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I was supposed to write a simple java app which finds all x's and y's that satisfy x^2 - ny^2 = k, where x, y >= 0 and x, y <= k Other conditions are : (x,y) are Natural numbers and 'n' is a square number. At first, I tried to work the equation out on paper and it was easy to figure out solutions for prime numbers, that only have two divisors, but I couldn't understand how could you figure out all of the variables I mentioned above when 'k' can be any Natural number you can think of without using brute force. After a little research, I used this approach. Set 'Pn' has all 'x's' and 'y's' that satisfy the equation x^2 - ny^2 = k, while x and y are Natural numbers and n is a square number. Set 'A' has divisors of k (i) and quotients (j) that come from dividing k by divisor (i) which is inside set 'A'. We don't take all of the divisors of k (for ex. k = 9009), our largest divisor of 'k' inside set 'A' is the one closest to the square root of 'k'. A square root of 9009 is approximately 95, closest number to 95 which divides 9009 is 91, therefore, the divisors of k, that are inside set 'A' are: i{1, 3, 7, 9, 11, 13, 21, 33, 39, 63, 77, 91}; Now, let's divide 9009 by all of the divisors above and get a final set 'A', that has all of the factors needed in further calculations: { (1; 9009) , (3; 3003), (7; 1287), (9;1001), (11;819), (13;693), (21;429), (33;273), (39;231), (63; 143), (77;117), (91;99) } (In the picture, I've missed (7; 1287) element, sorry). Now, let's figure out values of 'x' and 'y'. For example we'll only take first two elements from set 'A', and those are : { (1;9009), (3; 3003) }. x = ( j + i ) / 2; y = ( j - x ) / (square root of 'n', in our case n = 4, n = d^2, d = 3) = (j-x) / 3 Proof: x^2 - 9y^2 = (x+3y)(x-3y) = 9009, create a system of equations x + 3 y = j = 9009 x - 3 y = i = 1 combine x + 2y + x + (-2y) = 9010, 2x = 9010, x = 9010 / 2 = 4505 x = ( j + i ) / 2 = ( 9009 + 1 ) / 2 = 4505 we've found 'x', now, let's find 'y' 4505 + 3y = 9009 3y = 9009 - 4505 = 4504 y = 4504 / 3 = 1501.3333333333.... y = ( 9009 - 4505 ) / 3 = 1501.3333333333... These values cannot be included in our 'P9' set because 'y' is not a Natural number. Let's try out the second element in our 'A' set, which is (3, 3003) x = (3003 + 3) / 2 = 1503, y = (3003 - 1503) / 3 = 500, (1503)^2 - 9*(500)^2 = 2259009 - 2250000 = 9009 (x = 1503, y = 500) are the values that are suitable for 'P9' set. Interesting property: If you combine 9009 and 1 and divide it by n (n = 9) you won't get a Natural number, and the x and y values are not inside 'P9' set whereas you do the same operation with 3003 and 3, you'll get the opposite result: ( 9009 + 1 ) / 9 = 1001.11111 - > not a Natural number, 'x' and 'y' do not belong to 'P9' (3003 + 3 ) / 9 = 334 - Natural number, 'x' and 'y' belong to 'P9' Let's try a different value of 'k' and 'n': k = 16, n = 4. x^2 - 4y^2 = 16; i { 1, 2, 3 ,4} A{ (1, 16), (2, 8), (4, 4) } Let's use that "interesting property" 16+1 / n = 17 / 4 = 4.25 - not a Natural number, 2+8 / n = 10 / 4 = 2.5 - not a Natural number, 4+4 / 4 = 2 - Natural number, x = ( j + i ) / 2 = ( 4 + 4 ) / 2 = 4, y = ( j - x ) / (square root of 'n') = (4-4) / 2 = 0, 4^2 - 4*(0)^2 = 16, (x = 4, y = 0) are inside P4 set. Hope it made some sense, I've added java code inside a text file too. My question wasn't correct in the first place, but I still appreciate your contribution, thank you. equation.txt

Hello everyone, Im new in this forum and in advance I would like to apologise for possibly posting this thread in a wrong place. I am learning mathematics and I came across this problem that I can't find a solution for: x^2 - 4^2 = 1000000 , find all possible integer solutions for x and y. Being an amateur mathematician I tried to understand this problem using system of equations technique and instead of million i used a prime number (5) as an anwser and turned this equation into difference of squares: (x+2y)(x-2y) = 5; 5 is prime, has two divisors -> 5, 1; (x+2y) = 5 (x-2y) = 1 2y - 2y = 0 (y is out) 2x = 6 x = 3; 3+2y = 5 2y = 2 y = 1; Anwser : x = 3, y = 1; I tried to do similar thing with a million: (x+2y)(x-2y) = 1000000 let's say (x+2y) = 1000 and (x-2y) = 1000 (x+2y)= 1000 (x-2y) = 1000 2x = 2000 x=1000 y = 0; But how do you find other solutions to this equation ? I found these anwsers using brute force algorithm in java: [[31258, 15621], [12520, 6240], [6290, 3105], [2600, 1200], [1450, 525], [1000, 0]] is there a formula for finding all of the Integer possibilities? Where should I look ? Thank you