2011 UNCO Math Contest II Problems/Problem 4

Problem

Let $A = \left\{ 2,5,10,17,\cdots,n^2+1,\cdots\right\}$ be the set of all positive squares plus $1$ and $B = \left\{101, 104, 109, 116,\cdots,m^2 + 100,\cdots\right\}$ be the set of all positive squares plus $100$ .

(a) What is the smallest number in both $A$ and $B$ ?

(b) Find all numbers that are in both $A$ and $B$ .

Solution

The smallest number in both sets is 101. This is because it is both 1^2+100 and 10^2+1. Subtracting 10-1 gives us 9. This means that all other numbers in both sets must be less than 9 apart, because squares get further apart the bigger they get. Following a geometric sequence gives us 9,3,1 so the squares must be that far apart. This means the next term is 325, because it is 15^2+100 and 18^2+1. There is only one left, which is 2501, because it is 49^2+100 and 50^2+1.

2011 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2011 UNCO Math Contest II Problems/Problem 4

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