Hi,
I was working on a problem that involved 3 squares in arithmetic progression when I came upon this strange pattern that I can't explain...
if we let a, b and c equal 1, 5 and 7 respectively then we have a^2, b^2, c^2 as an arithmetic progression of 3 squares 1, 25, 49
Keeping a as 1 and checking all possible values of b up to 20,000,000 we get the following values for a, b and c that produce 3 squares in arithmetic progression...
Taking the above values of b we find that
29 = 6 * 5 - 1
169 = 6 * 29 - 5
985 = 6 * 169 - 29
5741 = 6 * 985 - 169
33461 = 6 * 5741 - 985
195025 = 6 * 33461 - 5741
1136689 = 6 * 195025 - 33461
6625109 = 6 * 1136689 - 195025
i.e. each successive value of b can be calculated from previous ones. Am I missing something blindingly obvious? (btw, it is the same for the values of c).
Thanks
I was working on a problem that involved 3 squares in arithmetic progression when I came upon this strange pattern that I can't explain...
if we let a, b and c equal 1, 5 and 7 respectively then we have a^2, b^2, c^2 as an arithmetic progression of 3 squares 1, 25, 49
Keeping a as 1 and checking all possible values of b up to 20,000,000 we get the following values for a, b and c that produce 3 squares in arithmetic progression...
a 1 1 1 1 1 1 1 1 1 | b 5 29 169 985 5741 33461 195025 1136689 6625109 | c 7 41 239 1393 8119 47321 275807 1607521 9369319 |
Taking the above values of b we find that
29 = 6 * 5 - 1
169 = 6 * 29 - 5
985 = 6 * 169 - 29
5741 = 6 * 985 - 169
33461 = 6 * 5741 - 985
195025 = 6 * 33461 - 5741
1136689 = 6 * 195025 - 33461
6625109 = 6 * 1136689 - 195025
i.e. each successive value of b can be calculated from previous ones. Am I missing something blindingly obvious? (btw, it is the same for the values of c).
Thanks