This week’s puzzle, as previously posted:

Simply stated, this week’s puzzle is to find the non-unique, two-digit cumulative differences between the squares of descending triplets of arithmetic sequences.

An arithmetic progression is a sequence of numbers in order that have a common difference; for example, the numbers 4, 7, 10, 13, 16, etc., are an arithmetic progression starting with 4 having a common difference of 3. One cumulative difference between the squares of descending triplets of this arithmetic sequence would be (13

^{2}– 10^{2}– 7^{2}) = 169 – 100 – 49 = 20; another would be (16^{2}– 13^{2}– 10^{2}) = 256 – 169 – 100 = -13 and a third would be (10^{2}– 7^{2}– 4^{2}) = 100 – 49 – 16 = 35.For all such arithmetic progressions, some of these cumulative differences between the squares of descending triplets are positive, some are negative and some are exactly zero. Some of these cumulative differences, like 115, can be arrived at through two different sequences (144

^{2}– 115^{2}– 86^{2}and 30^{2}– 23^{2}– 16^{2}), and some like 124 can only be arrived at in one sequence (78^{2}– 62^{2}– 46^{2}).For your chance at this week’s $50 ThinkGeek Gift Certificate, tell me how many of these cumulative differences between the squares of descending triplets in arithmetic progressions have two digits (i.e., 10-99) and can be arrived at through two or more sequences? Additionally, what are they? (They are finite, and can be listed.)

The “trick” for this week’s puzzle was looking at the differences between the squares of descending triplets of arithmetic sequence and recognizing them as something that can be factored. If you consider the highest of the three terms in the sequence to be n and the constant differences to be a, then the cumulative differences between the squares of the descending terms can be expressed as:

(n)^{2} – (n-a)^{2} – (n-2a)^{2}

or

-5a^{2} + 6an – n^{2}

At this point, you can “brute force” a solution with Excel, ranging across values of a and n, keeping both a and n positive. (Due to the way that negatives square, if you allow negative a and n values, the counts simply double.) At this point, the answer sets can vary, depending upon what constraints you put around the terms with respect to negative numbers.

If we keep all terms in the sequence positive, there are 18 values that have multiple solutions:

15(3 times), 27(2 times), 32(2), 35(3), 36(2), 39(2), 51(2), 55(3), 60(3), 63(4), 64(2), 75(3), 84(3), 87(2), 91(3), 95(2), 96(3), 99(4).

If we allow negative terms in the sequence (i.e., 4^{2} – 1^{2} – (-2)^{2} = 11), the count goes up to 39 values:

11(2 times), 12(2 times), 15(4), 19(2), 20(2), 23(2), 27(4), 28(2), 31(2), 32(2), 35(4), 36(3), 39(4), 43(2), 44(2), 47(2), 48(2), 51(4), 52(2), 55(4), 59(2), 60(4), 63(6), 64(3), 67(2), 68(2), 71(2), 75(6), 76(2), 79(2), 80(2), 83(2), 84(4), 87(4), 91(4), 92(2), 95(4), 96(4), 99(6).

Congratulations to **Randy Slavey**, the winner of this week’s $50 ThinkGeek Gift Certificate.

A list of all 2-digit solutions (including negative and zero sequence terms) appears below the fold for checking your own answers. Thanks to everyone that submitted a solution!

Complete list of potential responses:

