n libre es demasiado

Motivación:

a + n\,b = k(n + 1)

a + n\,b \equiv 0 \pmod{n + 1}

a + (-1)b \equiv 0 \pmod{n + 1}

a - b \equiv 0 \pmod{n + 1}

Si elegimos n + 1 = a - b, siempre hay uno que lo cumple.

a + (a - b - 1)b = a + a\,b - b^2 - b = (a - b) + (a - b)b = (a - b)(b + 1)

Por eso es una condición inútil 🙂

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