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, 2010

Number theory Difficulty 5.4 AIME, harder Prove it Ukraine

Find all natural numbers $n$ for which among the numbers $n, n+1, n+2, \dots, n^2$ there exist 4 pairwise distinct numbers $a, b, c, d$ such that equality $ab = cd$ holds. Justify the answer.

Solution

Answer: $n \ge 3$ .

If $6n \le n^2$ or equivalently $n \ge 6$ , then after putting $a = n$ , $c = 2n$ , $d = 3n$ , $b = 6n$ we will have $ab = cd$ . So every $n \ge 6$ satisfies the statement.

In the case $n = 5$ we have numbers $5, 6, 7, \dots, 25$ . Considering $a = 6$ , $b = 20$ , $c = 8$ , $d = 15$ we have the desired result.

In the case $n = 4$ we have numbers $4, 5, 6, \dots, 16$ . Considering $a = 4$ , $b = 15$ , $c = 6$ , $d = 10$ we have the desired result.

In the case $n = 3$ we can put $a = 3$ , $b = 8$ , $c = 4$ , $d = 6$ .

In the case $n = 1, 2$ in the set $n, n+1, n+2, \dots, n^2$ at most three numbers, so we can't find four pairwise distinct numbers.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.