From 1 to 100

Number Theory Level 3

True or False?

There exist distinct positive integers a a and b b such that a 100 + b 100 a^{100}+b^{100} is divisible by all of the following numbers: a + b , a 2 + b 2 , a 3 + b 3 , a 4 + b 4 , … , a 99 + b 99 , a 100 + b 100 . a+b, a^2+b^2, a^3+b^3, a^4+b^4, \ldots, a^{99}+b^{99}, a^{100}+b^{100}.

True False

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

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1 solution

Áron Bán-Szabó
Jul 19, 2017

True.

Let { a = ( 1 + 2 ) ( 1 + 2 2 ) ( 1 + 2 3 ) ( 1 + 2 4 ) ⋯ ( 1 + 2 99 ) b = 2 a \begin{cases} a=(1+2)(1+2^2)(1+2^3)(1+2^4)\cdots (1+2^{99}) \\ b=2a \end{cases}

For every 1 ≤ k ≤ 99 1\leq k\leq 99 integer a k + b k = a k + 2 k ∗ a k = a k ( 1 + 2 k ) a^k+b^k=a^k+2^k*a^k=a^k(1+2^k)

Since for every k k , 1 + 2 k 1+2^k is a divisor of a a , a 100 a k ∗ ( 1 + 2 k ) = a 100 − k 1 + 2 k \dfrac{a^{100}}{a^k*(1+2^k)}=\dfrac{a^{100-k}}{1+2^k}

From that it is clear, that ( a , b a, b ) is a suitable number pair.

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