I solved it by using someones pseudocode and to be honest I don't completely understand how it works

projecteuler/064/main.cpp

@@ -0,0 +1,76 @@
+/*
+Find the number of continued fractions for the square roots of n <= 10000 which have an odd period
+*/
+
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int gcd(int a, int b){
+    if(b == 0){
+        return a;
+    }
+    return gcd(b, a % b);
+}
+
+int gcd(int a, int b, int c){
+    return gcd(a, gcd(b, c));
+}
+
+// Pseudocode found in the video on continued fraction of square roots of integers https://www.youtube.com/watch?v=GFJsU9QsytM
+
+bool check(int n){
+    float x = sqrt((float)n);
+
+    if(x != sqrt(n)){
+        vector<int> cf;
+        int a, b = 1, c = 1, d = 0;
+        int bn, cn, dn, g;
+        int b1, c1, d1;
+
+        for(int i = 0; ; ++i){
+            a = floor((floor(b * x) + d) / c);
+            cf.push_back(a);
+
+            bn = b*c;
+            cn = b*b*n - d*d - a*a*c*c + 2*a*c*d;
+            dn = a*c*c - c*d;
+
+            g = gcd(bn, cn, dn);
+
+            b = bn / g;
+            c = cn / g;
+            d = dn / g;
+
+            if(i == 0){
+                b1 = b;
+                c1 = c;
+                d1 = d;
+            } else if(b1 == b && c1 == c && d1 == d){
+                break;
+            }
+        }
+
+        return cf.size() % 2 == 0;
+    }
+
+    return false;
+}
+
+int main(){
+
+    cout << "Hello this is Patrick" << endl;
+    auto start = chrono::high_resolution_clock::now();
+
+    int result = 0;
+
+    for(int n = 1; n <= 10000; ++n){
+        result += check(n);
+    }
+
+    cout << result << endl;
+
+    auto duration = chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - start);
+    cout << (float)duration.count()/1000 << endl;
+    return 0;
+}