Product of distances

Let a 1 , a 2 , a 3 , a 4 a_1,a_2,a_3,a_4 be distinct integers.

If 1 i < j 4 a i a j \displaystyle\prod _ {1\le i < j\le4} {|a_i - a_j|} is a perfect square,

then what is the minimum value of 1 i < j 4 a i a j ? \displaystyle\sum _ {1\le i < j\le4} {|a_i - a_j|} ?


Try my set here


The answer is 26.

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1 solution

Haosen Chen
Jul 19, 2018

WLOG 0 = a 1 < a 2 < a 3 < a 4 0=a_{1}<a_{2}<a_{3}<a_{4} ,since we only care about the difference between them.

Then the condition becomes that a 2 a 3 a 4 ( a 3 a 2 ) ( a 4 a 2 ) ( a 4 a 3 ) a_{2}a_{3}a_{4}(a_{3}-a_{2})(a_{4}-a_{2})(a_{4}-a_{3}) is a perfect square,

and what we want to minimize becomes f = 3 a 4 + a 3 a 2 f=3a_{4}+a_{3}-a_{2} .

It's easy to find that ( a 2 , a 3 , a 4 ) = ( 3 , 5 , 8 ) (a_{2},a_{3},a_{4})=(3,5,8) satisfies the condition and gives us f = 26 f=26 .

When a 4 9 a_{4}\ge 9 ,we have f 3 a 4 + 1 28 f\ge 3a_{4}+1\ge 28 .So we only need to check the case when a 4 = 3 , 4 , 5 , 6 , 7 , 8. a_{4}=3,4,5,6,7,8.

Well,it's not hard.Just pay your patience to check one by one.

I guess that there might be a more elegant solution to this beautiful problem.So please share with us if you've found one.

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