2023 IOQM/Problem 1

Problem

Let $n$ be a positive integer such that $1 \leq n \leq 1000$ . Let $M_n$ be the number of integers in the set

$X_n = \left\{\sqrt{4n + 1}, \sqrt{4n + 2}, \ldots, \sqrt{4n + 1000}\right\}$ . Let $a = \max\{M_n : 1 \leq n \leq 1000\}$ , and $b = \min\{M_n : 1 \leq n \leq 1000\}$ .

Find $a - b$ .

Solution 1(Spacing of squares)

If for any integer $n$ , if $\sqrt{n}$ is an integer this means $n$ is a perfect square. Now the problem reduces to finding the difference between maximum and minimum no. of perfect squares in the numbers: $4n+1, 4n+2 .... 4n+1000.$ There are 1000 numbers here.

The idea is that for the same range of numbers, the no. of perfect squares becomes rarer when the numbers become larger.

For example, there are 3 perfect squares between 1 and 10 but none between 50 and 60.

Of course we will prove this,

Claim: The distance between 2 consecutive perfect squares gets larger as they get bigger

Proof: Let the 2 consecutive perfect squares be $m^{2}$ and $(m+1)^{2}$ . Now distance between them (Number of numbers between them) is $(m+1)^{2}$ - $m^{2}$ -1 = $2m$ . So, we notice as $m$ gets larger(i.e. the perfect squares get larger as $m^{2}$ and $(m+1)^{2}$ get larger as m does), $2m$ (distance between $m^{2}$ and $(m+1)^{2}$ ) also does, which means that the distance between the consecutive squares get larger as the consecutive squares get larger (As $2m$ is the measure of distance between them). Hence, this proves our claim.

Now, if the distance between 2 perfect squares increases as they get larger, this suffices to prove that perfect squares get rarer as no.s get larger.

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From this list where perfect squares are highlighted RED (Except 1 which is Blue), it's evident that perfect squares get rarer as numbers get bigger and bigger.

⇒ The maximum value of $M_n$ occurs when $n$ is minimum and the minimum value of $M_n$ occurs when $n$ is maximum.

Minimum value of $n$ = 1 So, the numbers are 5, 6...1004. there are 29 perfect squares here, so

$a$ = $max.$ ( $M_n$ )= $29$

Maximum value of $n$ = 1000 So, the numbers are 4001, 4002...5000. there are 7 perfect squares here, so $b$ = $min.$ ( $M_n$ )= $7$

⇒ $a-b= 29-7 = \boxed{22}$

~SANSGANKRSNGUPTA

Solution 2

All of the sets from $1 \leq n \leq 1000$ will have 1000 elements the most the number of square numbers will be in $X_1$ and least number of perfect squares in $X_{1000}$ . So $a=M_1$ and $b=M_{1000}$ . This is because the gaps between square number is less when n is greater then when n is less. By checking for $X_1$ there would be squares from { $3^2, 4^2,..., 31^2$ }, a total of 29 numbers while in $X_{1000}$ there would be squares from { $64^2, 65^2,..., 70^2$ } a total of 7 numbers, so $a=29$ and $b=7$ , giving us $\boxed{a-b=\textbf{22}}$