Maths Olympiad Prep

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, 2011

Number theory Difficulty 6.1 National olympiad Prove it China

Determine, with proof, whether there is any odd integer n3n \ge 3 and nn distinct prime numbers p1,p2,,pnp_1, p_2, \dots, p_n, such that all pi+pi1p_i + p_{i-1} (i=1,2,,ni=1, 2, \dots, n, and pn1=p1p_{n-1} = p_1) are perfect squares?

Solution

Suppose that there exists odd integer n3n \ge 3 and nn distinct prime numbers p1,p2,,pnp_1, p_2, \dots, p_n satisfying the given condition.
If all p1,p2,,pnp_1, p_2, \dots, p_n are odd, then all the sums pi+pi+1p_i + p_{i+1} are multiples of 44, so the prime numbers p1,p2,,pnp_1, p_2, \dots, p_n modulo 44 appear to be 11 and 33 alternatively, and it contradicts to the fact that nn is odd.
If one of p1,p2,,pnp_1, p_2, \dots, p_n is 22, then without loss of generality, we may assume that p1=2p_1 = 2. As both p1+p2p_1 + p_2 and pn+p1p_n + p_1 are perfect squares and both are odd, it follows that p2p_2 and pnp_n are congruent to 33 modulo 44. Similar to the discussion in the first case, we know that the primes p2,p3,,pnp_2, p_3, \dots, p_n modulo 44 appear to be 11 and 33 alternatively, so n1n-1 is odd, which is a contradiction.
Hence, there are no odd integer n3n \ge 3 and nn primes satisfying the given conditions.

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