Olympiad Maths Prep

Track / Stage 6 / 187 of 400 #1187 of 2000

Problem 1187

National olympiad, first round
Number theory Difficulty 6.3 Prove it

Example 1 Let a,b,na, b, n be given positive integers, and it is known that for any k∈N∗(k≠b)k \in \mathbf{N}^{*}(k \neq b), we have (b−k)∣(a−kn)(b-k) \mid\left(a-k^{n}\right). Prove: a=bna=b^{n}.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Notice that, for any k∈N∗(k≠b)k \in \mathbf{N}^{*} (k \neq b), we have
bn−kn=(b−k)(bn−1+bn−2k+⋯+kn−1),b^{n}-k^{n}=(b-k)\left(b^{n-1}+b^{n-2} k+\cdots+k^{n-1}\right),

thus (b−k)∣(bn−kn)(b-k) \mid\left(b^{n}-k^{n}\right). Combining this with (b−k)∣(a−kn)(b-k) \mid\left(a-k^{n}\right), we get
(b−k)∣((a−kn)−(bn−kn)),(b-k) \mid\left(\left(a-k^{n}\right)-\left(b^{n}-k^{n}\right)\right),

which means (b−k)∣(a−bn)(b-k) \mid\left(a-b^{n}\right).
Taking k=b+1+∣a−bn∣k=b+1+\left|a-b^{n}\right|, we have
−(1+∣a−bn∣)∣(a−bn),-\left(1+\left|a-b^{n}\right|\right) \mid\left(a-b^{n}\right),

Thus, by Theorem 1 (4), we know a−bn=0a-b^{n}=0, and the proposition is proved.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.