Question

If x and y are prime numbers where unity equals the the square of the prime number "x" minus the twice of the square of the prime number y , then find the prime numbers x and y

Answer 1

@Kainui

Answer 2

this was the question i was working on

Answer 3

x^2-2y^2=1 Ohhhh I see. This is similar to that other one you were working on a week or two ago that one was great fun but it was p^2-q^2=r and what are all the prime numbers that satisfy this equation I believe, right?

Answer 4

I have a bad memory lol , i don't remember

Answer 5

I was gonna say if it was simply x^2-y^2=1 then we can pretty quickly guess x=3 and y=2 haha.

Answer 6

Yeah hehe

Answer 7

y must be 2 though.

Answer 8

for this question as well. Interesting.

Answer 9

then x = 3

Answer 10

x^2 -1 = 2y^2

(x-1)(x+1) = 2y^2

clearly x needs to have an odd parity and (3, 2) is one parituclar solution

For a general method for finding all the solutions of any equation of form `x^2-ny^2=1`, you need to study Pell's equation :
http://en.wikipedia.org/wiki/Pell's_equation

Answer 11

I did it without pell's equation

Answer 12

wait I had guessed the right answer I had been thinking

3^2-2^3=1

Answer 13

how exactly you did it and are you sure (3,2) is the only possible solution ?

Answer 14

Trial and error works for small powers and small number though , but good in MCQ's

Answer 15

@rational yes

Answer 16

good, make me feel sure too :) cuz i don't see yet how (3,2) is the only possible solution :/