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Beginner Hint 1 (AI-translated)

\n\n- Unlike one-p-rsa, the exponent `e` is changed from `65537` to `65538`.\n- As a result, the naive solution for one-p-rsa no longer works.\n- First, think about why the original approach stops working.\n\n

\n\n

Beginner Hint 2 (AI-translated)

\n\n- In one-p-rsa, `p-1` and `e` were coprime, so it was possible to compute the inverse of `e` modulo `p-1`.\n- This challenge is different. Since `e` factors as `2 * 3^2 * 11 * 331`, `gcd(e, p-1)` is even. It also has about a 1/2 chance to be a multiple of 3, because `p-1 mod 3` is either 0 or 1.\n- In this case, how can we solve it?\n- By leveraging [SageMath](https://github.com/alpacahack/resources/blob/main/resources/SageMath.md), you can solve it more easily.\n\n

\n","descriptionJa","[one-p-rsa](/daily/challenges/one-p-rsa) がやられたようだな……\n\n

初心者向けヒント1

\n\n- one-p-rsa とは異なり、 `e` が `65537` から `65538` になっています。\n- これにより one-p-rsa の素朴な解法では解けなくなっているようです。\n- まずは、なぜ解けなくなったかを考えてみましょう。\n\n

\n\n

初心者向けヒント2

\n\n- one-p-rsa では `p-1` と `e` が互いに素だったので、法 `p-1` での `e` の逆元を求めることが可能でした。\n- この問題では違います。 `e` は `2 * 3^2 * 11 * 331` と素因数分解できるため、 `gcd(e,p-1)` は偶数です。また 3 の倍数になる確率も `p-1 mod 3` が 0 or 1 であることから約 1/2 もあります。\n- このような場合、どうすれば解くことができるでしょうか。\n- [SageMath](https://github.com/alpacahack/resources/blob/main/resources/SageMath.md) を活用するといくらか楽に解けます。\n\n

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