IMO shortlist 1984/10 solution

25th IMO 1984 shortlist

Problem 10

Prove that the product of five consecutive positive integers cannot be the square of an integer.

Solution

One of the numbers is not divisible by 2 or 3. If it is divisible by a prime p, then p ≥ 5, so p does not divide any of the other numbers, so the number must be a square. So one of the numbers is b² = (6c±1)² = 12c(3c±1)+1 = 1 mod 24, since c(3c±1) is even. The only 5 consecutive positive integers which include two squares are 1,2,3,4,5, whose product is not a square (because the difference between any two squares other than 1,4 exceeds 4). So none of the other four numbers (apart from b²) is a square and hence none of the others is ±1 mod 6. So the numbers must be 24k, 24k+1 = b², 24k+2, 24k+3, 24k+4. But now 24k+4 = 4(6k+1), so 6k+1 must be a square and hence 24k+4 is a square. Contradiction.

25th IMO shortlist 1984