Solve : $x^2-92 y^2=1$ As some of you might know,this is Brahmagupta's equation . How to find solution for this ? 
I mean integral solution? How to solve it using programming ? 

I tried something like $x^2=1+92y^2$

$x=\sqrt{1+92y^2}$

Use brute force approach to check for every y ? Is there any better answer ?
I think the most common systematic way to solve this type of problem is using continued fractions. I'll reiterate @Quimey's suggestion to refer to Wikipedia for Pell's Equation
and specifically the section "Fundamental solution via continued fractions" and the Lenstra paper cited there.

In this case as a periodic continued fraction
$$
\sqrt{92} = [9;1,1,2,4,2,1,1,18,1,1,2\ldots] = 9+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{4+\cfrac{1}{2+\cdots}}}}}
$$
and after 7 terms we get the approximation $\sqrt{92}\simeq 1151/120$ which gives the fundamental solution.

As for implementing an algorithm for finding the continued fraction for square roots, you should be able to find resources online with a search, but can start by taking a look at this question.
def pr_sol(maxval,N,ee,ff):
    count=0
    aa,bb=0,1
    gg=N*ff
    while count&lt;maxval:
        if bb**2 - 1 ==N*aa**2:
            print(bb,'^2-1=',N,'*',aa,'^2')
        else:
            print(bb,'^2+1=',N,'*',aa,'^2')
        anew=ee*aa+ff*bb
        bnew=gg*aa+ee*bb
        aa=anew
        bb=bnew
        count+=1
  return 

pr_sol(20,92,1151,120) #print first 20 solutions when N=92
&quot;&quot;&quot; the smallest solution is 1151,120 where
1151 ^2-1= 92 * 120 ^2  (from wikipedia).
(I'm not sure if it's the same as wikipedia recursion,
but in this case the following recursion applies:
a_{t+1}=1151*a_{t} +120*b_{t}
b_{t+1}=120*92*a_{t} +1151*b_{t}
(in general you can generate such a recursion, which gives 
values for x^2-1=Ny^2  AND x^2+1=Ny^2 
(but x^2 ==91 mod 92 has no solutions so
there are no sol to x^2+1=Ny^2 ))

The first few solutions are:
1 ^2-1= 92 * 0 ^2  (a,b)=( 0 , 1 ) 
1151 ^2-1= 92 * 120 ^2 (a,b)=( 120 , 1151 )
2649601 ^2-1= 92 * 276240 ^2 (a,b)=( 276240 , 2649601 ) 
6099380351 ^2-1= 92 * 635904360 ^2 
14040770918401 ^2-1= 92 * 1463851560480 ^2 
32321848554778751 ^2-1= 92 * 3369785656320600 ^2
I've just been doing some investigation into these equations
and found this recursion. (which has some advantage. For 
example N=313 has very large first solution to x^2-1=Ny^2 but 
you can generate the numbers from the much smaller solution:
126862368^2 +1 =313 *  7170685^2
and the recursion: 
a_{t+1}= 126862368 a_t+7170685 b_t
b_{t+1}=313*7170685a_t+126862368  b_t
so 
pr_sol(20,313,126862368,7170685) 
prints first 20 solutions for N=313 
(note they alternate between x^2+1=313y^2 and x^2-1=313y^2)

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