The set of numbers {4,7,9,10,12,14} contains three arithmetic sequences of three terms, {4,7,10}, {10,12,14}, and {4,9,14}. Find a set of six perfect squares with that property. All the perfect squares should be greater than 1 and none should be equal to each other.
This is how I find a basic one, which leads to as many solutions as you want.
Letīs call three squares in A.P., A^2, B^2 and C^2.
We have 2*B^2 = A^2 + C^2.
Making A = m  n, and C = m + n, we achieve:
2*B^2 = 2*m^2 + 2*n^2 > B^2 = m^2 + n^2
So B, m and n are pythagorean triples.
Since "1" is disallowed as square, letīs make m=8 and n=6, which gives:
A^2 = (m  n)^2 = 2^2 = 4
B^2 = (m^2 + n^2) = 8^2 + 6^2 = 64 + 36 = 100
C^2 = (m + n)^2 = 14^2 = 196
So the first three are
4, 100 and 196.
If we now choose the 100 for being the first of the second set (but we canīt use m=8 and n=6 anymore) and since the first term is (mn)^2, the values (30, 40, 50) obtained from the first pythagorean triple (3, 4, 5), give us m  n = 10. So the first term of the second set will be 100, and thatīs what we need. Thus, for m = 40, and n = 30, we achive the second set: (40  30)^2 = 10^2 = 100, 30^2 + 40^2 =
2500, and (40 + 30)^2 = 70^2 =
4900.To obtain the third set, we use the 196 as being the first one in the set. The second must be 2500 or 4900.
If the second is 2500, we have (m  n) = 14 and m^2 + n^2 = 2500, and if the second is 4900, we have (m  n) = 14 and m^2 + n^2 = 4900.
Solving both cases, we find that only the later works, giving m = 56 and n = 42, which means that the last square is (56 + 42)^2 = 98^2 =
9604. If neither 2500 or 4900 worked, we can make another iteration, starting with a new triple for the second set or, if needed, with a new triple for the first set.
Thus, one (of infinite solutions) is :
(
4, 100, 196, 2500, 4900, 9604), being the sets (4, 100, 196), (100, 2500, 4900), and (196, 4900, 9604).
I believe that a deep study could find expressions to the 6 squares, all functions of a given pair (m,n), and then, no "iterations" needed.
Edited on November 18, 2005, 12:59 pm

Posted by pcbouhid
on 20051118 12:49:25 