problem_123: project euler solution 120

Name: Prime square remainders

Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and
let r be the remainder when (pn−1)^n + (pn+1)^n is divided by pn^2.

For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
The least value of n for which the remainder first exceeds 10^9 is 7037.

Find the least value of n for which the remainder first exceeds 10^10.

Reference: https://projecteuler.net/problem=123

Fixes: #2695

Author: Ravi Kandasamy Sundaram

Diff for: project_euler/problem_123/__init__.py

@@ -0,0 +1,85 @@
+"""
+"https://projecteuler.net/problem=123"
+
+Name: Prime square remainders
+
+Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and
+let r be the remainder when (pn−1)^n + (pn+1)^n is divided by pn^2.
+
+For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
+The least value of n for which the remainder first exceeds 10^9 is 7037.
+
+Find the least value of n for which the remainder first exceeds 10^10.
+
+
+Solution:
+
+n=1: (p-1) + (p+1) = 2p
+n=2: (p-1)^2 + (p+1)^2
+     = p^2 + 1 - 2p + p^2 + 1 + 2p  (Using (p+b)^2 = (p^2 + b^2 + 2pb),
+                                           (p-b)^2 = (p^2 + b^2 - 2pb) and b = 1)
+     = 2p^2 + 2
+n=3: (p-1)^3 + (p+1)^3  (Similarly using (p+b)^3 & (p-b)^3 formula and so on)
+     = 2p^3 + 6p
+n=4: 2p^4 + 12p^2 + 2
+n=5: 2p^5 + 20p^3 + 10p
+
+As you could see, when the expression is divided by p^2.
+Except for the last term, the rest will result in the remainder 0.
+
+n=1: 2p
+n=2: 2
+n=3: 6p
+n=4: 2
+n=5: 10p
+
+So it could be simplified as,
+    r = 2pn when n is odd
+    r = 2   when n is even.
+"""
+
+
+def sieve() -> int:
+    """
+    Returns a prime number generator using sieve method.
+    """
+    D = {}
+    q = 2
+    while True:
+        p = D.pop(q, None)
+        if p:
+            x = p + q
+            while x in D:
+                x += p
+            D[x] = p
+        else:
+            D[q * q] = q
+            yield q
+        q += 1
+
+
+def square_remainder(p, n) -> int:
+    """
+    Returns the square remainder with a simplified expression.
+    """
+    return 2 * p * n
+
+
+def solution(limit: int = 1e10) -> int:
+    """
+    Returns the least value of n for which the remainder first exceeds 10^10.
+    """
+    primes = sieve()
+
+    n = 1
+    while True:
+        p = next(primes)
+        if square_remainder(p, n) > limit:
+            return n
+        # Ignore the next prime as the reminder will be 2.
+        next(primes)
+        n += 2
+
+
+if __name__ == "__main__":
+    print(solution())