.
To me, this was an easy one. I figured that x=3and y=2 fast. Thanks!
You're welcome!
I did it in my head.
Good morning! If they are prime numbers it is easy. x must to be odd. If you use mod4, cleary you see that y is even as we do not have squares gongruent to 3 mod4 for odd numbers, only congurents to 1. And then y can be only 2. By the way, where to study pell's equation? If you replace x^2 for a and y^2 for b we have that a=9+ 2K and b= 4 + k ; k integer. Is it possiple that a and b are both perfect squares for any k0?
thankyou sir
0:35 "in this particular equation, we're looking for prime solutions" - that came out of nowhere. How did you establish that we would only look for prime solutions?
That was the plan but I didn't make it clear before the start
That's why I found (1,0) , (3,2) as solutions before hearing that sentence
Thnx for pells equation i want it
Pell's Equations are fun!
what is the method called?
is it correct using of congruence
It should be
2y^2 = (X+1)(x-1)
If y=2 then easy to show LHS factors as 4*2 so X=3.
If y is not 2 then (x-1)=2 and (X+1)=y^2. However this gives the same solution.
What about (1,0)?
It works but we are looking for prime numbers
Thank you for the explanation. Please Teacher, could you help me with this following exercise? Solve the equation in the integers: x^2- 5y^2= 5.
You're welcome!
Good question. We can solve this by using the following
x^2 - 5y^2 = 5 ... (x, y) = (5, 2)
x^2 - 5y^2 = 1 ... (x, y) = (9, 4)
x + y*sqrt(5) = (9 + 4sqrt(5))^n*(5+2sqrt(5)) ... n is an integer
Also check this out:
www.wolframalpha.com/input?i2d=true&i=Power%5Bx%2C2%5D-5Power%5By%2C2%5D%3D5+solve+for+integers
@@SyberMath thank you very much teacher
Helped me a lot
x^2=412y-3 can you make a video for this or somenthing like one side is linear and other side is quadratic diophantine equantions?
Sure! Why not?
x^2 = 412·y - 3 x^2 (mod 412) = -3 (mod 412). Thus you only need to consider x = 0, 1, ..., 206 (mod 412) and check the solutions to the above equation of modular arithmetic. For x = 0, ..., 14 (mod 412), there is trivially no answer, because 14^2 < 206. Equations where one side is linear and the other is quadratic can always be reduced to solving elementary quadratic equations in modular arithmetic.
@@angelmendez-rivera351 That is correct!
I did x-1*x+1 and then factored 2y^2 to get the prime solutions
Can you elaborate?
x^2 - 2*(y^2) = 1
x^2 - 1 = 2*(y^2)
(x - 1)(x + 1) = 2*(y^2)
Since y is prime, there are two ways to factor this. 2*y and y, or 2 and y^2.
First version. 2*y > y, and x + 1 > x - 1, so 2*y = x+1 and y = x-1.
2*y - y = (x + 1) - (x - 1)
y = 2
Putting it back into x^2 - 2*(y^2) = 1, we get x = 3.
Second version. y^2 > 2 for any prime y. Since x+1 > x-1, we get y^2 = x+1 and 2 = x-1.
Since 2 = x-1, we get 3 = x. y^2 = 4, so y = 2. Same solution.
can you please answer me what is the solution for the other case when x^2-2y^2=-1 i found that 5 and 7 are solutions but i cant prove that it is the only solution
Thanks but how do you explain that 17²-2*12² = 1?? So (17;12) is another solution for the equation, isnt it ?? Maybe I made a mistake
You're welcome! That's another solution but in this case we are looking for prime numbers. I will make a separate video for the general case...
Since, x and y are both even, so (-3,-2) also solution for given problem
Can we solve this way. X^2- 2y^2=1 or (x)^2-(√2y)^2=(x+✓2y)(x-√2y)=1or (x+√2y)=+-1 and (x_√2y)=+-1 as well. solving this we get x=+-1 and y=0
I over complicated this one so much
(capital x = x^2, same for y)
X - 2Y = 1
(x + 2y)(x - 2y) = 1
X - 2Y = 1
2Y = X - 1
Y = (X - 1)/2
x - 2y = 1/(x + 2y)
x -(X - 1) = 1/(x + X - 1)
-X + x + 1 = 1/(X + x - 1)
(-X + x + 1)(X + x - 1) = 1
-x^4 -x^3 + x^2 + x^3 + x^2 -x +x^2 +x -1 = 1
-x^4 + 3x^2 - 1 = 1
3X -(X + i)(X - i) = 1
3X -X(1 + i)(1 - i) = 1
X(3 - (1 + i)(1 - i)) = 1
X(3 -2) = 1
X = 1
x = +-1
1 - 2Y = 1
2Y = 0
Y = 0
X=1,y=0 is also a solution set
yes but we are looking for prime solutions
This equation is not a Pell's equation because he not discovered it, Euler Mistakenly named his name for this equation
Brother can you please take this equation for a video
Its a iit jee question and I can't figure it out.
Q;- integrate x^2/{1-x^4√(1+x4)} dx
Please make a vid on it brother ❤️
Will try
@@SyberMath great 🤩
Is 1-x^4 in parentheses like this?
integrate x^2/{(1-x^4)√(1+x4)} dx
@@SyberMath yes onlyx^2 is numerator and rest are denominator 🙏
Positif integers or all integers ??????
Some integers 😁
Nice and easy problem 😍😍
2X-4y=1
X=2+1=3
X=3
y=X-1=2
y=2
only even prime is 2? wow is the proof long? can you summerize it?
I'm not sure if I can summerize it but maybe I can winterize it!
😉😊😁
To be serious, the only factors of a prime number are itself and 1, by definition. All even numbers have 2 as a factor, by definition. So an even prime number must have itself equal to 2. Q.E.D.
x=3 and y=2. Without pen
I got curious as to what the other natural solutions might be and found out that (3,2) is actually the only one.
Yes!
@@SyberMath No.
Another method is to work in Z[sqrt(2)]
Can you elaborate?
👍
X is 3, y is 2
That's right!
Solved in first 20 seconds ( I am just an average from india)
@@tushar1198 No, you are not average! 😁
over the integer there are
6 couples (1.0) (-1,0) (3,2) (-3,-2) (-3,2) (2,-3)
There are infinite number of integer solutions. For every solution (x;y), (4y^2+1;2xy) is also a solution, you can try this with (3;2) for example and find out that (17;12) is also a solution
Or it dosent matter
It's so mediocre.
Why?