(14,3) : 14^{2} – 11^{2} – 8^{2} = 11

(4,3) : 4^{2} – 1^{2} – -2^{2} = 11

(4,2) : 4^{2} – 2^{2} – 0^{2} = 12

(8,2) : 8^{2} – 6^{2} – 4^{2} = 12

(19,4) : 19^{2} – 15^{2} – 11^{2} = 15

(5,2) : 5^{2} – 3^{2} – 1^{2} = 15

(5,4) : 5^{2} – 1^{2} – -3^{2} = 15

(7,2) : 7^{2} – 5^{2} – 3^{2} = 15

(6,2) : 6^{2} – 4^{2} – 2^{2} = 16

(24,5) : 24^{2} – 19^{2} – 14^{2} = 19

(6,5) : 6^{2} – 1^{2} – -4^{2} = 19

(13,3) : 13^{2} – 10^{2} – 7^{2} = 20

(5,3) : 5^{2} – 2^{2} – -1^{2} = 20

(29,6) : 29^{2} – 23^{2} – 17^{2} = 23

(7,6) : 7^{2} – 1^{2} – -5^{2} = 23

(12,3) : 12^{2} – 9^{2} – 6^{2} = 27

(34,7) : 34^{2} – 27^{2} – 20^{2} = 27

(6,3) : 6^{2} – 3^{2} – 0^{2} = 27

(8,7) : 8^{2} – 1^{2} – -6^{2} = 27

(18,4) : 18^{2} – 14^{2} – 10^{2} = 28

(6,4) : 6^{2} – 2^{2} – -2^{2} = 28

(39,8) : 39^{2} – 31^{2} – 23^{2} = 31

(9,8) : 9^{2} – 1^{2} – -7^{2} = 31

(11,3) : 11^{2} – 8^{2} – 5^{2} = 32

(7,3) : 7^{2} – 4^{2} – 1^{2} = 32

(10,3) : 10^{2} – 7^{2} – 4^{2} = 35

(10,9) : 10^{2} – 1^{2} – -8^{2} = 35

(44,9) : 44^{2} – 35^{2} – 26^{2} = 35

(8,3) : 8^{2} – 5^{2} – 2^{2} = 35

(23,5) : 23^{2} – 18^{2} – 13^{2} = 36

(7,5) : 7^{2} – 2^{2} – -3^{2} = 36

(9,3) : 9^{2} – 6^{2} – 3^{2} = 36

(11,10) : 11^{2} – 1^{2} – -9^{2} = 39

(17,4) : 17^{2} – 13^{2} – 9^{2} = 39

(49,10) : 49^{2} – 39^{2} – 29^{2} = 39

(7,4) : 7^{2} – 3^{2} – -1^{2} = 39

(12,11) : 12^{2} – 1^{2} – -10^{2} = 43

(54,11) : 54^{2} – 43^{2} – 32^{2} = 43

(28,6) : 28^{2} – 22^{2} – 16^{2} = 44

(8,6) : 8^{2} – 2^{2} – -4^{2} = 44

(13,12) : 13^{2} – 1^{2} – -11^{2} = 47

(59,12) : 59^{2} – 47^{2} – 35^{2} = 47

(16,4) : 16^{2} – 12^{2} – 8^{2} = 48

(8,4) : 8^{2} – 4^{2} – 0^{2} = 48

(14,13) : 14^{2} – 1^{2} – -12^{2} = 51

(22,5) : 22^{2} – 17^{2} – 12^{2} = 51

(64,13) : 64^{2} – 51^{2} – 38^{2} = 51

(8,5) : 8^{2} – 3^{2} – -2^{2} = 51

(33,7) : 33^{2} – 26^{2} – 19^{2} = 52

(9,7) : 9^{2} – 2^{2} – -5^{2} = 52

(15,14) : 15^{2} – 1^{2} – -13^{2} = 55

(15,4) : 15^{2} – 11^{2} – 7^{2} = 55

(69,14) : 69^{2} – 55^{2} – 41^{2} = 55

(9,4) : 9^{2} – 5^{2} – 1^{2} = 55

(16,15) : 16^{2} – 1^{2} – -14^{2} = 59

(74,15) : 74^{2} – 59^{2} – 44^{2} = 59

(10,4) : 10^{2} – 6^{2} – 2^{2} = 60

(10,8) : 10^{2} – 2^{2} – -6^{2} = 60

(14,4) : 14^{2} – 10^{2} – 6^{2} = 60

(38,8) : 38^{2} – 30^{2} – 22^{2} = 60

(11,4) : 11^{2} – 7^{2} – 3^{2} = 63

(13,4) : 13^{2} – 9^{2} – 5^{2} = 63

(17,16) : 17^{2} – 1^{2} – -15^{2} = 63

(27,6) : 27^{2} – 21^{2} – 15^{2} = 63

(79,16) : 79^{2} – 63^{2} – 47^{2} = 63

