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Algebra Difficulty 6.7 National olympiad Prove it Japan

Let a1,a2,a3,a4,a5,a6,a7a_1, a_2, a_3, a_4, a_5, a_6, a_7 be distinct positive integers. Find the minimum possible value of a7a1|a_7 - a_1| assuming that the sequence a1,2a2,3a3,4a4,5a5,6a6,7a7a_1, 2a_2, 3a_3, 4a_4, 5a_5, 6a_6, 7a_7 is an arithmetic progression. A sequence x1,x2,,x7x_1, x_2, \dots, x_7 is called an arithmetic progression if x2x1=x3x2==x7x6x_2 - x_1 = x_3 - x_2 = \dots = x_7 - x_6.

Solution

360
The assumption implies that there holds iai=a1+(i1)(2a2a1)ia_i = a_1 + (i-1)(2a_2 - a_1) for any integer 2i72 \le i \le 7. Subtracting a1a_1 from both sides and dividing by ii, one obtains
aia1=iaiia1i=(i1)(2a2a1)(i1)a1i=2(i1)(a2a1)i. a_i - a_1 = \frac{ia_i - ia_1}{i} = \frac{(i-1)(2a_2 - a_1) - (i-1)a_1}{i} = \frac{2(i-1)(a_2 - a_1)}{i}.
Since ii and i1i-1 are coprime, 2(a2a1)2(a_2 - a_1) is divided by ii for any 2i72 \le i \le 7, thus divided by 420420. By the assumption a2a10a_2 - a_1 \ne 0, thus 2(a2a1)420|2(a_2 - a_1)| \ge 420. Hence
a7a1=26(a2a1)767420=360. |a_7 - a_1| = \left| \frac{2 \cdot 6 \cdot (a_2 - a_1)}{7} \right| \ge \frac{6}{7} \cdot 420 = 360.
On the other hand, the sequence (a1,a2,a3,a4,a5,a6,a7)=(420,210,140,105,84,70,60)(a_1, a_2, a_3, a_4, a_5, a_6, a_7) = (420, 210, 140, 105, 84, 70, 60) satisfies the assumption and a7a1=360|a_7 - a_1| = 360, thus the answer is 360360.

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