Algorithm for integer solutions of a circle?

I am trying to search for integer solutions to the equation:

y^2 + x^2 = 2n^2

If I search this in wolfram alpha, they are all found almost immediately even for very large n. When I implemented a brute force approach it was very slow:

def psearch(n, count):
    for x in range(0, n):
        for y in range(0, n):
            if x*x + y*y == 2*n**2:
                print(x,y)
                count += 1
    return count

So I assume there is a much faster way to get all of the integer solutions to the equation above. How can I do this in python so that it will have much lower runtime?

Note: I have seen this question however it is about finding lattice points within a circle not the integer solutions to the equation of the circle. Also I am interested in finding the specific solutions not just the number of solutions.

Edit: I am still looking for something an order of magnitude faster. Here is an example: n=5 should have 12 integer solutions to find what those should be search this equation on Wolfram alpha.

Edit 2: @victor zen gave a phenomenal answer to the problem. Can anyone think of a way to optimize his solution further?

Answer

In your algorithm, you’re searching for all possible y values. This is unnecessary. The trick here is to realize that

y^2 + x^2 = 2n^2

directly implies that

y^2 = 2n^2-x^2

so that means you only have to check that 2n^2-x^2 is a perfect square. You can do that by

y2 = 2*n*n - x*x 
#check for perfect square
y = math.sqrt(y2)
if int(y + 0.5) ** 2 == y2:
    #We have a perfect square.

Also, in your algorithm, you are only checking x values up to n. This is incorrect. Since y^2 will always be positive or zero, we can determine the highest x value we need to check by setting y^2 to its lowest value (i.e 0). Consequentially, we need to check all integer x values satisfying

x^2 <= 2n^2

which reduces to

abs(x) <= sqrt(2)*n.

Combine this with the optimization of only checking the top quadrant, and you have an optimized psearch of

def psearch(n):
    count = 0
    top = math.ceil(math.sqrt(2*n*n))
    for x in range(1, top):
        y2 = 2*n*n - x*x 
        #check for perfect square
        y = math.sqrt(y2)
        if int(y + 0.5) ** 2 == y2:
            count+=4
    return count