A Quick and Easy Diophantine Equation

This video is about solving an interesting Diophantine Equation
If you need to post a picture of your solution or idea:
#NumberTheoryProblems #DiophantineEquations

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[TC]

This is a really underrated channel, I hope to see you blow up in the math community sometime. I've been watching your videos for a while, and only now noticed the number of subscribers after noticing that there were only 70 likes on this video. I'm telling you, you'll get at least 20k by March

[richardryan]

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.

[infintiy_math]

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.

[MelomaniacEarth]

DO U DAILY WORK ON THIS STUFF IS INSANE FOR EVERYTHING

[joaquingutierrez]

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

[RexxSchneider]

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.

[Rudxain]

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"

[cav]

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.

[smailmoudoub]

(1, 0) is also a solution. In fact, this équation have an infinity of solutions.

[danielvieira]

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.

[crazyhitman]

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

[aashsyed]

best video! love it :)
very cool
i hope you get way more popular way faster

[italixgaming]

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.

[davidseed]

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.

[nasrullahhusnan]

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)

[Ssilki_V_Profile]

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.

[laifiyazid]

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

[elkincampos]

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 and

Then (+-x_n, +-y_n) are solutions where n\in Z.

[alihachicha]

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

[ramaprasadghosh]

(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)