Problem:
Find the least positive integer N such that the set of 1000 consecutive integers beginning with 1000⋅N contains no square of an integer.
Solution:
Assume that integer x satisfies x2≤1000⋅N<1000⋅N+1000≤ (x+1)2. Observe that x2 and (x+1)2 differ by 2x+1. Because (x+1)2−x2 must exceed 1000, 2x+1>1000, and thus x≥500.
Let x=500+a, for some nonnegative integer a. Then x2=250000+ 1000a+a2 and (x+1)2=x2+1001+2a. Because x2 is greater than a multiple of 1000 by the amount of a2 and (x+1)2 is 1001+2a greater than x2, the requirement will be satisfied by the least value of a such that a2<1000 and a2+1001+2a≥2000 or equivalently a2+2a−999≥0. The least such positive integer value of a is 31.
Thus x2=250000+31000+961=281961<282000 and (x+1)2= 281961+1001+62=283024≥283000. The answer is therefore 282.
The problems on this page are the property of the MAA's American Mathematics Competitions