IMO 2001 SL N4

Origin: VIE | Category: Number Theory

Problem

Let p \geq5 be a prime number. Prove that there exists an integer a with 1 \leqa \leqp −2 such that neither ap−1 −1 nor (a + 1)p−1 −1 is divisible by p2.

Solution

Let C be the set of those a \in{1, 2, . . ., p−1} for which ap−1 \equiv1 (mod p2). At first, we observe that a, p −a do not both belong to C, regardless of the value of a. Indeed, by the binomial formula, (p −a)p−1 −ap−1 \equiv−(p −1)p ap−2 ̸\equiv0 (mod p2). As a consequence we deduce that |C| \leqp−1 2 . Further, we observe that p −k \inC ⇔k \equivk(p −k)p−1 (mod p2), i.e., p −k \inC ⇔k \equivk(kp−1 −(p −1)p kp−2) \equivkp + p (mod p2). (1) Now assume the contrary to the claim, that for every a = 1, . . . , p −2 one of a, a+1 is in C. In this case it is not possible that a, a+1 are both in C, for then p−a, p−a−1 ̸\inC. Thus, since 1 \inC, we inductively obtain that 2, 4, . . . , p−1 ̸\inC and 1, 3, 5, . . ., p−2 \inC. In particular, p−2, p−4 \inC, which is by (1) equivalent to 2 \equiv2p + p and 4 \equiv4p + p (mod p2). However, squaring the former equality and subtracting the latter, we ob- tain 2p+1p \equivp (mod p2), or 4 \equiv1 (mod p), which is a contradiction unless p = 3. This finishes the proof.