diff --git a/math/pivot_integer.cpp b/math/pivot_integer.cpp
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--- /dev/null
+++ b/math/pivot_integer.cpp
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+/**
+ * @file pivot_integer.cpp
+ * @brief Find the pivot integer where the sum of numbers from 1 to x equals the sum from x to n.
+ *
+ * This algorithm finds an integer x such that:
+ *     1 + 2 + ... + x = x + (x+1) + ... + n
+ *
+ * The mathematical solution:
+ * Total sum = n * (n + 1) / 2
+ * Pivot x exists only if total sum is a perfect square.
+ *
+ * @author Aditya
+ */
+
+#include <iostream>
+#include <cmath>
+
+/**
+ * @brief Function to compute the pivot integer.
+ * @param n The upper limit of the range [1, n].
+ * @return int The pivot integer or -1 if no pivot exists.
+ */
+int pivot_integer(int n) {
+    long long total = 1LL * n * (n + 1) / 2;
+    long long x = sqrt(total);
+
+    if (x * x == total)
+        return (int)x;
+
+    return -1;
+}
+
+/**
+ * @brief Main function for demonstration.
+ */
+int main() {
+    int n = 8;
+    std::cout << "Pivot integer for n = " << n << " is: " << pivot_integer(n) << std::endl;
+
+    return 0;
+}