Question

How can I solve equation x^2 - 4y - 1 = 0, where x and y are natural numbers.

Answer 1

by natural numbers you refer to 1?

Answer 2

By natural numbers I mean the set of all positive integers.

Answer 3

I found some solutions. But is there any general formula for x and y?

Answer 4

For example x = 3, y = 2 is a solution.

Answer 5

x^2-4y-1=0
4y=x^2-1
x>0,y>0.x&y are natural numbers
\[y=\frac{ x ^{2}-1 }{4 }\]
Least positive value of x which satisfies the above condition is x=3
for x=3,y=(3*3-1)/4=2
add multiple of 4 in x=3 ,we get infinite values of y