Solve problem 123.

python/e123.py

@@ -0,0 +1,40 @@
+from lib_prime import prime_nth
+
+
+def varray(i):
+    return i * 2 + 1
+
+
+def rem(i):
+    n = varray(i)
+    p = prime_nth(n)
+    return (pow(p - 1, n) + pow(p + 1, n)) % (p * p)
+
+
+def bin_search(lo, hi, target):
+    if hi - lo == 1:
+        return varray(hi)
+    i = lo + ((hi - lo) // 2)
+    r = rem(i)
+
+    # print(f"{i=:<6} {r=}")
+    if r < target:
+        return bin_search(i, hi, target)
+    elif r > target:
+        return bin_search(lo, i, target)
+    assert False
+
+
+def euler_123():
+    """
+    I found out that when n is even, the remainder is always 2. When the
+    remainder is odd, the series grows monotonically. That means we can use a
+    binary search on odd values to find the solution.
+    """
+    return bin_search(1000, 30000, 10**10)
+
+
+if __name__ == "__main__":
+    solution = euler_123()
+    print("e123.py: " + str(solution))
+    assert(solution == 21035)