sudarshan wrote:The second problem (easier version) reduces to finding all positive numbers "a" such that:
2*a^2+1 is a perfect square
2*a^2-1 is a perfect square
For any such a:
for the first case, y=2*a^2, x=a*sqrt(2*a^2+1)
for the second case, y=2*a^2-1, x=a*sqrt(2*a^2-1)
Now if only I knew how to find all those "a" values in a general way.
<The pattern (0.0)(1,1),(6,8),(35,49)(204,288) .... etc...>
But why is this pattern this way?
Now the pattern, say for x is given by x_n = 6*x_(n-1) - x_(n-2)
In other words, if first value of x=0, and second value = 1 , then the
next one is 6*1 - 0 =6
Next value is 6*6 -1 = 35
Next one is 6*35 - 6 = 204
Next one is 6*204 - 35
And so on..
(This is not hard to prove, and I will show it fairly soon later in the post - proof can be seen in any good book too - look up "Pell's equation" (the correct name ought to be Brhamgupta equation) also there are problems here in Brf dhaga similar to this where the proof is given ).
The interesting anecdote - some one asked Ramanujan a similar problem while he was taking a walk - Ramanujan, of course, answered it immediately. The answer was many digits long yet he answered it almost right-away- without even a paper or pen. His friend was really surprised (his friend was also quite a good Indian mathematician who was also studying in Cambridge and heard that problem) to see this. He asked how did Ramanujan do it so quickly - (The standard method using Pell's equation does require more than a few seconds) .. Ramanujan answered - " He knew the answer is related to continued fraction of sqrt(2).
Now sqrt(2) = 1 / (2+1 /( 2+ 1/ + ---) or:[url]<see graphics here>[/url]or wiki for article on continued fraction. (eg: https://en.wikipedia.org/wiki/Square_root_of_2#Continued_fraction_representation
Or IOW: the approximation for sqrt(2) by expanding this continued fraction, you get the fractions ..
are 1/1, 3/2, 7/5, 17/12, 41/29, 99/50
Each fraction in the series is a better approx of sqrt(s)
Now if you multiplied denominator and numerator (eg 3*2=6, 7*5=35, 17*12 you get successive values of x)
So above is the continued fraction method.
(Intuitively - as Ramanujan will say . you if you want 2*a^2 +-1 = b^2 the ratio of b and a is nearly sqrt(2)..so if working with integers, the continued fraction of sqrt(2) is the way to go).
Now if you want explicit formula here choose any integer nx = ((3+2 √ 2)^n - (3 - 2 √ 2)^n) / (4 √ 2)
y = ((3+2 √ 2) ^n + (3 - 2 √ 2) ^n - 2)/ 4
It's easy to put in the computer, (easy to verify that x and y both are integers and has the required property).
Hope people find it fun,