> Hi, > > 5x^2 + 20 = y^2 > > I can get valid integer values for x and y using > brute force (1, 5), (4, > 10), (11, 25) etc. > > I've been going through the work on this page... > > http://www-history.mcs.st-andrews.ac.uk/HistTopics/Pel > l.html > > which covers the work of Brahmagupta and Bhaskara II > on Pell type equations. > I can't seem to figure if this is any use for > calculating integer roots for > the above equation. > > Any help would be greatly appreciated. >
Put your equation as y^2 - 5*x^2 = 20. You have the initial solution y=5, x=11.
A solution to y^2 - 5*x^2 = 1 is y=9,x=4. So a list of solutions to your eqn is obtained by choosing some positive integer n and "multiplying out" the expression (5 - sqrt(5))*(9-4*sqrt(5))^n. Then put y=integer part and x=integer before the sqrt(5).
First of these is y=65, x=29.
Note there are more than the solutions obtained this way.