(9,6) : 9^{2} – 3^{2} – -3^{2} = 63

(12,4) : 12^{2} – 8^{2} – 4^{2} = 64

(21,5) : 21^{2} – 16^{2} – 11^{2} = 64

(9,5) : 9^{2} – 4^{2} – -1^{2} = 64

(18,17) : 18^{2} – 1^{2} – -16^{2} = 67

(84,17) : 84^{2} – 67^{2} – 50^{2} = 67

(11,9) : 11^{2} – 2^{2} – -7^{2} = 68

(43,9) : 43^{2} – 34^{2} – 25^{2} = 68

(19,18) : 19^{2} – 1^{2} – -17^{2} = 71

(89,18) : 89^{2} – 71^{2} – 53^{2} = 71

(10,5) : 10^{2} – 5^{2} – 0^{2} = 75

(10,7) : 10^{2} – 3^{2} – -4^{2} = 75

(20,19) : 20^{2} – 1^{2} – -18^{2} = 75

(20,5) : 20^{2} – 15^{2} – 10^{2} = 75

(32,7) : 32^{2} – 25^{2} – 18^{2} = 75

(94,19) : 94^{2} – 75^{2} – 56^{2} = 75

(12,10) : 12^{2} – 2^{2} – -8^{2} = 76

(48,10) : 48^{2} – 38^{2} – 28^{2} = 76

(21,20) : 21^{2} – 1^{2} – -19^{2} = 79

(99,20) : 99^{2} – 79^{2} – 59^{2} = 79

(10,6) : 10^{2} – 4^{2} – -2^{2} = 80

(26,6) : 26^{2} – 20^{2} – 14^{2} = 80

(104,21) : 104^{2} – 83^{2} – 62^{2} = 83

(22,21) : 22^{2} – 1^{2} – -20^{2} = 83

(11,5) : 11^{2} – 6^{2} – 1^{2} = 84

(13,11) : 13^{2} – 2^{2} – -9^{2} = 84

(19,5) : 19^{2} – 14^{2} – 9^{2} = 84

(53,11) : 53^{2} – 42^{2} – 31^{2} = 84

(109,22) : 109^{2} – 87^{2} – 65^{2} = 87

(11,8) : 11^{2} – 3^{2} – -5^{2} = 87

(23,22) : 23^{2} – 1^{2} – -21^{2} = 87

(37,8) : 37^{2} – 29^{2} – 21^{2} = 87

(114,23) : 114^{2} – 91^{2} – 68^{2} = 91

(12,5) : 12^{2} – 7^{2} – 2^{2} = 91

(18,5) : 18^{2} – 13^{2} – 8^{2} = 91

(24,23) : 24^{2} – 1^{2} – -22^{2} = 91

(14,12) : 14^{2} – 2^{2} – -10^{2} = 92

(58,12) : 58^{2} – 46^{2} – 34^{2} = 92

(11,6) : 11^{2} – 5^{2} – -1^{2} = 95

(119,24) : 119^{2} – 95^{2} – 71^{2} = 95

(25,24) : 25^{2} – 1^{2} – -23^{2} = 95

(25,6) : 25^{2} – 19^{2} – 13^{2} = 95

(11,7) : 11^{2} – 4^{2} – -3^{2} = 96

(13,5) : 13^{2} – 8^{2} – 3^{2} = 96

(17,5) : 17^{2} – 12^{2} – 7^{2} = 96

(31,7) : 31^{2} – 24^{2} – 17^{2} = 96

(12,9) : 12^{2} – 3^{2} – -6^{2} = 99

(124,25) : 124^{2} – 99^{2} – 74^{2} = 99

(14,5) : 14^{2} – 9^{2} – 4^{2} = 99

(16,5) : 16^{2} – 11^{2} – 6^{2} = 99

(26,25) : 26^{2} – 1^{2} – -24^{2} = 99

(42,9) : 42^{2} – 33^{2} – 24^{2} = 99

### About Judd Schorr

Judd is a life long IT and math geek, currently adding value in the area of digital analytics. Dad to both Max and Nora, who frequently star in the puzzles he writes for GeekDad, his dear wife Allison has learned to tolerate the constant puns.

I went for the brute force excel approach. Then promptly forgot to send in my entry . D’oh! Quite pleased I ended up with the right answer though.