FA-5526 / Number theory / Open access

Radical distinct prime product · case 01

Repeated prime powers are retained in the radical.

Verified by executionVariant 1 · 4 checks per implementationDownload source bundle ↓JSON ↗

ROOT CAUSE

Repeated prime powers are retained in the radical.

VERIFIED REPAIR

Apply the specified mathematical contract directly, preserving all terms and boundary cases: return math.prod(p for p in range(2,n+1) if n%p==0 and all(p%d for d in range(2,math.isqrt(p)+1)))

Unsuccessful approach: Multiplying all divisors counts composite factors and repeated prime contributions.

Case contract

Integer inputs; n is positive unless a zero or negative fixture explicitly extends that operation. A modulus is greater than one; p is prime; exponent k may be signed when an inverse exists. Radical distinct prime product. Exact operational definition: math.prod(p for p in range(2,n+1) if n%p==0 and all(p%d for d in range(2,math.isqrt(p)+1)))

Why this case matters

Small exact fixtures expose this error without platform timing, external services, or probabilistic observations. Number theory results depend on the stated convention.

1 / The failure

Exit 1

"""Failure Map reference implementation. Python standard library only."""
import json
import math
import calendar
import statistics
import itertools
from fractions import Fraction
from datetime import date, datetime, timedelta, timezone
from decimal import Decimal, ROUND_HALF_UP, ROUND_DOWN, ROUND_CEILING, ROUND_FLOOR

N = 1
observations = []
def solve(n):
    return n
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('fixture 1: (1,)', solve(*(1,)), 1)
check('fixture 2: (12,)', solve(*(12,)), 6)
check('fixture 3: (8,)', solve(*(8,)), 2)
check('fixture 4: (15,)', solve(*(15,)), 15)
print(json.dumps({"observations": observations, "passed": all(x["passed"] for x in observations)}, ensure_ascii=False))
raise SystemExit(0 if all(x["passed"] for x in observations) else 1)

Boundary fixture	Actual	Expected	Outcome
fixture 1: (1,)	`1`	`1`	Passed
fixture 2: (12,)	`12`	`6`	Failed
fixture 3: (8,)	`8`	`2`	Failed
fixture 4: (15,)	`15`	`15`	Passed

SHA-256 / 72e0f2113902e4c8e57cad448d5e0575e230a4bfccb23f64a9c4c30bb3624454

2 / The unsuccessful fix

Exit 1

"""Failure Map reference implementation. Python standard library only."""
import json
import math
import calendar
import statistics
import itertools
from fractions import Fraction
from datetime import date, datetime, timedelta, timezone
from decimal import Decimal, ROUND_HALF_UP, ROUND_DOWN, ROUND_CEILING, ROUND_FLOOR

N = 1
observations = []
def solve(n):
    return math.prod(p for p in range(2,n+1) if n%p==0)
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('fixture 1: (1,)', solve(*(1,)), 1)
check('fixture 2: (12,)', solve(*(12,)), 6)
check('fixture 3: (8,)', solve(*(8,)), 2)
check('fixture 4: (15,)', solve(*(15,)), 15)
print(json.dumps({"observations": observations, "passed": all(x["passed"] for x in observations)}, ensure_ascii=False))
raise SystemExit(0 if all(x["passed"] for x in observations) else 1)

Boundary fixture	Actual	Expected	Outcome
fixture 1: (1,)	`1`	`1`	Passed
fixture 2: (12,)	`1728`	`6`	Failed
fixture 3: (8,)	`64`	`2`	Failed
fixture 4: (15,)	`225`	`15`	Failed

SHA-256 / b03713ce475d3d4d90eb004ff8799e9051025155f0e0018e9297ab99bfd445a6

3 / The verified repair

Exit 0

"""Failure Map reference implementation. Python standard library only."""
import json
import math
import calendar
import statistics
import itertools
from fractions import Fraction
from datetime import date, datetime, timedelta, timezone
from decimal import Decimal, ROUND_HALF_UP, ROUND_DOWN, ROUND_CEILING, ROUND_FLOOR

N = 1
observations = []
def solve(n):
    return math.prod(p for p in range(2,n+1) if n%p==0 and all(p%d for d in range(2,math.isqrt(p)+1)))
def check(label, actual, expected):
    observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('fixture 1: (1,)', solve(*(1,)), 1)
check('fixture 2: (12,)', solve(*(12,)), 6)
check('fixture 3: (8,)', solve(*(8,)), 2)
check('fixture 4: (15,)', solve(*(15,)), 15)
print(json.dumps({"observations": observations, "passed": all(x["passed"] for x in observations)}, ensure_ascii=False))
raise SystemExit(0 if all(x["passed"] for x in observations) else 1)

Boundary fixture	Actual	Expected	Outcome
fixture 1: (1,)	`1`	`1`	Passed
fixture 2: (12,)	`6`	`6`	Passed
fixture 3: (8,)	`2`	`2`	Passed
fixture 4: (15,)	`15`	`15`	Passed

SHA-256 / 0f894bf65b56285e3b0608bb056b33dc7a081abb55af2d21a14004a1f4fd492f

Verification & scope

This reproducer isolates one failure mechanism. Results cover the supplied fixtures. Variants within a family share a test contract and should remain grouped when constructing evaluation splits. Related mechanisms with a shared evaluation_group must also remain together; these controlled models are not independent production incidents.

Observations recorded using Python 3.12.14 at 2026-09-29T14:37:50.479709+00:00.

Case digest / 6d5c55127784fa0ee0f42a936760161ebd1c313f01fdca8d3a1667d05c61b1cb