eulerbooks/notebooks/problem0075.ipynb

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    "# [Singular Integer Right Triangles](https://projecteuler.net/problem=75)\n",
    "\n",
    "This problem is eerily similar to [problem 39](https://projecteuler.net/problem=39). We'll reuse our generator from that problem:"
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    "from itertools import count\n",
    "\n",
    "def primitive_pythagorean_triplets(max_perim):\n",
    "    for m in count(2):\n",
    "        if 2*m^2 + 2*m > max_perim:\n",
    "            break\n",
    "\n",
    "        for n in range(1, m):\n",
    "            if not ((m % 2) != (n % 2)) or gcd(m, n) != 1:\n",
    "                continue\n",
    "            \n",
    "            a = m^2 - n^2\n",
    "            b = 2*m*n\n",
    "            c = m^2 + n^2\n",
    "            \n",
    "            if a + b + c > max_perim:\n",
    "                break\n",
    "            \n",
    "            yield (a, b, c)"
   ]
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   "source": [
    "And we'll reuse the main loop from that problem, too!"
   ]
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    "from collections import Counter\n",
    "\n",
    "limit = 1500000\n",
    "lengths = Counter()\n",
    "\n",
    "for (a, b, c) in primitive_pythagorean_triplets(limit):\n",
    "    for k in count(1):\n",
    "        L = k * (a + b + c)\n",
    "        if L > limit:\n",
    "            break\n",
    "        \n",
    "        lengths[L] += 1"
   ]
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    "The only differences in this problem are our maximum perimeter, and that we're looking for perimeters that can only be made from one Pythagorean triplet."
   ]
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    {
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      "text/plain": [
       "161667"
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     "execution_count": 3,
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   "source": [
    "len({k for (k, v) in lengths.items() if v == 1})"
   ]
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  {
   "cell_type": "markdown",
   "id": "84d83e3b",
   "metadata": {},
   "source": [
    "## Relevant sequences\n",
    "* Perimeters with one Pythagorean triple: [A098714](https://oeis.org/A098714)\n",
    "\n",
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
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