100 points. Python 3

import math,itertools,bisect

lookup=[0]*(5*10**6+1)
for v,u in itertools.combinations(range(1,int((5*10**6)**0.5)+1,2),2):
        if u>v and math.gcd(u,v)==1:
            s=u**2+u*v
            for i in range(s,5*10**6+1,s):
                lookup[i]+=1
ans=[i for i in range(5*10**6+1) if lookup[i]==1]
t=int(input().strip())
for _ in range(t):
    n=int(input().strip())
    count=bisect.bisect_right(ans,n)
    print(count)

ALERT:

Please don't use Pythagorean Triplets Generate algorithm in Geeks for geeks. Those codes are wrong. Please use this StackExchange instead

you can find my java solution here

any one can send code for this answer

Lazy solution to dumb test cases: precalculate all the counts up to N=5e6 then return the corresponding value for each test case. You really should put more effort (or at least some effort) into reasonable/decent/interesting requirements.