Is this really solvable?
x 2 − 8 y 2 = 1 given x and y are positive integers that satisfy the equation above, find the 5th least value of x .
The answer is 3363.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
well i know this.what if x2-8y2=3 solve this
Log in to reply
There would be no solutions as x 2 − 8 y 2 ≡ 0 , 1 , 4 m o d ( 8 ) . This can be obtain by considering that 8 y 2 ≡ 0 m o d ( 8 ) and by testing values of x from 0 to 7.
Log in to reply
However, this is a particular case so I will provide an example for x 2 − 8 y 2 = 4 : x 2 = 8 y 2 + 4 ⇒ 4 ∣ x 2 l e t x = 2 k . . . . k 2 − 8 y 2 = 1 Then we can use the results above. Generally, there exist many equation of the form x 2 − N y 2 = k that can be reduced to x ′ 2 + N y ′ 2 = 1 by using modular arithmetic or parity (even/odd).
well ,could you please explain what if in the above question,the result is not 1,but 3
Log in to reply
Its generalized Pell's equation x 2 − N y 2 = k and you can search to find different solutions for it.
Log in to reply
how, what if my equation is x2-8y2=3
Log in to reply
@Kaustubh Miglani – I already mentioned that in my comment. That's generalized Pell's equation for k = 3 . There are lots of solutions for it. Look for example .
Log in to reply
@Kazem Sepehrinia – could u please pleaseplease exlain.i am not at all able to understand it.please
Log in to reply
@Kaustubh Miglani – I'll try to provide a solution Asdfgh.
Log in to reply
@Curtis Clement – please be quick.i would be really thankful
Firstly, I will reduce this Pell's equation as follows: l e t y ′ = 2 y : . . . x 2 − 2 ( y ′ ) 2 = 1 . Now we must find the "fundamental" (first) solution which is (x, y') = (3,2). From there we can use a recurrence relation for x 2 − 2 y 2 = 1 (without proof): x k + 1 = a x k + 2 b y k ′ . . . . y k + 1 ′ = b x k + a y k ′ w h e r e a 2 − 2 b 2 = 1 Using (a,b) = (3,2) gives: x k + 1 = 3 x k + 4 y k ′ . . . y k + 1 = 2 x k + 3 y k ′ Using these recurrence yields: x 2 = 3 x 1 + 4 y 1 ′ = 1 7 a n d y 2 = 2 x 1 + 3 y 1 ′ = 1 2 . Continuing this process yields x 5 = 3 3 6 3