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Ukrainian Mathematical Olympiad

Number theory Difficulty 5.3 AIME, harder Prove it Ukraine

Prove that for any positive integer nn the number 5n(5n+1)−6n(3n+2n)5^n (5^n + 1) - 6^n (3^n + 2^n) is divisible by 9191.

Solution

Позначимо значення даного виразу через ana_n. Маємо:
an=25n+5n−18n−12n=(25n−18n)−(12n−5n)==(25−18)(25n−1+25n−2⋅18+⋯+25⋅18n−2+18n−1)−−(12−5)(12n−1+12n−2⋅5+⋯+12⋅5n−2+5n−1). \begin{aligned} a_n &= 25^n + 5^n - 18^n - 12^n = (25^n - 18^n) - (12^n - 5^n) = \\ &= (25 - 18)(25^{n-1} + 25^{n-2} \cdot 18 + \dots + 25 \cdot 18^{n-2} + 18^{n-1}) - \\ &\quad -(12 - 5)(12^{n-1} + 12^{n-2} \cdot 5 + \dots + 12 \cdot 5^{n-2} + 5^{n-1}). \end{aligned}
Оскільки 25−18=12−5=725 - 18 = 12 - 5 = 7, то ana_n ділиться без остачі на 77. Аналогічно,
an=25n+5n−18n−12n=(25n−12n)−(18n−5n)==(25−12)(25n−1+25n−2⋅12+⋯+25⋅12n−2+12n−1)−−(18−5)(18n−1+18n−2⋅5+⋯+18⋅5n−2+5n−1), \begin{aligned} a_n &= 25^n + 5^n - 18^n - 12^n = (25^n - 12^n) - (18^n - 5^n) = \\ &= (25 - 12)(25^{n-1} + 25^{n-2} \cdot 12 + \dots + 25 \cdot 12^{n-2} + 12^{n-1}) - \\ &\quad -(18 - 5)(18^{n-1} + 18^{n-2} \cdot 5 + \dots + 18 \cdot 5^{n-2} + 5^{n-1}), \end{aligned}
тому ana_n ділиться без остачі на 1313. Оскільки 77 і 1313 взаємно прості, то ana_n ділиться без остачі на 7⋅13=917 \cdot 13 = 91.

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