# I have got a very messy solution of pell's equation. What to do?

I was having free time, and was having no books around, so I tried to find a solution to Pell's equation by myself to pass time. But then I really got a strange result regarding the integer solutions of the equation. x^2 – 1 = ny^2 -------- (1) So, (x/y)^2 – (1/y)^2 = n Taking (1/y)^2 as k, this turns out the equation as (x^2)/k = n + k So, x^2 = k^2 + nk Thus, K^2 + nk – x^2 = 0 By the quadratic formula, the first value of k turns out to be imaginary, but inspecting the second value of k, and converting it in the form of y^2, it turns out to be, y^2 = 2/(n + √(n^2 + 4x)) and eventually, ny^2 + (y^2)( √(n^2 + 4x)) = 3 substituting value of ny^2 from (1), we get x^2 – 1 + (y^2)( √(n^2 + 4x))=2 This implies, x^2 + y^2(√(n^2+4x))=3 Now taking √(n^2 + 4x) as z , and also noticing that minimum value of both n and x is 1, the only value of z which fits the equation is 5, and for that x must be equal to 1, so the equation turns out to be, 1 + y√5 = 3 But in this case y cannot be an integer, so it eventually proves that pell’s equation has no integer solutions.

What is wrong with my proof ?

Note by Siddharth Kumar
5 years, 11 months ago

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Okay thnx I got it. I HAVE FINALLY GOT A SOLUTION BY MYSELF!

- 5 years, 11 months ago

In that case, can you edit your discussion and write it up properly so that the rest of us can read and understand what you're saying?

Please double check that you didn't make any careless calculation mistakes. There were several in your post, even allowing for errors carried forward.

Staff - 5 years, 11 months ago

You can start by fixing this step, and then work out the rest and be careful with your algebra:

"(x/y)^2 – (1/y)^2 = n Taking (1/y)^2 as k, this turns out the equation as (x^2)/k = n + k"

Specifically, note that if 1/y^2 = k, then (x/y)^2 = x^2 * k not x^2/k.

- 5 years, 11 months ago

plzz help me

- 5 years, 11 months ago

Staff - 5 years, 11 months ago

@Siddharth K. Can you type your proof in LaTex, I'm confused

- 5 years, 11 months ago

You made the mistake on the 4th line by writing $$\frac{x^2}{y^2}$$ = $$\frac{x^2}{k}$$, rather than the correct substitution, $$\frac{x^2}{y^2}$$ = $$x^{2}$$k. If you interested in learning how the equation is derived and why it works then visit the wiki page I created: https://brilliant.org/wiki/quadratic-diophantine-equations-pells-equation/ Also hover your mouse over my equations to see how I formatted them - that should help :)

- 3 years, 11 months ago